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All questions of Differential Equations for Mathematics Exam
The order of equation p tan y + q tan x = sec2z is- a)1
- b)2
- c)0
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The order of equation p tan y + q tan x = sec2z is
a)
1
b)
2
c)
0
d)
none of these
|
Praveen Sharma answered |
It is linear PDE with x andz as independent variables and y as dependent variable
Consider the differential equation y" + 6y' + 25y - 0 with initial condition y(0) = 0. Then, the general solution of the IVP is- a)e-3x (A cos 4x + B sin 4x)
- b)Be-3x sin 4x
- c)Ae-4x sin 3x
- d)e-4x (A cos 3x + B sin 3x)
Correct answer is option 'B'. Can you explain this answer?
Consider the differential equation y" + 6y' + 25y - 0 with initial condition y(0) = 0. Then, the general solution of the IVP is
a)
e-3x (A cos 4x + B sin 4x)
b)
Be-3x sin 4x
c)
Ae-4x sin 3x
d)
e-4x (A cos 3x + B sin 3x)
|
Chirag Verma answered |
we have (D2 + 6D + 25)y = 0
So, y =e-3x [A cos 4x + B sin 4x]
but y(0) = 0 implies A = 0
Hence, y = Be-3x sin 4x
So, y =e-3x [A cos 4x + B sin 4x]
but y(0) = 0 implies A = 0
Hence, y = Be-3x sin 4x
For a partial differential equation, in a function φ (x, y) and two variables x, y, what is the form obtained after separation of variables is applied?- a)Φ (x, y) = X(x) + Y(y)
- b)Φ (x, y) = X(x) - Y(y)
- c)Φ (x, y) = X(x) / Y(y)
- d)Φ (x, y) = X(x)Y(y)
Correct answer is option 'D'. Can you explain this answer?
For a partial differential equation, in a function φ (x, y) and two variables x, y, what is the form obtained after separation of variables is applied?
a)
Φ (x, y) = X(x) + Y(y)
b)
Φ (x, y) = X(x) - Y(y)
c)
Φ (x, y) = X(x) / Y(y)
d)
Φ (x, y) = X(x)Y(y)
|
Veda Institute answered |
The method of separation of variables relies upon the assumption that a function of the form,
Φ (x, y) = X(x)Y(y)
will be a solution to a linear homogeneous partial differential equation in x and y. This is called a product solution and provided the boundary conditions are also linear and homogeneous this will also satisfy the boundary conditions.
Φ (x, y) = X(x)Y(y)
will be a solution to a linear homogeneous partial differential equation in x and y. This is called a product solution and provided the boundary conditions are also linear and homogeneous this will also satisfy the boundary conditions.
If general solution of the differential equation ay'" + by" + cy’ + dy = 0 is linearly spanned by ex, sinx and cosx, then which one of the following holds?- a)a + b - c - d = 0
- b)a + b + c + d = 0
- c)a - b + c - d = 0
- d)a - b - c - d=0
Correct answer is option 'B'. Can you explain this answer?
If general solution of the differential equation ay'" + by" + cy’ + dy = 0 is linearly spanned by ex, sinx and cosx, then which one of the following holds?
a)
a + b - c - d = 0
b)
a + b + c + d = 0
c)
a - b + c - d = 0
d)
a - b - c - d=0
|
|
Chirag Verma answered |
Consider the differential equation ( x + y + 1) dx + (2x + 2y + 1) dy = 0. Which of the following statements is true?- a)The differential equation is linear
- b)The differential equation is exact
- c)ex + y is an integrating factor of the differential equation
- d)A suitable substitution transforms the differentiable equation to the variable separable form
Correct answer is option 'D'. Can you explain this answer?
Consider the differential equation ( x + y + 1) dx + (2x + 2y + 1) dy = 0. Which of the following statements is true?
a)
The differential equation is linear
b)
The differential equation is exact
c)
ex + y is an integrating factor of the differential equation
d)
A suitable substitution transforms the differentiable equation to the variable separable form
|
Dronacharya Institute answered |
To determine which statement is true about the given differential equation, let's analyze each option:
a) The differential equation is linear:
A linear differential equation is one in which the dependent variable and its derivatives appear only to the first power and are not multiplied together. In the given equation, both x and y appear to the first power, but they are multiplied together in the terms (xy)dx and (2xy)dy. Therefore, the given differential equation is not linear.
b) The differential equation is exact:
A differential equation is exact if it can be written in the form M(x, y)dx + N(x, y)dy = 0, where M and N are functions of x and y, and ∂M/∂y = ∂N/∂x. Let's check if this condition holds for the given equation:
M(x, y) = x + y
N(x, y) = 2x + 2y
∂M/∂y = 1
∂N/∂x = 2
Since ∂M/∂y is not equal to ∂N/∂x, the given differential equation is not exact.
