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All questions of Differential Equations for Electrical Engineering (EE) Exam
Which of the following is not a standard method for finding the solutions for differential equations?- a)Variable Separable
- b)Homogenous Equation
- c)Orthogonal Method
- d)Bernoulli’s Equation
Correct answer is option 'C'. Can you explain this answer?
Which of the following is not a standard method for finding the solutions for differential equations?
a)
Variable Separable
b)
Homogenous Equation
c)
Orthogonal Method
d)
Bernoulli’s Equation
|
Sanya Agarwal answered |
The following are the different standard methods used in finding the solution of a differential equation:
- Variable Separable
- Homogenous Equation
- Non-homogenous Equation reducible to Homogenous Equation
- Exact Differential Equation
- Non-exact Differential Equation that can be made exact with the help of integrating factors
- Linear First Order Equation
- Bernoulli’s Equation
over the triangle is 
The general solution of (x2 D2 – xD), y= 0 is :
- a) y = C1 + C2 ex
- b) y = C1 + C2 x2
- c)y = C1 x + C2 x2
- d)y = C1 + C2 x
Correct answer is option 'C'. Can you explain this answer?
The general solution of (x2 D2 – xD), y= 0 is :
a)
y = C1 + C2 ex
b)
y = C1 + C2 x2
c)
y = C1 x + C2 x2
d)
y = C1 + C2 x
|
Akshara Rane answered |
+ 4xD + 3y = 0) can be found by assuming a solution of the form y = e^(mx).
Substituting this into the differential equation gives:
x^2(m^2)e^(mx) + 4xme^(mx) + 3e^(mx) = 0
Dividing through by e^(mx) gives:
m^2x^2 + 4mx + 3 = 0
This is a quadratic equation in m, which can be solved using the quadratic formula:
m = (-4x ± √(16x^2 - 4(3)(x^2))) / (2x^2)
Simplifying this gives:
m = (-2 ± √1) / x
m1 = -1/x, m2 = -3/x
Therefore, the general solution is a linear combination of the two solutions:
y = c1e^(-x) + c2e^(-3x)
where c1 and c2 are arbitrary constants determined by initial or boundary conditions.
Substituting this into the differential equation gives:
x^2(m^2)e^(mx) + 4xme^(mx) + 3e^(mx) = 0
Dividing through by e^(mx) gives:
m^2x^2 + 4mx + 3 = 0
This is a quadratic equation in m, which can be solved using the quadratic formula:
m = (-4x ± √(16x^2 - 4(3)(x^2))) / (2x^2)
Simplifying this gives:
m = (-2 ± √1) / x
m1 = -1/x, m2 = -3/x
Therefore, the general solution is a linear combination of the two solutions:
y = c1e^(-x) + c2e^(-3x)
where c1 and c2 are arbitrary constants determined by initial or boundary conditions.
The following differential equation has 
a)degree = 2, order = 1b)degree = 1, order = 2 c)degree = 4, order = 3 d)degree = 2, order = 3Correct answer is option 'B'. Can you explain this answer?
|
Anmol Basu answered |
Order of highest derivative = 2
Hence, most appropriate answer is (b)
Hence, most appropriate answer is (b)
The general solution of the differential equation
- a)tan y − cot x = c(c is a constant
- b)tan x − cot y = c(c is a constant)
- c)tan y + cot x = c(c is a constant)
- d)tan x + cot Y = c(c is a constant)
Correct answer is option 'C'. Can you explain this answer?
The general solution of the differential equation
a)
tan y − cot x = c(c is a constant
b)
tan x − cot y = c(c is a constant)
c)
tan y + cot x = c(c is a constant)
d)
tan x + cot Y = c(c is a constant)
|
Pioneer Academy answered |
sec2ydy = cosec2x.dx
tan y = −cot x + c
tan y + cot x = c
A function n(x) satisfies the differential equation 
where L is a constant. The boundary conditions are: n(0)=K and n(∞)= 0. The solution to this equation is - a)n(x) = K exp(x/L)
- b)n(x) = K exp( x / √L )
- c)n(x) = K2 exp(-x/L)
- d)n(x) = K exp( -x/L)
Correct answer is option 'D'. Can you explain this answer?
A function n(x) satisfies the differential equation where L is a constant. The boundary conditions are: n(0)=K and n(∞)= 0. The solution to this equation is
where L is a constant. The boundary conditions are: n(0)=K and n(∞)= 0. The solution to this equation is a)
n(x) = K exp(x/L)
b)
n(x) = K exp( x / √L )
c)
n(x) = K2 exp(-x/L)
d)
n(x) = K exp( -x/L)
|
|
Sanvi Kapoor answered |
For finite solution c1 = 0
What is the area common to the circles r = a and r = 2a cos θ? - a)0.524 a2
- b)0.614 a2
- c)1.047 a2
- d)1.228 a2
Correct answer is option 'D'. Can you explain this answer?
What is the area common to the circles r = a and r = 2a cos θ?
a)
0.524 a2
b)
0.614 a2
c)
1.047 a2
d)
1.228 a2
|
Srishti Chopra answered |
Area common to circles r = a
And r = 2a cos θ is 1.228 a2
Which of the following equations cannot be solved by using the method of separation of variables?- a)Laplace Equation
- b)Helmholtz Equation
- c)Alpha Equation
- d)Biharmonic Equation
Correct answer is option 'C'. Can you explain this answer?
Which of the following equations cannot be solved by using the method of separation of variables?
a)
Laplace Equation
b)
Helmholtz Equation
c)
Alpha Equation
d)
Biharmonic Equation
|
Vertex Academy answered |
The method of separation of variables is used to solve a wide range of linear partial differential equations with boundary and initial conditions, such as:
- Heat equation
- Wave equation
- Laplace equation
- Helmholtz equation
- Biharmonic equation
The solution of dy/dx = y2 with initial value y(0) = 1 is bounded in the interval - a)−∞ ≤ x ≤ ∞
- b)−∞ ≤ x ≤ 1
- c)x < 1, x > 1
- d)−2 ≤ x ≤ 2
Correct answer is option 'C'. Can you explain this answer?
The solution of dy/dx = y2 with initial value y(0) = 1 is bounded in the interval
a)
−∞ ≤ x ≤ ∞
b)
−∞ ≤ x ≤ 1
c)
x < 1, x > 1
d)
−2 ≤ x ≤ 2
|
|
Sanya Agarwal answered |
⇒ from the Integrating the equation, we get
⇒ at x = 0, y = 1, hence -1/1 = c ⇒ c = -1
x ≠ 1, at (x = 1) y will be undefined
x > 1, x < 1
The solutions of the equation 3yy’ + 4x = 0 represents a:- a)Family of circles
- b)Family of ellipses
- c)Family of Parabolas
- d)Family of hyperbolas
Correct answer is option 'B'. Can you explain this answer?
The solutions of the equation 3yy’ + 4x = 0 represents a:
a)
Family of circles
b)
Family of ellipses
c)
Family of Parabolas
d)
Family of hyperbolas
|
|
Vertex Academy answered |
3yy’ + 4x = 0
3ydy = - 4x dx
On integration both the sides:
Where c is the constant of integration.
