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All questions of Unit 7: Differential Equations for Grade 9 Exam

The solution of the initial value problem  edy/dx = x + 1, y(0) = 3 is :
  • a)
    y = (x – 1)log(x + 1) + 3
  • b)
    y = (x + 1)log(x + 1) -x+3
  • c)
    y = (x + 1)log(x + 1) + 4
  • d)
    y = (x + 1)log(x + 1) – 1
Correct answer is option 'B'. Can you explain this answer?

Preeti Iyer answered
 edy/dx = x + 1
⇒ dy/dx = log(x+1)
⇒ dy = log(x+1)dx
Integrating both sides, we get
⇒ ∫dy =  ∫log(x+1)dx
y = log(x + 1)  ∫1dx - ∫[d/dx{log(x + 1)} ∫1dx]dx
y = xlog(x + 1) -  ∫1/(x+1)xdx
y = xlog(x + 1) -  ∫(1 - 1/(x + 1)dx
y = xlog(x + 1) -  ∫dx +  ∫1/(x+1)dx
y = xlog(x + 1) - x + log|x + 1| + c
y = (x + 1)log|x + 1| - x + c
Here c is 3
y = (x + 1)log|x + 1| - x + 3 
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The differential equation,is a:
  • a)
    second order, third degree equation.
  • b)
    second order, second degree equation. 
  • c)
    third order, third degree equation.
  • d)
    second order, first degree equation
Correct answer is option 'D'. Can you explain this answer?

Aashu answered
I assume you know the basics of order and degree.. And i am going to tell only why degree is 2... Since a polynomial must have power as natural no. Not a fraction... So we square both side and then expand.. You will find power of highest order(d^2y/dx^2) is 2... It is called degree.. So degree is 2

Low Haemoglobin causes________.
  • a)
    Malaria
  • b)
    Anaemia
  • c)
    Cancer
  • d)
    Cholera
Correct answer is option 'B'. Can you explain this answer?

Priyanka Sharma answered
Hemoglobin, the substance that gives color to red blood cells, is the substance that allows for the transport of oxygen throughout the body. Low hemoglobin levels lead to anemia, which causes symptoms like fatigue and trouble breathing.

The differential equation  is a:
  • a)
    Third order, third degree equation
  • b)
    Second order, second degree equation
  • c)
    Second order, first degree equation
  • d)
    Second order, third degree equation
Correct answer is option 'B'. Can you explain this answer?

Suresh Iyer answered
3*(d2y/dx2) = [1+(dy/dx)2]3/2
On squaring both side,
9*(d2y/dx2)2 = [1+(dy/dx)2]3
The order of the equation is 2. The power of the term determining the order determines the degree.
So, the degree is also 2.

The solution of the differential equation is :
  • a)
    2y = x2(log x + 1) + c
  • b)
  • c)
  • d)
    y = x2(log x + 1) + c
Correct answer is option 'B'. Can you explain this answer?

Poonam Reddy answered
dy/dx = x logx
=> ∫dy = ∫x logx dx
y = logx . x2/2 - ∫1/x . x2/2 dx + c
y = x2/2 log x -½ ∫x dx + c
y = x2/2 log x - ½ . x2/2 + c
y = x2/2 log x - x2/4 + c

The integrating factor of differential equation is :
  • a)
    -tan x
  • b)
    sec2x
  • c)
    -sec2x
  • d)
    tan x
Correct answer is option 'B'. Can you explain this answer?

Vikas Kapoor answered
 dy/dx+2ytanx=sinx
This is in the form of dy/dx + py = θ
where p=2tanx,θ=sinx
∴ finding If e∫pdx = e∫2tanxdx 
=e(2log secx)
=e(log sec2x)
=sec2x

The solution of the differential equation x dy = (2y + 2x4 + x2) dx is:​
  • a)
    y = x4 + x2 log |x2 + cx2
  • b)
    y = x3 + 2x + cx2
  • c)
    y = x3 + x log |x| + cx
  • d)
    y = x4 + x2 log |x| + cx2
Correct answer is option 'D'. Can you explain this answer?