c) ex^y is an integrating factor of the differential equation:
An integrating factor is a function that can be multiplied to a differential equation to make it exact. In this case, the integrating factor would need to be able to transform the given equation into an exact one. However, as we determined in the previous option, the given equation is not exact. Therefore, ex^y is not an integrating factor.
d) A suitable substitution transforms the differential equation to the variable separable form:
Variable separable form means that the equation can be written in the form f(x)dx = g(y)dy, where f(x) and g(y) are functions of x and y, respectively. To determine if a suitable substitution can transform the given equation into this form, let's make the substitution x = u and y = v, where u and v are functions of a new variable t (u = u(t) and v = v(t)). Substituting these into the given equation, we have:
(uv)du + (2uv)dv = 0
Factoring out uv, we get:
uv(du + 2dv) = 0
Since this equation can be separated into f(u)du = -2g(v)dv, where f(u) = uv and g(v) = -v, we can conclude that a suitable substitution transforms the differential equation to the variable separable form.
Therefore, the correct answer is option d) A suitable substitution transforms the differential equation to the variable separable form.
a) The differential equation is linear:
A linear differential equation is one in which the dependent variable and its derivatives appear only to the first power and are not multiplied together. In the given equation, both x and y appear to the first power, but they are multiplied together in the terms (xy)dx and (2xy)dy. Therefore, the given differential equation is not linear.
b) The differential equation is exact:
A differential equation is exact if it can be written in the form M(x, y)dx + N(x, y)dy = 0, where M and N are functions of x and y, and ∂M/∂y = ∂N/∂x. Let's check if this condition holds for the given equation:
M(x, y) = x + y
N(x, y) = 2x + 2y
∂M/∂y = 1
∂N/∂x = 2
Since ∂M/∂y is not equal to ∂N/∂x, the given differential equation is not exact.
c) ex^y is an integrating factor of the differential equation:
An integrating factor is a function that can be multiplied to a differential equation to make it exact. In this case, the integrating factor would need to be able to transform the given equation into an exact one. However, as we determined in the previous option, the given equation is not exact. Therefore, ex^y is not an integrating factor.
d) A suitable substitution transforms the differential equation to the variable separable form:
Variable separable form means that the equation can be written in the form f(x)dx = g(y)dy, where f(x) and g(y) are functions of x and y, respectively. To determine if a suitable substitution can transform the given equation into this form, let's make the substitution x = u and y = v, where u and v are functions of a new variable t (u = u(t) and v = v(t)). Substituting these into the given equation, we have:
(uv)du + (2uv)dv = 0
Factoring out uv, we get:
uv(du + 2dv) = 0
Since this equation can be separated into f(u)du = -2g(v)dv, where f(u) = uv and g(v) = -v, we can conclude that a suitable substitution transforms the differential equation to the variable separable form.
Therefore, the correct answer is option d) A suitable substitution transforms the differential equation to the variable separable form.
If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if g(x, 0) = x2, then the general solution of the differential equation is
- a)2x3 + 2xy + y2 = c
- b)2x3 + 6xy + 3y2 = c
- c)2x + 2xy + y2 = c
- d)x2 + 2xy + y2 = c
Correct answer is option 'B'. Can you explain this answer?
If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if g(x, 0) = x2, then the general solution of the differential equation is
a)
2x3 + 2xy + y2 = c
b)
2x3 + 6xy + 3y2 = c
c)
2x + 2xy + y2 = c
d)
x2 + 2xy + y2 = c
|
|
Chirag Verma answered |
If the general solutions of a differential equation are (y + c)2 = cx, where c is an arbitrary constant, then the order and degree of differential equation is - a)1, 2
- b)2, 1
- c)1, 3
- d) None of these
Correct answer is option 'A'. Can you explain this answer?
If the general solutions of a differential equation are (y + c)2 = cx, where c is an arbitrary constant, then the order and degree of differential equation is
a)
1, 2
b)
2, 1
c)
1, 3
d)
None of these
|
|
Chirag Verma answered |
There will only one constant in the first-order differential equation. Differentiating the given equation.
Putting the value of c in Eq. (1) and simplifying we will get a first-order and second-degree equation. Hence, (A) is the correct answer.
Consider the differential equation 2 cos (y2)dx - xy sin (y2)dy = 0- a)ex is an integrating factor
- b)e-x is an integrating factor
- c)3x is an integrating factor
- d)x3 is an integrating factor
Correct answer is option 'D'. Can you explain this answer?
Consider the differential equation 2 cos (y2)dx - xy sin (y2)dy = 0
a)
ex is an integrating factor
b)
e-x is an integrating factor
c)
3x is an integrating factor
d)
x3 is an integrating factor
|
Kanika Verma answered |
Given: The differential equation 2 cos (y^2)dx - xy sin (y^2)dy = 0
To solve this differential equation, we can use the method of integrating factors. An integrating factor is a function that we can multiply to the entire equation to make it exact.