Thus, the solution of given differential equation represents the family of Ellipses.
Where c is the constant of integration.
Thus, the solution of given differential equation represents the family of Ellipses.
with the condition that y = 1 at x = 1, is
Consider the following differential equation:
Which of the following is the solution of the above equation (c is an arbitrary constant)?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Consider the following differential equation:
Which of the following is the solution of the above equation (c is an arbitrary constant)?
a)
b)
c)
d)
|
Engineers Adda answered |
Given differential eqaution is,
Let, y = v × x
Let, y = v × x
dy = vdx + xdv
By substituting values of Y and dY in equation 1, we get
Integrating both sides
Integrating both sides
2 log x = log |sec v| - log v + log c
The differential equation
is solving by the method of variation of parameters, where the complementary function is given by- y = c1y1(x) + c2y2(x)- a)The value of y1(x) = e3x
- b)The value of y2(x) = e3x
- c)The value of y2(x) = xe3x
- d)
Correct answer is option 'D'. Can you explain this answer?
The differential equationis solving by the method of variation of parameters, where the complementary function is given by- y = c1y1(x) + c2y2(x)
is solving by the method of variation of parameters, where the complementary function is given by- y = c1y1(x) + c2y2(x)a)
The value of y1(x) = e3x
b)
The value of y2(x) = e3x
c)
The value of y2(x) = xe3x
d)
|
|
Engineers Adda answered |
Concept:
Calculation:
Calculation:
Given:
(D2 – 6D + 9)y = e3x/x2
Auxiliary equation is
(D2 – 6D + 9) = 0
⇒ (D – 3)2 = 0
C.F. = (C1 + C2x) e3x
C.F = C1(e3x) + C2(xe3x)
Every Cauchy sequence is- a)Unbounded
- b)Bounded
- c)Infinite
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Every Cauchy sequence is
a)
Unbounded
b)
Bounded
c)
Infinite
d)
None of these
|
|
Sanya Agarwal answered |
Basic properties of Cauchy sequences:
(i) Every convergent sequence is a Cauchy sequence,
(ii) Every Cauchy sequence of real (or complex) numbers is bounded,
(iii) If in a metric space, a Cauchy sequence possessing a convergent subsequence with limit is itself convergent and has the same limit.
∴ Every Cauchy sequence is bounded
A solution which does not contain any arbitrary constants is called a general solution.- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
A solution which does not contain any arbitrary constants is called a general solution.
a)
True
b)
False
|
Rahul Chauhan answered |
The answer is True.
Explanation:
In mathematics and engineering, a solution that does not contain any arbitrary constants is referred to as a general solution. This means that the solution represents all possible solutions to a given problem or equation, without any additional constraints or specific values.
Definition of a General Solution:
A general solution is a solution in which all possible solutions to a problem or equation are included. It does not contain any arbitrary constants, which means that it represents the complete set of solutions without any additional restrictions or specific values.
Characteristics of a General Solution:
1. Contains No Arbitrary Constants: A general solution does not involve any arbitrary constants. It represents the complete set of solutions without imposing any specific values or restrictions.
2. Represents All Possible Solutions: A general solution encompasses all possible solutions to a problem or equation. It includes all valid solutions without any additional constraints.
3. Allows for Flexibility: Since a general solution does not involve any arbitrary constants, it allows for flexibility in determining specific solutions. It provides a framework within which specific solutions can be derived by assigning suitable values to the constants.
4. Can Be Modified: A general solution can be modified or refined by adding specific constraints or conditions. By introducing additional information or requirements, a more specific solution can be obtained.
5. Applies to a Range of Problems: The concept of a general solution is applicable to various mathematical and engineering problems. It is a fundamental concept in differential equations, linear algebra, and other branches of mathematics.
In conclusion, a general solution is a solution that represents all possible solutions to a problem or equation without any arbitrary constants. It offers flexibility and can be modified or refined to obtain more specific solutions.
Explanation:
In mathematics and engineering, a solution that does not contain any arbitrary constants is referred to as a general solution. This means that the solution represents all possible solutions to a given problem or equation, without any additional constraints or specific values.
Definition of a General Solution:
A general solution is a solution in which all possible solutions to a problem or equation are included. It does not contain any arbitrary constants, which means that it represents the complete set of solutions without any additional restrictions or specific values.
Characteristics of a General Solution:
1. Contains No Arbitrary Constants: A general solution does not involve any arbitrary constants. It represents the complete set of solutions without imposing any specific values or restrictions.
2. Represents All Possible Solutions: A general solution encompasses all possible solutions to a problem or equation. It includes all valid solutions without any additional constraints.
3. Allows for Flexibility: Since a general solution does not involve any arbitrary constants, it allows for flexibility in determining specific solutions. It provides a framework within which specific solutions can be derived by assigning suitable values to the constants.
4. Can Be Modified: A general solution can be modified or refined by adding specific constraints or conditions. By introducing additional information or requirements, a more specific solution can be obtained.
5. Applies to a Range of Problems: The concept of a general solution is applicable to various mathematical and engineering problems. It is a fundamental concept in differential equations, linear algebra, and other branches of mathematics.
In conclusion, a general solution is a solution that represents all possible solutions to a problem or equation without any arbitrary constants. It offers flexibility and can be modified or refined to obtain more specific solutions.
Which of the following is the property of error function?- a)erf (0) = 1
- b)erf (∞) = 1
- c)erf (0) = ∞
- d)erf (∞) = 0
Correct answer is option 'B'. Can you explain this answer?
Which of the following is the property of error function?
a)
erf (0) = 1
b)
erf (∞) = 1
c)
erf (0) = ∞
d)
erf (∞) = 0
|
|
Maulik Joshi answered |
1) erf(0) = 0
2) erf(-x) = -erf(x)
3) erf(x) = 1 - erfc(x)
4) erf(inf) = 1
5) erf(-inf) = -1
The correct answer is a) erf(0) = 0.
2) erf(-x) = -erf(x)
3) erf(x) = 1 - erfc(x)
4) erf(inf) = 1
5) erf(-inf) = -1
The correct answer is a) erf(0) = 0.
One dimensional wave equation is- a)
- b)
- c)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
One dimensional wave equation is
a)
b)
c)
d)
None of these
|
|
Pioneer Academy answered |
Concept:
Wave equation:
It is a second-order linear partial differential equation for the description of waves (like mechanical waves).
The Partial Differential equation is given as,
For One-Dimensional equation,
where, A = α2, B = 0, C = -1
Put all the values in equation (1)
∴ 0 - 4 (α2)(-1)
4α2 > 0.
So, this is a one-dimensional wave equation.
Additional Information
having A = α2, B = 0, C = 0
Put all the values in equation (1), we get
0 - 4(α2)(0) = 0, therefore it shows parabolic function.
So, this is a one-dimensional heat equation.
having A = 1, B = 0, C = 1
Put all the values in equation (1), we get 0 - 4(1)(1) = -4, therefore it shows elliptical function.
So, this is a two-dimensional heat equation.
Wave equation:
It is a second-order linear partial differential equation for the description of waves (like mechanical waves).