Sai Kulkarni answered
xdy = (2y + 2x4 + x2)dx
→ dy/dx − (2x)y = 2x3 + x
This differential is of the form y′+P(x)y=Q(x) which is the general first order linear differential equation, where P(x) and Q(x) are continuous function defined on an interval.
The general solution for this is y∙I.F = ∫I.F × Q(x)dx
Where I.F = e∫P(x)dx is the integrating factor of the differential equation.
I.F = e∫P(x)dx
= e∫−2/xdx
= e(−2∙lnx)
= eln(x−2)
= x−2
Thus y(1/x2) = ∫1/x2(2x3 + x)dx
=∫(2x + 1/x)dx
= x2 + lnx + C
⟹ y = x4 + x2lnx + c

The solution of the differential equation is :
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Aryan Khanna answered
y/x cos y/x dx − (x/y sin y/x + cosy/x) dy = 0
⇒dy/dx = (y/x cos y/x)/(x/y sin y/x + cos y/x)
This is a homogeneous differential equation.
Putting y = vx and dy/dx = v+xdv/dx, we have v+xdv/dx = (v2cosv)/(sinv+vcosv)
⇒ xdv/dx = (v2cosv)/(sinv+vcosv) - v
⇒x dv/dx = (v2 cosv - v2 cosv - vsinv )/(sinv+vcosv) 
⇒ x dv/dx = - [v sinv/(sinv + vcosv)]
⇒ ∫[(sinv + vcosv)/v sinv]dv = ∫dx/x
Integrating both the sides, we get
∫(cot v + 1/v)dv = - ln(x) + c
ln(sin v) + ln(v) = -ln(x) + c
ln(sin(y/x) + ln(y/x) + ln(x) = c
ln(y/x sin y/x * x) + c
y sin(y/x) = c

The solution of the equation x3dx + (y + 1)2 dy = 0
  • a)
    12[x4 - (y +1)3] = c
  • b)
    12[x4 + (y +1)3] = c
  • c)
    3x4 + 4(y +1)3 = c
  • d)
    3x4 + 3(y +1)3 = c
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
x3dx + (y + 1)dy = 0
=> d(x4/4) + d((y + 1)3/3) = 0
=> d(x4/4 + ((y + 1)3/3)) = 0
=> (x4/4 + ((y + 1)3/3)) = constt
=> 12 *(x4/4 + ((y + 1)3/3)) = 12 * constt
=> 3x4 + 4(y + 1)3 = constt

The solution of the differential equation is :
  • a)
    x3 + 10y3 = x2
  • b)
    x3 + y3 = 10x2
  • c)
    x3 + y2 = 10x3
  • d)
    x3 -10y3 = x2
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
dy/dx = (x2 + 3y2)/2xy………….(1)
Let y = vx
dy/dx = v + xdv/dx
Substitute the value of y and dy/dx in (1)
v + x dv/dx = (1+3v2)/2v
x dv/dx = (1+3v2)/2v - v
x dv/dx = (1 + 3v2 - 2v2)/2
x dv/dx = (1+ v2)/2v
2v/(1+v2) dv = dx/x…………(2)
Integrating both the sides
∫2v/(1+v2) dv = ∫dx/x
Put t = 1 + v2
dt = 2vdv
∫dt/t  = ∫dx/x
=> log|t| = log|x| + log|c|
=> log|t/x| = log|c|
t/x = +- c
(1+v2)/x = +-c
(1 + (y2)/(x2))/x = +-c
x2 + y2 = Cx3……….(3)
y(1) = 3
1 + 9 = c(1)3
c = 10
From eq(3), we get x
3
+ y2 = 10x3

The solution of the differential equation is :
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Notes Wala answered
dy/dx + y/x = x2
differential equation is in the form : dy/dx + Py = Q
P = 1/x      Q = x2
I.F = e∫P(x)dx
I.F = e∫1/x dx
I.F = e[log x] 
I.F = x
y I.F = ∫(Q * I.F) dx + c
yx = ∫x2 * x * dx + c
yx = ∫x3 dx + c
xy = [x4]/4 + c 