Step 1: Check if the equation is exact:
An equation of the form M(x, y)dx + N(x, y)dy = 0 is exact if and only if ∂M/∂y = ∂N/∂x.
Let's find the partial derivatives:
∂M/∂y = -4ycos(y^2)
∂N/∂x = -ysin(y^2)
Since ∂M/∂y is not equal to ∂N/∂x, the given differential equation is not exact.
Step 2: Find the integrating factor:
To find the integrating factor, we can use the formula: integrating factor = e^(∫(∂N/∂x - ∂M/∂y)/N dx).
In this case, ∂M/∂y - ∂N/∂x = -4ycos(y^2) - (-ysin(y^2)) = -4ycos(y^2) + ysin(y^2).
We need to find the integral of (-4ycos(y^2) + ysin(y^2))/(-xy sin(y^2)) dx.
Simplifying the expression, we get (∫(4cos(y^2)/y - sin(y^2))/x dx.
The integral (∫4cos(y^2)/y dx) can be solved using the substitution method, and the integral (∫sin(y^2)/x dx) can be solved using the logarithmic method.
After finding the integral, we get the integrating factor as x^4/y^4.
Step 3: Multiply the integrating factor to the equation:
Multiplying the integrating factor x^4/y^4 to the given differential equation, we get:
(2x^3/y^4)cos(y^2)dx - (x^5/y^3)sin(y^2)dy = 0
This equation is exact, and we can solve it by finding the potential function or by using other methods.
Therefore, the correct answer is option D) x^3 is an integrating factor.
To solve this differential equation, we can use the method of integrating factors. An integrating factor is a function that we can multiply to the entire equation to make it exact.
Step 1: Check if the equation is exact:
An equation of the form M(x, y)dx + N(x, y)dy = 0 is exact if and only if ∂M/∂y = ∂N/∂x.
Let's find the partial derivatives:
∂M/∂y = -4ycos(y^2)
∂N/∂x = -ysin(y^2)
Since ∂M/∂y is not equal to ∂N/∂x, the given differential equation is not exact.
Step 2: Find the integrating factor:
To find the integrating factor, we can use the formula: integrating factor = e^(∫(∂N/∂x - ∂M/∂y)/N dx).
In this case, ∂M/∂y - ∂N/∂x = -4ycos(y^2) - (-ysin(y^2)) = -4ycos(y^2) + ysin(y^2).
We need to find the integral of (-4ycos(y^2) + ysin(y^2))/(-xy sin(y^2)) dx.
Simplifying the expression, we get (∫(4cos(y^2)/y - sin(y^2))/x dx.
The integral (∫4cos(y^2)/y dx) can be solved using the substitution method, and the integral (∫sin(y^2)/x dx) can be solved using the logarithmic method.
After finding the integral, we get the integrating factor as x^4/y^4.
Step 3: Multiply the integrating factor to the equation:
Multiplying the integrating factor x^4/y^4 to the given differential equation, we get:
(2x^3/y^4)cos(y^2)dx - (x^5/y^3)sin(y^2)dy = 0
This equation is exact, and we can solve it by finding the potential function or by using other methods.
Therefore, the correct answer is option D) x^3 is an integrating factor.
Orthogonal trajectories of the family of curves (x - 1)2 + y2 + 2ax = 0 are the solution of the differential equation
- a)
- b)
- c)x2 - y2 - 1 - 2xydy/dx = 0(d)
- d)
Correct answer is option 'A'. Can you explain this answer?
Orthogonal trajectories of the family of curves (x - 1)2 + y2 + 2ax = 0 are the solution of the differential equation
a)
b)
c)
x2 - y2 - 1 - 2xydy/dx = 0(d)
d)
|
|
Chirag Verma answered |
we have (x - 1)2 + y2 + 2ax = 0 ...(i)
Differentiating w.r.t. x, we get
2(x - 1) + 2y y' + 2a = 0
implies
Putting expression of a in eq. (i), we get
(x - 1)2 + y2 - 2x((x - 1) + yy') = 0
implies x2 - 2x + 1 + y2 - 2x2 + 2x + 2xyy' = 0
Orthogonal trajectories are given by
Differentiating w.r.t. x, we get
2(x - 1) + 2y y' + 2a = 0
implies
Putting expression of a in eq. (i), we get
(x - 1)2 + y2 - 2x((x - 1) + yy') = 0
implies x2 - 2x + 1 + y2 - 2x2 + 2x + 2xyy' = 0
Orthogonal trajectories are given by
If y = In (sin (x + a)) + b, where a and b are constants, is the primitive, then the corresponding lowest order differential equation is- a)y" = - (1 + (y')2)
- b)y" =y2 - (y')2
- c)y" = 1 + (y')2
- d)y" =y' + y2
Correct answer is option 'A'. Can you explain this answer?