The Partial Differential equation is given as,
For One-Dimensional equation,
where, A = α2, B = 0, C = -1
Put all the values in equation (1)
∴ 0 - 4 (α2)(-1)
4α2 > 0.
So, this is a one-dimensional wave equation.
Additional Information
having A = α2, B = 0, C = 0
Put all the values in equation (1), we get
0 - 4(α2)(0) = 0, therefore it shows parabolic function.
So, this is a one-dimensional heat equation.
having A = 1, B = 0, C = 1
Put all the values in equation (1), we get 0 - 4(1)(1) = -4, therefore it shows elliptical function.
So, this is a two-dimensional heat equation.
Consider the differential equation
with y(1) = 2π. There exists a unique solution for this differential equation when t belongs to the interval- a)(–2, 2)
- b)(–10, 10)
- c)(–10, 2)
- d)(0, 10)
Correct answer is option 'A'. Can you explain this answer?
Consider the differential equationwith y(1) = 2π. There exists a unique solution for this differential equation when t belongs to the interval
with y(1) = 2π. There exists a unique solution for this differential equation when t belongs to the intervala)
(–2, 2)
b)
(–10, 10)
c)
(–10, 2)
d)
(0, 10)
|
|
Pioneer Academy answered |
It is in the standard form of first order linear equation.
Integrating factor
Solution of differential equation is:
If t = ±9, solution doesn’t exist.
Solution of differential equation is:
If t = ±9, solution doesn’t exist.
Hence t ≠ ±9.
From the options, (-2, 2) doesn’t consists of ±9, hence, (-2, 2) is the correct.
The expression 
for the volume of a cone is equal to- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The expression for the volume of a cone is equal to
for the volume of a cone is equal toa)
b)
c)
d)
|
Nitin Joshi answered |
Choices (a) and (b) be correct because
for the volume of a cone is equal to
Which one of the following is the general solution of the first order differential equation dy/dx = (x + y − 1)2, where x, y are real?- a)y = 1 + x + tan–1 (x + c), where c is a constant.
- b)y = 1 + x + tan(x + c), where c is a constant.
- c)y = 1 – x + tan–1 (x + c), where c is a constant.
- d)y = 1 – x + tan(x + c), where c is a constant.
Correct answer is option 'D'. Can you explain this answer?
Which one of the following is the general solution of the first order differential equation dy/dx = (x + y − 1)2, where x, y are real?
a)
y = 1 + x + tan–1 (x + c), where c is a constant.
b)
y = 1 + x + tan(x + c), where c is a constant.
c)
y = 1 – x + tan–1 (x + c), where c is a constant.
d)
y = 1 – x + tan(x + c), where c is a constant.
|
Meghana Desai answered |
General Solution of the First Order Differential Equation
To find the general solution of the first order differential equation dy/dx = (x + y - 1)^2, we can rewrite the equation as:
dy/(x + y - 1)^2 = dx
Integrating both sides, we get:
Integration of the Differential Equation
∫ dy/(x + y - 1)^2 = ∫ dx
Substitute u = x + y - 1
Let u = x + y - 1. Then, du = dx + dy. Substituting these into the integral, we have:
∫ du/u^2 = ∫ dx
Integrate both sides
Integrating both sides, we get:
-1/u = x + C
where C is the constant of integration.
Substitute back y = u - x + 1
Substitute u = x + y - 1 back into the equation, we get:
-1/(x + y - 1) = x + C
Solving for y, we get:
y = 1 - x + tan(x + C)
Final General Solution
Therefore, the general solution of the first order differential equation dy/dx = (x + y - 1)^2 is:
y = 1 - x + tan(x + C), where C is a constant.
So, the correct answer is option 'D'.
The following partial differential equation is defined for u:u (x, y)
The set auxiliary conditions necessary to solve the equation uniquely, is - a)three initial conditions
- b)three boundary conditions
- c)two initial conditions and one boundary condition
- d)one initial conditions and two boundary conditions
Correct answer is option 'D'. Can you explain this answer?
The following partial differential equation is defined for u:u (x, y)
The set auxiliary conditions necessary to solve the equation uniquely, is
The set auxiliary conditions necessary to solve the equation uniquely, is
a)
three initial conditions
b)
three boundary conditions
c)
two initial conditions and one boundary condition
d)
one initial conditions and two boundary conditions
|
|
Pioneer Academy answered |
Given:
∵ y ≥ 0 ⇒ It can be replaced with ‘t’.
This is a 1-D Heat equation. It measures temperature distribution in a uniform rod.
The general solution is u = f(x, t)
Auxiliary solutions include both initial and boundary conditions.
(1) Number of initial conditions = Highest order of time derivative in partial differential = 1
(2) The number of boundary conditions:
To solve this partial differential equation, it needs to be integrated twice that will introduce two arbitrary constants.
Hence 2 boundary conditions and 1 initial condition are required to solve this Partial differential equation.
∵ y ≥ 0 ⇒ It can be replaced with ‘t’.
This is a 1-D Heat equation. It measures temperature distribution in a uniform rod.
The general solution is u = f(x, t)
Auxiliary solutions include both initial and boundary conditions.
(1) Number of initial conditions = Highest order of time derivative in partial differential = 1
(2) The number of boundary conditions:
To solve this partial differential equation, it needs to be integrated twice that will introduce two arbitrary constants.Hence 2 boundary conditions and 1 initial condition are required to solve this Partial differential equation.
A particle undergoes forced vibrations according to the law xn(t) + 25x(t) = 21 sin t. If the particle starts from rest at t = 0, find the displacement at any time t > 0. - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A particle undergoes forced vibrations according to the law xn(t) + 25x(t) = 21 sin t. If the particle starts from rest at t = 0, find the displacement at any time t > 0.
a)
b)
c)
d)
|
|
Sanya Agarwal answered |
Concept:
The solution of the linear differential equation is of the form:
y = C.F + P.I
where C.F is the complementary function and P.I is the particular integral.
The 2nd order linear differential equation in the symbolic form is represented as:
When X = Sin(ax)
f(-a2) is calculated by replacing D2 with a2 in f(D)
In the case of a forced damped system, the vibration equation:
The solution of x is (CF + PI)Where CF is complimentary function and PI is particular integral
But after some time CF becomes zero.
∴ x = P.I
Calculation:
Given:
xn(t) + 25x(t) = 21 sin t
The above equation can be written as:
D2 + 25x = 21 × Sin (t)
Here, a = 1, So putting -a2 = D2 = -1 in the above equation
∴ The displacement at any time t > 0 = 
Which of the following is a type of Iterative method of solving non-linear equations?- a)Graphical method
- b)Interpolation method
- c)Trial and Error methods
- d)Direct Analytical methods
Correct answer is option 'B'. Can you explain this answer?
Which of the following is a type of Iterative method of solving non-linear equations?
a)
Graphical method
b)
Interpolation method
c)
Trial and Error methods
d)
Direct Analytical methods
|
|
Poulomi Patel answered |
Interpolation method
The interpolation method is a type of iterative method used for solving non-linear equations. It involves the use of interpolation techniques to approximate the solution of the equation.