The solution of the differential equation is :
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Om Desai answered
The given differential equation may be written as
dy/dx + (2x/(x2+1)y = √x2+4/(x2+1)  ... (i)
This is of the form dy/dx + Py = Q,
 where P=2x/(x2+1) and Q=(√x2+4)/(x2+1)
Thus, the given differential equation is linear.
IF=e(∫Pdx)
= e(∫2x(x2+1)dx)
= e(log(x2+1) = (x2+1)
So, the required solution is given by
y × IF = ∫{Q×IF}dx + C,
i.e., y(x2+1)=∫(√x2+4)/(x2+1)×(x2+1)dx
⇒y(x2+1)=∫(√x2+4)dx
=1/2x (√x2+4) +1/2 × (2)2 × log|x+(√x2+4)| + C
=1/2x (√x2+4) + 2log|x+(√x2+4) + C.
Hence, y(x2+1) = 1/2x(√x2+4) + 2log|x+√x2+4| + C is the required solution.

Formation of the differential equation corresponding to the ellipse major axis 2a and minor axis 2b is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
Equation of ellipse :
 x2/a2 + y2/b2 = 1
Differentiation by x,
2x/a2 + (dy/dx)*(2y/b2) = 0
dy/dx = -(b2/a2)(x/y)
-(b2/a^2) = (dy/dx)*(y/x) ----- eqn 1
Again differentiating by x,
d2y/dx2 = -(b2/a2)*((y-x(dy/dx))/y2)
Substituting value of -b2/a2 from eqn 1
d2y/dx2 = (dy/dx)*(y/x)*((y-x(dy/dx))/y2)
d2y/dx2 = (dy/dx)*((y-x*(dy/dx))/xy)
(xy)*(d2y/dx2) = y*(dy/dx) - x*(dy/dx)2
(xy)*(d2y/dx2) + x*(dy/dx)2- y*(dy/dx) = 0

The first order, first degree differential equation y’ = f(x,y)  is said to be homogeneous, if​
  • a)
    f(x , y) is a homogeneous function of degree zero.
  • b)
    f(x , y) is a homogeneous function of second degree.
  • c)
    f(x , y) is a homogeneous function of first degree.
  • d)
    f(x , y) is a homogeneous function of third degree.
Correct answer is option 'A'. Can you explain this answer?

Arshiya Mehta answered
The correct answer is option A, because a first-order, first-degree differential equation is said to be homogeneous if the function f(x,y) is a homogeneous function of degree zero.
A function f(x,y) is said to be homogeneous of degree n if it satisfies the equation:
f(λx, λy) = λ^n * f(x,y)
where λ is a scalar. This means that if we multiply both the independent and dependent variables by a constant λ, the value of the function is scaled by a factor of λ^n.
In the case of a first-order, first-degree differential equation, the function f(x,y) is a function of two variables: x and y. For this type of equation, f(x,y) is said to be homogeneous if it is a homogeneous function of degree zero, which means that it satisfies the equation:
f(λx, λy) = f(x,y)
In other words, if we multiply both the independent and dependent variables by a constant λ, the value of the function does not change. This means that f(x,y) does not depend on the magnitude of the variables x and y, but only on their relative values.
For example, if f(x,y) = xy, then f(λx, λy) = λxy = xy, so f(x,y) is a homogeneous function of degree zero. On the other hand, if f(x,y) = x^2 + y^2, then f(λx, λy) = λ^2x^2 + λ^2y^2 = x^2 + y^2, so f(x,y) is not a homogeneous function.
Therefore, the correct answer is option A, because a first-order, first-degree differential equation is said to be homogeneous if the function f(x,y) is a homogeneous function of degree zero.

Identify the type of differential equation
  • a)
    Variable separable
  • b)
    Homogeneous
  • c)
    Linear
  • d)
    Both Variable separable and Linear
Correct answer is option 'B'. Can you explain this answer?

Tejas Verma answered
(x−y)dy/dx = x + 2y
⇒ dy/dx = (x + 2y)/(x−y)
F (x,y) = (x + 2y)/(x−y)
F(Ax, Ay) = (Ax + 2Ay)/(Ax−Ay) 
= A(x + 2y)/A(x−y) 
= (x + 2y)/(x − y) = F (x,y)
Hence, the equation is homogenous.