If y = In (sin (x + a)) + b, where a and b are constants, is the primitive, then the corresponding lowest order differential equation is
a)
y" = - (1 + (y')2)
b)
y" =y2 - (y')2
c)
y" = 1 + (y')2
d)
y" =y' + y2
|
|
Chirag Verma answered |
we have y = In sin (x + a ) + b
implies y'= cot (x + a)
implies y" = - cosec2 (x + a)
implies y'= cot (x + a)
implies y" = - cosec2 (x + a)
General solution of pde given below is (y2+ z2+ x2 )p - 2xyq + 2xz = 0- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
General solution of pde given below is (y2+ z2+ x2 )p - 2xyq + 2xz = 0
a)
b)
c)
d)
|
|
Chirag Verma answered |
ANSWER :- a
Solution :- Lagrange auxiliary equations are :-
dx/(y^2 + z^2 - x^2) = -dy/(-2xy) = -dz/(2xz)
Taking the last two members, we get
dy/y = dz/z
Integrating log y = logz + logc1
=y/z = c1
Using ,multipliers x,y,z we get (xdx + ydy + zdz)/-x(x^2 + y^2 + z^2)
(xdx + ydy + zdz)/-x(x^2 + y^2 + z^2) = dx/(-2xz)
2(xdx + ydy + zdz)/(x^2 + y^2 + z^2) = dz/z
Integrating, log(x^2 + y^2 + z^2) = logz + logc2
(x^2 + y^2 + z^2) = zc2
Hence the required solution is f(c1,c2) = 0
= f(y/z, (x^2 + y^2 + z^2)/z) = 0
Let f, g : [-1, 1] → R, f (x) = x3, g(x) = x2 IxI Then
y" + py' + qy = 0 on [-1, 1]- a)f and g are linearly independent on [-1,1]
- b)f and g are linearly dependent on [-1, 1]
- c)f(x) g(x) - f(x) g (x) is NOT Identically zero on [-1, 1]
- d)There exist continuous function pix) and q(x) such that/ and g satisfy
Correct answer is option 'B'. Can you explain this answer?
Let f, g : [-1, 1] → R, f (x) = x3, g(x) = x2 IxI Then
y" + py' + qy = 0 on [-1, 1]
y" + py' + qy = 0 on [-1, 1]
a)
f and g are linearly independent on [-1,1]
b)
f and g are linearly dependent on [-1, 1]
c)
f(x) g(x) - f(x) g (x) is NOT Identically zero on [-1, 1]
d)
There exist continuous function pix) and q(x) such that/ and g satisfy
|
|
Chirag Verma answered |
Here, lxI = -x for x < 0 = x for x ≥ 0
So, g(x) = x3 for x ≥ 0
= -x3 for x < 0
and f(x) = x3
So, g(x) = x3 for x ≥ 0
= -x3 for x < 0
and f(x) = x3
Find the differentiation of x3 + y3 – 3xy + y2 = 0?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Find the differentiation of x3 + y3 – 3xy + y2 = 0?
a)
b)
c)
d)
|
|
Veda Institute answered |
Differentiation of x3 is 3x2
differentiation of y3 is 3y2 dy / dx
differentiation of -3xy is [-3y-3x(dy/dx)]
differentiation of y2 is 2y dy / dx
Hence,
differentiation of y3 is 3y2 dy / dx
differentiation of -3xy is [-3y-3x(dy/dx)]
differentiation of y2 is 2y dy / dx
Hence,
The general solution of y" - m2y = 0 is- a)cj sinh mx + c2 cosh mx
- b)c1 cos mx + c2 sin mx
- c)c1 cos mx + c2 sinh mx
- d)c1 sin mx + c2 cosh mx
Correct answer is option 'A'. Can you explain this answer?
The general solution of y" - m2y = 0 is
a)
cj sinh mx + c2 cosh mx
b)
c1 cos mx + c2 sin mx
c)
c1 cos mx + c2 sinh mx
d)
c1 sin mx + c2 cosh mx
|
|
Chirag Verma answered |
Correct Answer :- A
Explanation : y''- m2y = 0
r2 - m2 = 0
r = +-m
y = Aemx + Be-mx
Let A=C+D and B=D-C
= Cemx + Demx - Ce-mx + De-mx
= C(emx-e-mx) + D(emx+e-mx)
= Csinh(mx) + Dcosh(mx)
What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)2 + 1 = 0?- a)Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zero
- b)Because for any real function, the left-hand side of the equation becomes zero
- c)Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zero
- d)Because for any real function, the left-hand side of the equation becomes infinity
Correct answer is option 'C'. Can you explain this answer?