Iterative methods
Iterative methods are a class of numerical methods used to solve equations by making an initial guess and then refining it through a series of iterations until a desired level of accuracy is achieved. These methods are particularly useful for solving non-linear equations, where direct analytical methods may not be feasible.
Interpolation
Interpolation is a mathematical technique used to estimate unknown values based on known data points. It involves constructing a function that passes through the given data points and then using this function to approximate the value of the unknown variable.
Application in solving non-linear equations
In the context of solving non-linear equations, the interpolation method can be used to approximate the solution by constructing an interpolating function that passes through two known points and then finding the value of the unknown variable at the intersection of this function with the x-axis.
Steps in the interpolation method
1. Choose two initial guesses for the unknown variable, x.
2. Evaluate the function at the two initial guesses to obtain corresponding function values, f(x1) and f(x2).
3. Use interpolation techniques, such as linear interpolation or polynomial interpolation, to construct an interpolating function that passes through the points (x1, f(x1)) and (x2, f(x2)).
4. Find the value of x where the interpolating function intersects the x-axis, which represents an approximation of the solution to the non-linear equation.
5. Repeat steps 2-4 until the desired level of accuracy is achieved.
Advantages of the interpolation method
- It is a simple and intuitive method.
- It does not require knowledge of the derivative of the function.
- It can be easily implemented using computer software or calculators.
- It provides a systematic approach to approximate the solution of non-linear equations.
In conclusion, the interpolation method is a type of iterative method commonly used for solving non-linear equations. It involves constructing an interpolating function based on two initial guesses and using this function to approximate the solution. This method is advantageous due to its simplicity and ease of implementation.
The interpolation method is a type of iterative method used for solving non-linear equations. It involves the use of interpolation techniques to approximate the solution of the equation.
Iterative methods
Iterative methods are a class of numerical methods used to solve equations by making an initial guess and then refining it through a series of iterations until a desired level of accuracy is achieved. These methods are particularly useful for solving non-linear equations, where direct analytical methods may not be feasible.
Interpolation
Interpolation is a mathematical technique used to estimate unknown values based on known data points. It involves constructing a function that passes through the given data points and then using this function to approximate the value of the unknown variable.
Application in solving non-linear equations
In the context of solving non-linear equations, the interpolation method can be used to approximate the solution by constructing an interpolating function that passes through two known points and then finding the value of the unknown variable at the intersection of this function with the x-axis.
Steps in the interpolation method
1. Choose two initial guesses for the unknown variable, x.
2. Evaluate the function at the two initial guesses to obtain corresponding function values, f(x1) and f(x2).
3. Use interpolation techniques, such as linear interpolation or polynomial interpolation, to construct an interpolating function that passes through the points (x1, f(x1)) and (x2, f(x2)).
4. Find the value of x where the interpolating function intersects the x-axis, which represents an approximation of the solution to the non-linear equation.
5. Repeat steps 2-4 until the desired level of accuracy is achieved.
Advantages of the interpolation method
- It is a simple and intuitive method.
- It does not require knowledge of the derivative of the function.
- It can be easily implemented using computer software or calculators.
- It provides a systematic approach to approximate the solution of non-linear equations.
In conclusion, the interpolation method is a type of iterative method commonly used for solving non-linear equations. It involves constructing an interpolating function based on two initial guesses and using this function to approximate the solution. This method is advantageous due to its simplicity and ease of implementation.
Solve the following equation:
yexydx + (xexy + 2y)dy = 0- a)xexy + 2y2 = c
- b)xexy + y2 = c
- c)exy + 2y2 = c
- d)exy + y2 = c
Correct answer is option 'D'. Can you explain this answer?
Solve the following equation:
yexydx + (xexy + 2y)dy = 0
yexydx + (xexy + 2y)dy = 0
a)
xexy + 2y2 = c
b)
xexy + y2 = c
c)
exy + 2y2 = c
d)
exy + y2 = c
|
|
Sanya Agarwal answered |
The given equation is in the form of
M dx + N dy = 0
Here M = y exy
N = (x exy + 2y)
Hence the differential equation is an exact equation.
The solution is
Find the particular solution of the differential equation
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Find the particular solution of the differential equation
a)
b)
c)
d)
|
|
Engineers Adda answered |
Given differential equation is
auxiliary equation is (D − 2)2 = 0 ⇒ D = 2, 2
This problem can be solved by using the method of variation of parameters. Then
Given 
the value of y at x = 2 is ________ (round off to nearest integer) (Important - Enter only the numerical value in the answer)
Correct answer is between '162,170'. Can you explain this answer?
Given the value of y at x = 2 is ________ (round off to nearest integer) (Important - Enter only the numerical value in the answer)
the value of y at x = 2 is ________ (round off to nearest integer) (Important - Enter only the numerical value in the answer)|
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Sanya Agarwal answered |
Given the differential equation is,
By integrating with respect to ‘x’, we get
⇒ ln(20) = C
By integrating with respect to ‘x’, we get
⇒ ln(20) = C
Now, the differential equation becomes
At x = 2, y = 20 (e2 + 1) = 167.78
At x = 2, y = 20 (e2 + 1) = 167.78
Cauchy’s linear differential equation 
can be reduced to a linear differential equation with constant coefficient by using substitution- a)x = ez
- b)y = ez
- c)z = ex
- d)z = ey
Correct answer is option 'A'. Can you explain this answer?
Cauchy’s linear differential equation can be reduced to a linear differential equation with constant coefficient by using substitution
can be reduced to a linear differential equation with constant coefficient by using substitutiona)
x = ez
b)
y = ez
c)
z = ex
d)
z = ey
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Gate Gurus answered |
Concept:
Any linear equation of the following form:
is considered as Cauchy’s differential equation. The equation has variable coefficients so its solution becomes tedious but we can convert the above equation into the linear differential equation with constant coefficientsBy taking,
log x = z or x = ez
Proof:
log x = z
Taking differentiation on both sides we get,
Now, it can be solved by finding C.F and P.I just like we solve linear differential equations with constant coefficients.
Now, it can be solved by finding C.F and P.I just like we solve linear differential equations with constant coefficients.
General solution of the Cauchy-Euler equation
- a)y = c1x2 + c2x4
- b)y = c1x2 + c2x-4
- c)y = (c1 + c2 In x) x4
- d)y = c1x4 + c2x-4 In x
Correct answer is option 'C'. Can you explain this answer?
General solution of the Cauchy-Euler equation
a)
y = c1x2 + c2x4
b)
y = c1x2 + c2x-4
c)
y = (c1 + c2 In x) x4
d)
y = c1x4 + c2x-4 In x
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Gate Gurus answered |
Concept:
For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.
Calculation:
Calculation:
Given:
Put x = et
Put x = et
⇒ t = ln x
Now, the above differential equation becomes
Now, the above differential equation becomes
D(D – 1)y – 7Dy + 16y = 0
⇒ D2y – Dy – 7Dy + 16y = 0
⇒ (D2 – 8D + 16)y = 0
Auxiliary equation:
(D2 – 8 D + 16) = 0
⇒ D = 4
The solutions for the above roots of auxiliary equations are:
y(t) = (c1 + c2 t) e4t
⇒ y(x) = (c1 + c2 ln x) x4
The Wronskian of the differential equation
- a)ex
- b)-e3x
- c)e2x
- d)e5x
Correct answer is option 'B'. Can you explain this answer?