Write the order and degree of the given differential equation:
​
  • a)
    2,3
  • b)
    1,1
  • c)
    2,1
  • d)
    3, 1
Correct answer is option 'D'. Can you explain this answer?

Gaurav Kumar answered
Order = 3 and Degree = 1
The order of a differential equation is determined by the highest-order derivative; the degree is determined by the highest power on a variable. The higher the order of the differential equation, the more arbitrary constants need to be added to the general solution.

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931).
  • a)
    7.93%
  • b)
    8.93%
  • c)
    9.93%
  • d)
    6.93%
Correct answer is option 'D'. Can you explain this answer?

Anirban Desai answered
To solve this problem, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:
A is the final amount
P is the principal amount (initial amount)
r is the annual interest rate (as a decimal)
n is the number of times the interest is compounded per year
t is the number of years

In this case, the principal doubles itself in 10 years. So, the final amount A is 2 times the principal P.

2P = P(1 + r/n)^(nt)

Since we are given that the principal increases continuously, we can take the limit as n approaches infinity. This simplifies the formula to:

2 = e^(rt)

Where e is the base of natural logarithms and is approximately equal to 2.71828.

Now, we can take the natural logarithm of both sides to solve for r:

ln(2) = rt

Dividing both sides by t:

r = ln(2)/t

Given that t = 10 years, we can substitute this value into the equation:

r = ln(2)/10

Using the given value of loge2 = 0.6931, we can calculate r:

r = 0.6931/10 = 0.06931 = 6.93%

Therefore, the value of r is 6.93%.

So, the correct answer is option D.

Degree of a differential equation, when the equation is polynomial equation in y′ is
  • a)
    Highest (positive integral index) of the lowest order derivative in the given differential equation.
  • b)
    Lowest power (positive integral index) of the highest order derivative in the given differential equation.
  • c)
    Lowest power (positive integral index) of the lowest order derivative in the given differential equation.
  • d)
    Highest power (positive integral index) of the highest order derivative in the given differential equation.
Correct answer is option 'D'. Can you explain this answer?

Malavika Menon answered
The degree of a differential equation is determined by the highest power of the derivative(s) present in the equation.

When the equation is a polynomial equation in y, the degree is determined by the highest power of y in the equation. The derivative(s) of y do not affect the degree in this case.

The order and degree of the differential equation: (y”)2 + (y”)3 + (y’)4 + y5 = 0 is:​
  • a)
    2, 4
  • b)
    3, 5
  • c)
    2, 5
  • d)
    2, 3
Correct answer is option 'D'. Can you explain this answer?

Naina Bansal answered
Toolbox:
The highest order derivative present in the differential equation determines its order.
The highest power raise to the derivative determines its degree.

(y′′′)^2+(y′′)^3+(y′)^4+y^5 = 0
The order of the given equation is 3.
The degree corresponding to the highest order is 2

Find the differential equation of all the straight lines touching the circle x2 + y2 = r2.
  • a)
    r2 (1+(dy/dx))2
  • b)
    3r2 (1+(dx/dy))2
  • c)
    2r2 (1+(dy/dx))2
  • d)
    r2 (1+(dx/dy))2
Correct answer is option 'A'. Can you explain this answer?

Ajay Yadav answered
  1. Let y = mx + c be the equation of all the straight lines touching the circle.
    Given : The equation of the circle is x2 + y2 = r2----------> (1)
    The tangent to the circle is c2 = r2(1+m2)
    c = r√(1+m2)
    we know that y = mx + c---------->(2)
    y = mx + r√(1+m2) ---------->(3)
    y - mx = r√(1+m2)
    Differentiating wrt x we get dy/dx -m =0
    dy/dx = m
    Substituting this in equation (3)
    y - (dy/dx . x) = r√(1+(dy/dx)2)
    Squaring on both sides, we get
    y2 - (dy/dx . x)2 = [ r√(1+(dy/dx)2)]2
    [y - x(dy/dx)]2 = r2 (1+(dy/dx))2 is the required differential equation.
    Answer: The differential equation of all the straight lines touching the circle x2 + y2 = r2 is [y - x(dy/dx)]2 = r2 (1+(dy/dx))2​​​​​​
 

Differential equations are equations containing functions y = f(x), g(x) and
  • a)
    tangent of y at zero
  • b)
    maxima of y
  • c)
    derivatives of y
  • d)
    minima of y
Correct answer is option 'C'. Can you explain this answer?