What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)2 + 1 = 0?
a)
Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zero
b)
Because for any real function, the left-hand side of the equation becomes zero
c)
Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zero
d)
Because for any real function, the left-hand side of the equation becomes infinity
|
|
Veda Institute answered |
Given: (y’)2 + 1 = 0
Consider if y = 2x, then y’ = 2 and hence the left-hand side of the equation becomes 3 which is greater than 1. Therefore, the left-hand side of the equation will always be greater than, or equal to one and thus cannot be zero and hence the differential equation is not satisfied.
Consider if y = 2x, then y’ = 2 and hence the left-hand side of the equation becomes 3 which is greater than 1. Therefore, the left-hand side of the equation will always be greater than, or equal to one and thus cannot be zero and hence the differential equation is not satisfied.
What is the degree of the non-homogeneous partial differential equation,
- a)Degree-2
- b)Degree-1
- c)Degree-5
- d)Degree-0
Correct answer is option 'C'. Can you explain this answer?
What is the degree of the non-homogeneous partial differential equation,
a)
Degree-2
b)
Degree-1
c)
Degree-5
d)
Degree-0
|
|
Veda Institute answered |
Degree of an equation is defined as the power of the highest derivative present in the equation. Hence from the equation, 
the degree is 5.
with respect x?
is
What is the order of the non-homogeneous partial differential equation,
- a)Order-3
- b)Order-0
- c)Order-1
- d)Order-2
Correct answer is option 'D'. Can you explain this answer?
What is the order of the non-homogeneous partial differential equation,
a)
Order-3
b)
Order-0
c)
Order-1
d)
Order-2
|
|
Chirag Verma answered |
The order of an equation is defined as the highest order derivative present in the equation and hence from the equation,
it is clear that it of 2nd order.
The solution of the differential equation
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The solution of the differential equation
a)
b)
c)
d)
|
|
Veda Institute answered |
we have
and
Since,
So, differential equation is exact
Now,
and
So, solution of differential equation
and
Since,
So, differential equation is exact
Now,
and
So, solution of differential equation
Find the order of the difference equation Δ3yn – Δ2yn – Δyn = 3- a)4
- b)3
- c)2
- d)5
Correct answer is option 'B'. Can you explain this answer?
Find the order of the difference equation Δ3yn – Δ2yn – Δyn = 3
a)
4
b)
3
c)
2
d)
5
|
Sumit Sengupta answered |
Order of the Difference Equation
To find the order of the given difference equation, we need to look at the highest difference operator present in the equation.
Given Difference Equation:
Δ3yn - Δ2yn - Δyn = 3
Finding the Order:
- The highest difference operator present in the equation is the third-order difference operator (Δ3).
- Therefore, the order of the difference equation is 3.
So, the correct answer is option B) 3.
To find the order of the given difference equation, we need to look at the highest difference operator present in the equation.
Given Difference Equation:
Δ3yn - Δ2yn - Δyn = 3
Finding the Order:
- The highest difference operator present in the equation is the third-order difference operator (Δ3).
- Therefore, the order of the difference equation is 3.
So, the correct answer is option B) 3.
If z = 3xy + 4x2, what is the value of ∂z / ∂x?
- a)3y + 8x
- b)3x + 4x2
- c)3xy + 8x
- d)3y + 3x + 8x
Correct answer is option 'A'. Can you explain this answer?
If z = 3xy + 4x2, what is the value of ∂z / ∂x?
a)
3y + 8x
b)
3x + 4x2
c)
3xy + 8x
d)
3y + 3x + 8x
|
|
Chirag Verma answered |
Given: z = 3xy + 4x2
Using partial differentiation, we need to differentiate the function z with respect to x keeping y as constant. Thus, ∂z / ∂x = 3y + 8x.
Using partial differentiation, we need to differentiate the function z with respect to x keeping y as constant. Thus, ∂z / ∂x = 3y + 8x.
find the value of fy at (x, y) = (0, 1).- a)101
- b)0
- c)210
- d)-96
Correct answer is option 'D'. Can you explain this answer?
find the value of fy at (x, y) = (0, 1).
a)
101
b)
0
c)
210
d)
-96
|
|
Chirag Verma answered |
Using Euler theorem
xfx + yfy = n f(x, y)
Substituting x = 0; n = -96 and y = 1 we have
fy = -96. f(0, 1) = -96.(1⁄1)
= – 96.
xfx + yfy = n f(x, y)
Substituting x = 0; n = -96 and y = 1 we have
fy = -96. f(0, 1) = -96.(1⁄1)
= – 96.
The differential equation (2x2 + by2)dx + cxydy = 0 is made exact by multiplying the integrating factor 1/x2. Then the relation between and c is- a)2c = b
- b)b = c
- c)2b + c = 0
- d)b + 2c = 0
Correct answer is option 'C'. Can you explain this answer?