The Wronskian of the differential equation
a)
ex
b)
-e3x
c)
e2x
d)
e5x
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Sanya Agarwal answered |
If the given differential equation is in the form, y'' + py' + qy = x
Where, p, q, x are functions of x.
Then Wronskian is,
Here, y1 and y2 are the solutions of y'' + py' + qy = 0
Given differential equation is,
Auxiliary equation is,
D2 – 3D + 2 = 0
⇒ D = 1, 2
C.F. = c1 e2x + c2 ex
y1 = e2x, y2 = ex
Consider the initial value problem below. The value of y at x = In 2, (rounded off to 3 decimal places) is dy/dx = 2x - y, y(0) = 1 (Important - Enter only the numerical value in the answer)
Correct answer is between '0.8774,0.8952'. Can you explain this answer?
Consider the initial value problem below. The value of y at x = In 2, (rounded off to 3 decimal places) is dy/dx = 2x - y, y(0) = 1 (Important - Enter only the numerical value in the answer)
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Sanvi Kapoor answered |
Concept:
The standard form of a first-order linear differential equation is,
Where P and Q are the functions of x.
Where P and Q are the functions of x.
Integrating factor, IF = e∫Pdx
Now, the solution for the above differential equation is,
y(IF) = ∫IF.Qdx
Calculation:
By comparing the above differential equation with the standard differential equation,
P = 1, Q = 2x
Integrating factor, IF = e∫Pdx = e∫1dx = ex
Now, the solution is
y(ex) = ∫ex(2x)dx
yex = 2(xex − ex) + C
⇒ y = 2(x − 1) + Ce−x
y(0) = 1
⇒ 1 = 2 (0 – 1) + C
⇒ C = 3
Now, the solution becomes
y = 2x – 2 + 3e-x
At x = ln 2,
y = 2 ln 2 – 2 + 3 (0.5) = 0.8862
What is the complete solution for the equation x (y - z) p + y (z - x) q = z (x - y)?- a)ϕ (x + y + z, xyz) = 0
- b)ϕ (x + 2y + z, xz) = 0
- c)ϕ (2x + y + z, xyz) = 0
- d)ϕ (x + y + z, xy) = 0
Correct answer is option 'A'. Can you explain this answer?
What is the complete solution for the equation x (y - z) p + y (z - x) q = z (x - y)?
a)
ϕ (x + y + z, xyz) = 0
b)
ϕ (x + 2y + z, xz) = 0
c)
ϕ (2x + y + z, xyz) = 0
d)
ϕ (x + y + z, xy) = 0
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Nilesh Kapoor answered |
To solve the equation x(y - z)p y(z - x)q = z(x - y), we need to simplify the equation and find the values of x, y, and z that satisfy the equation.
Given equation: x(y - z)p y(z - x)q = z(x - y)
Simplifying the equation:
xy^2p - xz^2p + y^2zq - yx^2q = zx - zy
xy^2p - xz^2p + y^2zq - yx^2q - zx + zy = 0
Now let's break down the options and see which one satisfies the equation:
a) (x y z, xyz) = 0
This option suggests that the values of x, y, and z multiplied together should be equal to zero. However, this condition does not ensure that the given equation is satisfied. Therefore, option 'A' is not the correct solution.
b) (x 2y z, xz) = 0
This option suggests that the values of x and z multiplied together should be equal to zero. Again, this condition does not guarantee that the equation is satisfied. Therefore, option 'B' is not the correct solution.
c) (2x y z, xyz) = 0
This option suggests that the values of 2x, y, and z multiplied together should be equal to zero. Similar to the previous options, this condition does not lead to the solution of the equation. Therefore, option 'C' is not the correct solution.
d) (x y z, xy) = 0
This option suggests that the values of x, y, and z multiplied by xy should be equal to zero. However, this condition does not satisfy the equation. Therefore, option 'D' is not the correct solution.
Therefore, the correct solution for the equation x(y - z)p y(z - x)q = z(x - y) is option 'A', which states that (x y z, xyz) = 0.
Given equation: x(y - z)p y(z - x)q = z(x - y)
Simplifying the equation:
xy^2p - xz^2p + y^2zq - yx^2q = zx - zy
xy^2p - xz^2p + y^2zq - yx^2q - zx + zy = 0
Now let's break down the options and see which one satisfies the equation:
a) (x y z, xyz) = 0
This option suggests that the values of x, y, and z multiplied together should be equal to zero. However, this condition does not ensure that the given equation is satisfied. Therefore, option 'A' is not the correct solution.
b) (x 2y z, xz) = 0
This option suggests that the values of x and z multiplied together should be equal to zero. Again, this condition does not guarantee that the equation is satisfied. Therefore, option 'B' is not the correct solution.
c) (2x y z, xyz) = 0
This option suggests that the values of 2x, y, and z multiplied together should be equal to zero. Similar to the previous options, this condition does not lead to the solution of the equation. Therefore, option 'C' is not the correct solution.
d) (x y z, xy) = 0
This option suggests that the values of x, y, and z multiplied by xy should be equal to zero. However, this condition does not satisfy the equation. Therefore, option 'D' is not the correct solution.
Therefore, the correct solution for the equation x(y - z)p y(z - x)q = z(x - y) is option 'A', which states that (x y z, xyz) = 0.
If we use the Fourier transform 
to solve the partial differential equation 
in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕk(y) depend on y as yα and yβ. The values of α and β are - a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If we use the Fourier transform to solve the partial differential equation in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕk(y) depend on y as yα and yβ. The values of α and β are
to solve the partial differential equation in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕk(y) depend on y as yα and yβ. The values of α and β are a)
b)
c)
d)
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Sanvi Kapoor answered |
The Fourier transform, 
CALCULATION:
We have;
Now take the derivative of ϕ(x, y) we have;
Again derivative we have;
Now putting these values in equation 1) we have;
Now, by applying Cauchy's differential equation we have;
Again solving we have;
The roots are written as;
Hence option 3) is the correct answer.
CALCULATION:
We have;
Now take the derivative of ϕ(x, y) we have;
Again derivative we have;
Now putting these values in equation 1) we have;
Now, by applying Cauchy's differential equation we have;
Again solving we have;
The roots are written as;
Hence option 3) is the correct answer.
Solve: (x2D2 - 4xD + 6)y = x2
where D = d/dx - a)y = c1 x2 + c2 x2 - x2 logx2
- b)y = c1 x2 + c2 x3 - xlogx2
- c)y = c1 x2 + c2 x3 - x2logx2
- d)y = c1 x2 + c2 x2 - x logx2
Correct answer is option 'C'. Can you explain this answer?