The differential equation (2x2 + by2)dx + cxydy = 0 is made exact by multiplying the integrating factor 1/x2. Then the relation between and c is
a)
2c = b
b)
b = c
c)
2b + c = 0
d)
b + 2c = 0
|
|
Chirag Verma answered |
Multiplying the differential equation by 1/x2, we get 
It is exact
So,
implies
2b + c = 0
It is exact
So,
implies
2b + c = 0
The Wronskian of the function 
x |x| is zero for
- a)x = 0
- b)x > 0
- c)x < 0
- d)all x
Correct answer is option 'A'. Can you explain this answer?
The Wronskian of the function x |x| is zero for
x |x| is zero fora)
x = 0
b)
x > 0
c)
x < 0
d)
all x
|
Sagnik Dash answered |
Wronskian never changes its sign in a particular domain. If f1 and f2 are two solutions of the differential equation then W( f1,f2)(x) is either identically zero or never zero . since the wronskian is zero hence it's everywhere zero .hence option A
General solution of the differential equation 
is given by- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
General solution of the differential equation is given by
is given bya)
b)
c)
d)
|
|
Chirag Verma answered |
we have xdy = (y + xe-y/x)dx
implies xdy - ydx = xe-y/x dx
implies xdy - ydx = xe-y/x dx
Let k be real constant. The solution of the differential equations 
satisfies the relation- a)y - z = ke3x
- b)y + z = ke-x
- c)3y + z = ke3x
- d)3y + z = ke-3x
Correct answer is option 'C'. Can you explain this answer?
Let k be real constant. The solution of the differential equations satisfies the relation
satisfies the relationa)
y - z = ke3x
b)
y + z = ke-x
c)
3y + z = ke3x
d)
3y + z = ke-3x
|
|
Veda Institute answered |
we have dy/dx = 2y + z and ...(i)
dz/dx = 3y ...(ii)
from 3 (i) l (ii), we get
implies
on integrating, we get
implies ln(3y + z) = 3x + c
So, 3y + z = ke3x
dz/dx = 3y ...(ii)
from 3 (i) l (ii), we get
implies
on integrating, we get
implies ln(3y + z) = 3x + c
So, 3y + z = ke3x
The general integral of z(xp — yq) = y2 — x2 is- a)z2 = x2 + y2 + f(xy)
- b)z2 = x2 — y2 + f(xy)
- c)z2 = —x2 — y2 + f(xy)
- d)z2 = y2 — z2 + f(xy)
Correct answer is option 'C'. Can you explain this answer?
The general integral of z(xp — yq) = y2 — x2 is
a)
z2 = x2 + y2 + f(xy)
b)
z2 = x2 — y2 + f(xy)
c)
z2 = —x2 — y2 + f(xy)
d)
z2 = y2 — z2 + f(xy)
|
|
Chirag Verma answered |
ANSWER :- c
Solution :- z(xzx−yzy)=y^2−x^2
xzx−yzy = (y^2−x^2)z
dx/dt=x , letting x(0)=1 , we have x=e^t
dy/dt=−y , letting y(0)=y0 , we have y=y0e^(−t) = y0/x
dz/dt=(y^2−x^2)/z
=y^20e^(−2t) −e^(2t)/z ,
we have z^2=f(y0)−y^20e^(−2t)−e^(2t)
=f(xy)−y^2−x^2 ,
i.e. x2+y2+z2=f(xy)
z^2=- x^2 - y^2 + f(xy)
Consider 4xyz = pq + 2px2y + 2qxy2
(i) It can be reduced to Claerut's form by some suitable transformation.
(ii) z = ax + by + ab is complete integral
(iii) z = —x2y2 is singular solution choose correct code.- a)Only (i) , (ii) are correct
- b)Only (ii) and (iii) correct
- c)(i) and (iii) are correct
- d)all are correct
Correct answer is option 'C'. Can you explain this answer?
Consider 4xyz = pq + 2px2y + 2qxy2
(i) It can be reduced to Claerut's form by some suitable transformation.
(ii) z = ax + by + ab is complete integral
(iii) z = —x2y2 is singular solution choose correct code.
(i) It can be reduced to Claerut's form by some suitable transformation.
(ii) z = ax + by + ab is complete integral
(iii) z = —x2y2 is singular solution choose correct code.
a)
Only (i) , (ii) are correct
b)
Only (ii) and (iii) correct
c)
(i) and (iii) are correct
d)
all are correct
|
Ajay Karhade answered |
4xyz = pq + 2px
2
y + 2qxy
2
The general integral of the partial differential equation (y + z x) zx — (x + yz)Zy = x2 — y2 is- a)f(x2 + y2 + z2,xy + z ) = 0
- b)f(x2 + y2 - z2,x y + z ) = 0
- c)f(x2 - y2- z2,xy + z) = 0
- d)f(x2 + y2 + z2,xy - z ) = 0
Correct answer is option 'B'. Can you explain this answer?
The general integral of the partial differential equation (y + z x) zx — (x + yz)Zy = x2 — y2 is
a)
f(x2 + y2 + z2,xy + z ) = 0
b)
f(x2 + y2 - z2,x y + z ) = 0
c)
f(x2 - y2- z2,xy + z) = 0
d)
f(x2 + y2 + z2,xy - z ) = 0
|
Ishika Singhania answered |
To find the general integral of the partial differential equation (y z x) zx, we can integrate with respect to both x and z. Let's denote the unknown function as F(x,z):
∂F/∂x = yz + zx
Integrating both sides with respect to x:
F(x,z) = ∫ (yz + zx) dx
= yz*x + 0.5*z*x^2 + g(z)
where g(z) is an arbitrary function of z.
Now, let's differentiate F(x,z) with respect to z:
∂F/∂z = yx + g'(z)
Comparing this with the original partial differential equation, we can see that g'(z) must be equal to 0 for the equation to hold. Therefore, we can conclude that g(z) is a constant, let's call it C.
The general integral of the partial differential equation (y z x) zx is:
F(x,z) = yz*x + 0.5*z*x^2 + C
where C is an arbitrary constant.
∂F/∂x = yz + zx
Integrating both sides with respect to x:
F(x,z) = ∫ (yz + zx) dx
= yz*x + 0.5*z*x^2 + g(z)
where g(z) is an arbitrary function of z.
Now, let's differentiate F(x,z) with respect to z:
∂F/∂z = yx + g'(z)
Comparing this with the original partial differential equation, we can see that g'(z) must be equal to 0 for the equation to hold. Therefore, we can conclude that g(z) is a constant, let's call it C.
The general integral of the partial differential equation (y z x) zx is:
F(x,z) = yz*x + 0.5*z*x^2 + C
where C is an arbitrary constant.
We have
through the curve x2 + y2 = 2x, z = 0 isThe general solution of 
(Here c1 and c2 are arbitrary constants.)
- a)(c1 + c2x )e3x
- b)(c1 + c2 In x)x3
- c)(c1 + c2x )x2
- d) (c1 + c2 In x)ex2
Correct answer is option 'B'. Can you explain this answer?
The general solution of
(Here c1 and c2 are arbitrary constants.)
(Here c1 and c2 are arbitrary constants.)
a)
(c1 + c2x )e3x
b)
(c1 + c2 In x)x3
c)
(c1 + c2x )x2
d)
(c1 + c2 In x)ex2
|
|
Chirag Verma answered |
Let z = log x, Then
The differential equation changes to
implies
So,
The differential equation changes to
implies
So,
If y'1 (x) = 3y1(x) + 4y2(x) and y'2 (x) = 4y1(x) + 3y2(x), then y1(x) is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If y'1 (x) = 3y1(x) + 4y2(x) and y'2 (x) = 4y1(x) + 3y2(x), then y1(x) is
a)
b)
c)
d)
|
|
Veda Institute answered |
we have (D - 3)y1 - 4y2 = 0
4y2 + (3 - D)y1 = 0
Solving equations, we get
So,
4y2 + (3 - D)y1 = 0
Solving equations, we get
So,
The particular integral of the differential equation
y" + y' + 3y = 5 cos(2x + 3) is- a)2 cos(2x + 3) - sin (2x + 3)
- b)2 sin (2x + 3) + cos(2x + 3)
- c)sin (2x + 3) - 2 cos (2x + 3)
- d)2 sin (2x + 3) - cos (2x + 3)
Correct answer is option 'D'. Can you explain this answer?
The particular integral of the differential equation
y" + y' + 3y = 5 cos(2x + 3) is
y" + y' + 3y = 5 cos(2x + 3) is
a)
2 cos(2x + 3) - sin (2x + 3)
b)
2 sin (2x + 3) + cos(2x + 3)
c)
sin (2x + 3) - 2 cos (2x + 3)
d)
2 sin (2x + 3) - cos (2x + 3)
|
|
Chirag Verma answered |
we have (D2 + D + 3)y = 5 cos(2x + 3)
So,
So,
The orthogonal trajectories of the curves y2 = 3x3+x + c are
- a)2 tan-1 3x + 3 In |y| = k
- b)3 tan-1 3x + 2 In |y| = k
- c)3 tan-1 3x - 2 In |y| = k
- d)3 In |x| - 2 tan-1 3y = k
Correct answer is option 'A'. Can you explain this answer?