Solve: (x2D2 - 4xD + 6)y = x2
where D = d/dx
where D = d/dx
a)
y = c1 x2 + c2 x2 - x2 logx2
b)
y = c1 x2 + c2 x3 - xlogx2
c)
y = c1 x2 + c2 x3 - x2logx2
d)
y = c1 x2 + c2 x2 - x logx2
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Raghavendra Dasgupta answered |
To solve the given differential equation, we substitute the value of D in terms of d/dx and rearrange the equation to isolate y.
Given: (x^2D^2 - 4xD + 6)y = x^2
We are given that D = d/dx, so substituting this in the equation, we get:
(x^2(d/dx)^2 - 4x(d/dx) + 6)y = x^2
Simplifying the equation further, we have:
(x^2(d^2y/dx^2) - 4x(dy/dx) + 6)y = x^2
Expanding the equation, we get:
x^2(d^2y/dx^2)y - 4x(dy/dx)y + 6y = x^2
Rearranging the terms, we have:
x^2(d^2y/dx^2)y - 4x(dy/dx)y + (6y - x^2) = 0
Now, we can see that this is a homogeneous linear differential equation. So, let's assume y = x^m and substitute it in the equation.
(x^2d^2/dx^2)(x^m) - 4(xd/dx)(x^m) + (6x^m - x^2) = 0
Differentiating y = x^m twice, we get:
m(m-1)x^m-2 - 4m(x^m-1) + (6x^m - x^2) = 0
Expanding and combining like terms, we have:
m(m-1)x^m-2 - 4mx^m-1 + 6x^m - x^2 = 0
Now, let's divide the entire equation by x^m-2 to simplify further:
m(m-1) - 4mx + 6x^2/x^m-2 - x^2/x^m-2 = 0
Simplifying, we have:
m(m-1) - 4mx + 6x^2/x^2 - x^2/x^2 = 0
m(m-1) - 4mx + 6x^2 - x^2 = 0
m^2 - m - 4mx + 5x^2 = 0
Now, we can equate the coefficients of the like powers of x to get the values of m:
m^2 - m = 0 (coefficient of x^0)
-4m + 5 = 0 (coefficient of x^2)
Solving these equations, we get:
m = 0 or m = 1
m = 5/4
Therefore, the general solution to the differential equation is:
y = c1x^0 + c2x^1 + c3x^(5/4)
Simplifying further, we have:
y = c1 + c2x + c3x^(5/4)
This matches with the option C, which states that y = c1x^2 + c2x^3 - x^2log(x^2).
Hence, the correct answer is option C.
Given: (x^2D^2 - 4xD + 6)y = x^2
We are given that D = d/dx, so substituting this in the equation, we get:
(x^2(d/dx)^2 - 4x(d/dx) + 6)y = x^2
Simplifying the equation further, we have:
(x^2(d^2y/dx^2) - 4x(dy/dx) + 6)y = x^2
Expanding the equation, we get:
x^2(d^2y/dx^2)y - 4x(dy/dx)y + 6y = x^2
Rearranging the terms, we have:
x^2(d^2y/dx^2)y - 4x(dy/dx)y + (6y - x^2) = 0
Now, we can see that this is a homogeneous linear differential equation. So, let's assume y = x^m and substitute it in the equation.
(x^2d^2/dx^2)(x^m) - 4(xd/dx)(x^m) + (6x^m - x^2) = 0
Differentiating y = x^m twice, we get:
m(m-1)x^m-2 - 4m(x^m-1) + (6x^m - x^2) = 0
Expanding and combining like terms, we have:
m(m-1)x^m-2 - 4mx^m-1 + 6x^m - x^2 = 0
Now, let's divide the entire equation by x^m-2 to simplify further:
m(m-1) - 4mx + 6x^2/x^m-2 - x^2/x^m-2 = 0
Simplifying, we have:
m(m-1) - 4mx + 6x^2/x^2 - x^2/x^2 = 0
m(m-1) - 4mx + 6x^2 - x^2 = 0
m^2 - m - 4mx + 5x^2 = 0
Now, we can equate the coefficients of the like powers of x to get the values of m:
m^2 - m = 0 (coefficient of x^0)
-4m + 5 = 0 (coefficient of x^2)
Solving these equations, we get:
m = 0 or m = 1
m = 5/4
Therefore, the general solution to the differential equation is:
y = c1x^0 + c2x^1 + c3x^(5/4)
Simplifying further, we have:
y = c1 + c2x + c3x^(5/4)
This matches with the option C, which states that y = c1x^2 + c2x^3 - x^2log(x^2).
Hence, the correct answer is option C.
The solution to the differential equation f’’(x)+4f’(x)+4f(x)=0 is - a)f1(x) = e-2x
- b)f1(x) = e2x, f2(x) = e-2x
- c)f1(x) = e-2x, f2(x) = xe-2x
- d)f1(x) = e-2x, f2(x) = e-x
Correct answer is option 'C'. Can you explain this answer?
The solution to the differential equation f’’(x)+4f’(x)+4f(x)=0 is
a)
f1(x) = e-2x
b)
f1(x) = e2x, f2(x) = e-2x
c)
f1(x) = e-2x, f2(x) = xe-2x
d)
f1(x) = e-2x, f2(x) = e-x
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Anshu Patel answered |
Let y(x) = emx (m ≠ 0)be the trial soln .Auxiliary equation. m2 + 4m+ 4 = 0 ⇒(m+ 2)2 = 0
In particular, when A =1,B =1,then f(x) = (1 + x)e−2x
= e−2x + xe−2x
The solution of differential equation dx – (x + y + 1) dy = 0 is- a)(x + y + 2) ey = C
- b)(x – y - 2) ey = C
- c)(x + y + 2) e-y = C
- d)(x – y + 2) ey = C
Correct answer is option 'C'. Can you explain this answer?
The solution of differential equation dx – (x + y + 1) dy = 0 is
a)
(x + y + 2) ey = C
b)
(x – y - 2) ey = C
c)
(x + y + 2) e-y = C
d)
(x – y + 2) ey = C
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Vertex Academy answered |
Given differential equation is,
dx – (x + y + 1) dy = 0
Put (x + y + 1) = t
Now, the differential equation becomes
⇒ t - In (t + 1) = x + C1
⇒ (x + y + 1) – In (x + y + 2) = x + C1
⇒ y + 1 – C1 = In (x + y + 2)
⇒ In (x + y + 2) = y + k
⇒ (x + y + 2) = ey+k
⇒ (x + y + 2) e
= C
The solution of differential equation
will be ___________, where c1 and c2 are arbitrary constants.- a)y = c1x + c2x2
- b)y = c1 log x + c2x
- c)y = c1 + c2x
- d)y = c1x2 + c2x3
Correct answer is option 'A'. Can you explain this answer?
The solution of differential equationwill be ___________, where c1 and c2 are arbitrary constants.
will be ___________, where c1 and c2 are arbitrary constants.a)
y = c1x + c2x2
b)
y = c1 log x + c2x
c)
y = c1 + c2x
d)
y = c1x2 + c2x3
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Gate Gurus answered |
Concept:
Linear differential equations with variable coefficients can be reduced to linear differential equations with constant coefficients by suitable substitutions.
Euler Cauchy Homogeneous linear equation:
Calculation:
Calculation:
Given:
Euler Cauchy equation:
Now, the above differential equation becomes
Now, the above differential equation becomes
D (D – 1) y – 2 D y + 2 y = 0
⇒ D2 y – D y – 2 D y + 2 y = 0
⇒ (D2 – 3 D + 2) y = 0
Auxiliary equation:
(D2 – 3 D + 2) = 0
⇒ (D - 2) (D - 1) = 0
⇒ D = 2, 1
Now the roots are real and different, and the general solution of given equation is
y = c1 e2t + c2 et
⇒ y = c1x + c2x2
Solve
Which of the following is the solution of above equation, when x = 0?- a)Can be any constant
- b)Undefined
- c)Zero
- d)1
Correct answer is option 'B'. Can you explain this answer?
Solve
Which of the following is the solution of above equation, when x = 0?
a)
Can be any constant
b)
Undefined
c)
Zero
d)
1
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Sanya Agarwal answered |
Concept:
The given equation is multiplied with x to make it a equation of Cauchy Euler type.
The given equation is of Cauchy Euler type. Cauchy Euler equation is homogenous linear equation with variable coefficient.
The given equation is multiplied with x to make it a equation of Cauchy Euler type.
The given equation is of Cauchy Euler type. Cauchy Euler equation is homogenous linear equation with variable coefficient.
They can be reduced to linear equations with constant coefficients
Calculation:
Given:
Put x = et
log x = t
The given equation is written as
[x2D2 + xD]y = 0 ---(1)
Put xD = D’, x2D2 = D’(D’ - 1)
[D’(D’ - 1) + D’]y = 0
D’2 y = 0
The auxiliary equation is D’2 = 0
The roots of this equation are 0, 0
C.F = (At + B) e0t
y = (At + B)
Put t = log x
y = A log x + B
At x = 0; y is undefined as log 0 is undefined.
Which of the following is an example of non-linear differential equation?- a)y = mx + c
- b)x + x’ = 0
- c)x + x2 = 0
- d)x” + 2x = 0
Correct answer is option 'C'. Can you explain this answer?
Which of the following is an example of non-linear differential equation?
a)
y = mx + c
b)
x + x’ = 0
c)
x + x2 = 0
d)
x” + 2x = 0
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Sanya Agarwal answered |
For a differential equation to be linear the dependent variable should be of first degree. Since in equation x+x2=0, x2 is not a first power, it is not an example of linear differential equation.
What are the conditions called which are required for a signal to fulfil to be represented as Fourier series?- a)Dirichlet’s conditions
- b)Gibbs phenomenon
- c)Fourier conditions
- d)Fourier phenomenon
Correct answer is option 'A'. Can you explain this answer?
What are the conditions called which are required for a signal to fulfil to be represented as Fourier series?
a)
Dirichlet’s conditions
b)
Gibbs phenomenon
c)
Fourier conditions
d)
Fourier phenomenon
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Sanya Agarwal answered |
When the Dirichlet’s conditions are satisfied, then only for a signal, the fourier series exist. Fourier series is of two types- trigonometric series and exponential series.
The solution of the differential equation (dy/dx) = ky, y(0) = c is- a)x = ce-ky
- b)x = kecy
- c)y = cekx
- d)y = ce-kx
Correct answer is option 'C'. Can you explain this answer?
The solution of the differential equation (dy/dx) = ky, y(0) = c is
a)
x = ce-ky
b)
x = kecy
c)
y = cekx
d)
y = ce-kx
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Alok Iyer answered |
Solution:
The given differential equation is
(dy/dx) = ky
To solve this differential equation, we can separate the variables by moving all terms involving y to one side and all terms involving x to the other side:
(dy/y) = k(dx)
Next, we can integrate both sides of the equation. The integral of (dy/y) is ln|y|, and the integral of k(dx) is kx + C, where C is the constant of integration:
ln|y| = kx + C
To eliminate the absolute value, we can exponentiate both sides of the equation:
e^(ln|y|) = e^(kx + C)
This simplifies to:
|y| = e^(kx) * e^C
Since e^C is a constant, we can replace it with another constant, let's say A:
|y| = Ae^(kx)
Now, we can consider two cases for the absolute value:
Case 1: y > 0
In this case, the absolute value can be removed:
y = Ae^(kx)
Case 2: y < />
In this case, the absolute value becomes a negative sign:
-y = Ae^(kx)
Multiplying both sides by -1, we get:
y = -Ae^(kx)
Combining both cases, we can write the solution as:
y = Ce^(kx)
where C = A if y > 0, and C = -A if y < />
Given that y(0) = c, we can substitute this initial condition into the solution:
c = Ce^(k*0)
Since e^0 = 1, this simplifies to:
c = C
Therefore, the constant C is equal to c. Substituting this back into the solution, we get:
y = ce^(kx)
Hence, the correct solution to the given differential equation is option 'C': y = ce^(kx).
The given differential equation is
(dy/dx) = ky
To solve this differential equation, we can separate the variables by moving all terms involving y to one side and all terms involving x to the other side:
(dy/y) = k(dx)
Next, we can integrate both sides of the equation. The integral of (dy/y) is ln|y|, and the integral of k(dx) is kx + C, where C is the constant of integration:
ln|y| = kx + C
To eliminate the absolute value, we can exponentiate both sides of the equation:
e^(ln|y|) = e^(kx + C)
This simplifies to:
|y| = e^(kx) * e^C
Since e^C is a constant, we can replace it with another constant, let's say A:
|y| = Ae^(kx)
Now, we can consider two cases for the absolute value:
Case 1: y > 0
In this case, the absolute value can be removed:
y = Ae^(kx)
Case 2: y < />
In this case, the absolute value becomes a negative sign:
-y = Ae^(kx)
Multiplying both sides by -1, we get:
y = -Ae^(kx)
Combining both cases, we can write the solution as:
y = Ce^(kx)
where C = A if y > 0, and C = -A if y < />
Given that y(0) = c, we can substitute this initial condition into the solution:
c = Ce^(k*0)
Since e^0 = 1, this simplifies to:
c = C
Therefore, the constant C is equal to c. Substituting this back into the solution, we get:
y = ce^(kx)
Hence, the correct solution to the given differential equation is option 'C': y = ce^(kx).
A spherical naphthalene ball exposed to the atmosphere loses volume at a rate proportional to its instantaneous surface area due to evaporation. If the initial diameter of the ball is 2 cm and the diameter reduces to 1 cm after 3 months, the ball completely evaporates in - a)6 month s
- b)9 month s
- c)12 month s
- d)infinite time
Correct answer is option 'A'. Can you explain this answer?
A spherical naphthalene ball exposed to the atmosphere loses volume at a rate proportional to its instantaneous surface area due to evaporation. If the initial diameter of the ball is 2 cm and the diameter reduces to 1 cm after 3 months, the ball completely evaporates in
a)
6 month s
b)
9 month s
c)
12 month s
d)
infinite time
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Diya Patel answered |
By the given condition
The differential equation representing the family of circles touching y-axis at origin is:- a)Linear & of first order
- b)Linear & of second order
- c)Non-linear & of first order
- d)Non-linear & of second order
Correct answer is option 'C'. Can you explain this answer?