The orthogonal trajectories of the curves y2 = 3x3+x + c are
a)
2 tan-1 3x + 3 In |y| = k
b)
3 tan-1 3x + 2 In |y| = k
c)
3 tan-1 3x - 2 In |y| = k
d)
3 In |x| - 2 tan-1 3y = k
|
|
Chirag Verma answered |
Question should be
y2 = 3x3 + x + c
Differentiating w.r.t. x, we get
Replacing
the equation of orthogonal trajectory is given by
implies
Integrating both sides, we get
implies 2 tan-1 3x + 3 In |y| = k
y2 = 3x3 + x + c
Differentiating w.r.t. x, we get
Replacing
the equation of orthogonal trajectory is given byimplies
Integrating both sides, we get
implies 2 tan-1 3x + 3 In |y| = k
Let f : R2 → R be defined by
Which of the following statements holds regarding the continuity and the existence of partial derivatives of f at (0, 0)?
- a)Both partial derivatives of f exist at (0, 0) and f is not continuous at (0, 0)
- b)Both partial derivates of f exist at (0, 0) and f is continuous at (0, 0)
- c)One partial derivative of f does NOT exist at (0, 0) and f is continuous at (0, 0)
- d)One partial derivative of f does NOT exist at (0, 0) and f is NOT continuous at (0, 0)
Correct answer is option 'A'. Can you explain this answer?
Let f : R2 → R be defined by
Which of the following statements holds regarding the continuity and the existence of partial derivatives of f at (0, 0)?
Which of the following statements holds regarding the continuity and the existence of partial derivatives of f at (0, 0)?
a)
Both partial derivatives of f exist at (0, 0) and f is not continuous at (0, 0)
b)
Both partial derivates of f exist at (0, 0) and f is continuous at (0, 0)
c)
One partial derivative of f does NOT exist at (0, 0) and f is continuous at (0, 0)
d)
One partial derivative of f does NOT exist at (0, 0) and f is NOT continuous at (0, 0)
|
|
Chirag Verma answered |
By the definition of partial derivatives
If we moving along the curve y = mx2
⇒ f(x, y) is not continuous.
If we moving along the curve y = mx2
⇒ f(x, y) is not continuous.
What are the tangents to the curve x3 + y3 = 3axy at the origin?- a)x = 0, y = 0
- b)x = 0
- c)y = 0
- d)x = y
Correct answer is option 'A'. Can you explain this answer?
What are the tangents to the curve x3 + y3 = 3axy at the origin?
a)
x = 0, y = 0
b)
x = 0
c)
y = 0
d)
x = y
|
|
Chirag Verma answered |
Given: x3 + y3 = 3axy
To find the tangent to the curve at the origin, we need to equate the lowest degree term to 0.
Therefore, 3axy = 0, which gives x = 0 and y = 0 as two tangents to the curve at origin.
To find the tangent to the curve at the origin, we need to equate the lowest degree term to 0.
Therefore, 3axy = 0, which gives x = 0 and y = 0 as two tangents to the curve at origin.
Let y1(x) and y2(x) be linearly independent solution of the differential equation y" + P(x) y' + Q(x)y = 0, where P(x) and Q(x) are continuous functions on an interval I. Then y3(x) = ay1(x) + by2(x) and y4(x) = cy1(x) dy2(x) are linearly independent solutions of the given differential equations if- a)ad = bc
- b)ac = bd
- c)ad ≠ bc
- d)ac ≠ bd
Correct answer is option 'C'. Can you explain this answer?
Let y1(x) and y2(x) be linearly independent solution of the differential equation y" + P(x) y' + Q(x)y = 0, where P(x) and Q(x) are continuous functions on an interval I. Then y3(x) = ay1(x) + by2(x) and y4(x) = cy1(x) dy2(x) are linearly independent solutions of the given differential equations if
a)
ad = bc
b)
ac = bd
c)
ad ≠ bc
d)
ac ≠ bd
|
|
Chirag Verma answered |
y3(x) and y4(x) are linearly dependent if y3(x) = ky4(x) where k is any constant
Since, y1(x) and y2(x) are independent
So, a - kc = 0
and b - kd = 0
or
So, y3(x) and y4(x) are independent if
implies ad ≠ bc
Since, y1(x) and y2(x) are independent
So, a - kc = 0
and b - kd = 0
or
So, y3(x) and y4(x) are independent if
implies ad ≠ bc
Which of the following relations hold true for division rule of differentiation?- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Which of the following relations hold true for division rule of differentiation?
a)
b)
c)
d)
|
|
Veda Institute answered |
The division rule of differentiation for two functions is given by,
Solution of the differential equation 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Solution of the differential equation
a)
b)
c)
d)
|
|
Chirag Verma answered |
=> x2 lnx dy - xy dy=xy dx – y2 lny dx …….dividing by x2 y2 then
on integrating we get
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