The differential equation representing the family of circles touching y-axis at origin is:
a)
Linear & of first order
b)
Linear & of second order
c)
Non-linear & of first order
d)
Non-linear & of second order
|
|
Jay Sharma answered |
Understanding the Family of Circles
The family of circles that touch the y-axis at the origin can be represented by the general equation of a circle. The equation can be described as:
- (x - h)² + (y - k)² = r²
Here, for circles touching the y-axis at the origin, the center (h, k) will be (r, k) where r is the radius and k can take any value.
Equation of the Circles
The specific equation becomes:
- (x - r)² + (y - k)² = r²
This can be expanded and rearranged as:
- x² - 2xr + r² + y² - 2yk + k² = r²
This simplifies to:
- x² + y² - 2xr - 2yk + k² = 0
Deriving the Differential Equation
To find the differential equation, we differentiate this equation with respect to x:
- 2x + 2y(dy/dx) - 2r - 2(dy/dx)k = 0
Rearranging gives us a first-order differential equation in terms of dy/dx.
Nature of the Differential Equation
- The equation is non-linear because of the presence of the terms involving (dy/dx) in a multiplicative form.
- It is of first order since it involves the first derivative but not higher derivatives.
Conclusion
Thus, the correct classification of the differential equation representing the family of circles touching the y-axis at the origin is:
- Non-linear & of first order (Option C).
The family of circles that touch the y-axis at the origin can be represented by the general equation of a circle. The equation can be described as:
- (x - h)² + (y - k)² = r²
Here, for circles touching the y-axis at the origin, the center (h, k) will be (r, k) where r is the radius and k can take any value.
Equation of the Circles
The specific equation becomes:
- (x - r)² + (y - k)² = r²
This can be expanded and rearranged as:
- x² - 2xr + r² + y² - 2yk + k² = r²
This simplifies to:
- x² + y² - 2xr - 2yk + k² = 0
Deriving the Differential Equation
To find the differential equation, we differentiate this equation with respect to x:
- 2x + 2y(dy/dx) - 2r - 2(dy/dx)k = 0
Rearranging gives us a first-order differential equation in terms of dy/dx.
Nature of the Differential Equation
- The equation is non-linear because of the presence of the terms involving (dy/dx) in a multiplicative form.
- It is of first order since it involves the first derivative but not higher derivatives.
Conclusion
Thus, the correct classification of the differential equation representing the family of circles touching the y-axis at the origin is:
- Non-linear & of first order (Option C).
The ordinary differential equation
- a)linear and homogeneous
- b)linear and non-homogeneous
- c)non-linear and homogeneous
- d)non-linear and non-homogeneous
Correct answer is option 'B'. Can you explain this answer?
The ordinary differential equation
a)
linear and homogeneous
b)
linear and non-homogeneous
c)
non-linear and homogeneous
d)
non-linear and non-homogeneous
|
|
Vertex Academy answered |
Concept:
Identification of Non-linear Differential Equation:
If any differential equation consists at least one of the above properties, then it is called non-linear differential equation and if any differential equation is free from all the above properties, then it is a linear differential equation.
If any differential equation consists at least one of the above properties, then it is called non-linear differential equation and if any differential equation is free from all the above properties, then it is a linear differential equation.
Given:
It is free from all the four characteristics described in the table. Hence it is a linear differential equation.
Product of dependent and independent variable i.e. x and u is present. Hence it is a non-homogenous equation.
Singular solution of a differential equation is one that cannot be obtained from the general solution gotten by the usual method of solving the differential equation.- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
Singular solution of a differential equation is one that cannot be obtained from the general solution gotten by the usual method of solving the differential equation.
a)
True
b)
False
|
Neha Mukherjee answered |
Singular Solution of a Differential Equation
The given statement is true. A singular solution of a differential equation is one that cannot be obtained from the general solution obtained by the usual method of solving the differential equation.
General Solution
The general solution of a differential equation contains all possible solutions to the equation. It is obtained by integrating the differential equation and including an arbitrary constant.
Method of Solving Differential Equations
The usual method of solving a differential equation involves finding the general solution by integrating the equation and then applying initial or boundary conditions to determine the particular solution.
Singular Solution
A singular solution is a special solution that cannot be derived from the general solution. It arises when the arbitrary constant in the general solution takes on a specific value that satisfies additional conditions or constraints. These additional conditions are not accounted for in the general solution.
Example
Consider the differential equation: dy/dx = x/y
The general solution of this differential equation can be obtained by separating the variables and integrating both sides:
∫y dy = ∫x dx
y^2/2 = x^2/2 + C
Here, C is the arbitrary constant and represents the general solution.
However, suppose we have an additional condition that the solution must pass through the point (1, 1). Substituting these values into the general solution, we get:
(1)^2/2 = (1)^2/2 + C
1/2 = 1/2 + C
C = 0
The singular solution is obtained when the arbitrary constant C takes on a specific value of zero. Therefore, the singular solution to the differential equation is:
y^2/2 = x^2/2
This singular solution satisfies the additional condition and cannot be obtained from the general solution.
Conclusion
In conclusion, a singular solution of a differential equation is one that arises when the arbitrary constant in the general solution takes on a specific value that satisfies additional conditions or constraints. These singular solutions cannot be obtained from the general solution obtained by the usual method of solving the differential equation.
The given statement is true. A singular solution of a differential equation is one that cannot be obtained from the general solution obtained by the usual method of solving the differential equation.
General Solution
The general solution of a differential equation contains all possible solutions to the equation. It is obtained by integrating the differential equation and including an arbitrary constant.
Method of Solving Differential Equations
The usual method of solving a differential equation involves finding the general solution by integrating the equation and then applying initial or boundary conditions to determine the particular solution.
Singular Solution
A singular solution is a special solution that cannot be derived from the general solution. It arises when the arbitrary constant in the general solution takes on a specific value that satisfies additional conditions or constraints. These additional conditions are not accounted for in the general solution.
Example
Consider the differential equation: dy/dx = x/y
The general solution of this differential equation can be obtained by separating the variables and integrating both sides:
∫y dy = ∫x dx
y^2/2 = x^2/2 + C
Here, C is the arbitrary constant and represents the general solution.
However, suppose we have an additional condition that the solution must pass through the point (1, 1). Substituting these values into the general solution, we get:
(1)^2/2 = (1)^2/2 + C
1/2 = 1/2 + C
C = 0
The singular solution is obtained when the arbitrary constant C takes on a specific value of zero. Therefore, the singular solution to the differential equation is:
y^2/2 = x^2/2
This singular solution satisfies the additional condition and cannot be obtained from the general solution.
Conclusion
In conclusion, a singular solution of a differential equation is one that arises when the arbitrary constant in the general solution takes on a specific value that satisfies additional conditions or constraints. These singular solutions cannot be obtained from the general solution obtained by the usual method of solving the differential equation.
Chapter doubts & questions for Differential Equations - Engineering Mathematics for Electrical Engineering 2025 is part of Electrical Engineering (EE) exam preparation. The chapters have been prepared according to the Electrical Engineering (EE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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