Dipanjan Verma answered
Differential equations are mathematical equations that involve derivatives of an unknown function. They are widely used in various fields of science and engineering to model and analyze dynamic systems. In this question, we are asked to identify the correct option regarding the nature of the functions involved in differential equations.

Differential equations involve functions y = f(x) and g(x), where y represents the dependent variable and x represents the independent variable. These functions can have various properties and behaviors based on the given equation. Let's analyze each option to understand why option 'C' is the correct answer.

a) Tangent of y at zero:
The tangent of a function at a specific point represents the slope of the function at that point. This option suggests that the differential equation involves the tangent of y at zero. However, this is not a general characteristic of differential equations. Differential equations can involve tangents at any point, not just zero.

b) Maxima of y:
Maxima of a function refers to the points where the function reaches its highest values. While differential equations can involve maxima or minima, it is not a general property of all differential equations. Therefore, this option is not correct.

c) Derivatives of y:
This option suggests that the differential equations involve derivatives of y. This is the correct answer. Differential equations often involve derivatives of the unknown function y with respect to the independent variable x. By including derivatives, we can describe how the rate of change of y varies with respect to x.

d) Minima of y:
Minima of a function refers to the points where the function reaches its lowest values. Similar to option 'b', this is not a general property of all differential equations. While some differential equations may involve minima, it is not a defining characteristic.

In conclusion, the correct answer is option 'c' because differential equations commonly involve derivatives of the unknown function y. Including derivatives allows us to describe the rate of change and behavior of the function with respect to the independent variable.

General solution of a given differential equation
  • a)
    contains arbitrary constants depending on the order of the differential equation
  • b)
    contains exactly one arbitrary constant
  • c)
    contains exactly two arbitrary constants
  • d)
    does not contain arbitrary constants
Correct answer is option 'A'. Can you explain this answer?

Madhurima Mehta answered
The general solution of a given differential equation contains arbitrary constants depending on the order of the differential equation. Let's understand this in detail:

1. Differential Equation:
A differential equation is an equation that relates a function with its derivatives. It represents a relationship between the function and its rate of change.

2. Order of a Differential Equation:
The order of a differential equation is the highest order of the derivative present in the equation. For example, if the equation involves only the first derivative of the function, it is a first-order differential equation. Similarly, if it involves the second derivative, it is a second-order differential equation, and so on.

3. Particular Solution:
A particular solution is a specific solution of a differential equation that satisfies the given initial conditions.

4. General Solution:
The general solution of a differential equation is a family of solutions that contains all possible solutions of the equation. It includes a set of functions that satisfy the differential equation, along with arbitrary constants.

5. Arbitrary Constants:
Arbitrary constants are constants that can take any value. They are introduced in the general solution to account for the different solutions that satisfy the given differential equation.

6. Number of Arbitrary Constants:
The number of arbitrary constants in the general solution depends on the order of the differential equation.

- For a first-order differential equation, the general solution contains exactly one arbitrary constant.
- For a second-order differential equation, the general solution contains exactly two arbitrary constants.
- For higher-order differential equations, the general solution contains arbitrary constants equal to the order of the differential equation.

7. Importance of Arbitrary Constants:
The arbitrary constants in the general solution allow us to find the particular solution by substituting specific values for the constants. These values are determined by the given initial conditions or boundary conditions.

In conclusion, the general solution of a given differential equation contains arbitrary constants depending on the order of the differential equation. These arbitrary constants allow us to find particular solutions that satisfy the given initial or boundary conditions.

General Solution of (ex + e-x) dy - (ex - e-x) dx = 0
  • a)
    y = log(e2x+ e−x) + C
  • b)
    y = log(ex+ e−x) + C
  • c)
    y = (ex+ e−x) + C
  • d)
    y = log(e−2x+ e−x) + C
Correct answer is option 'B'. Can you explain this answer?

Samridhi Deshpande answered
General Solution of the given differential equation:
- To find the general solution of the given differential equation, start by rearranging the equation in the form of dy/dx.
- The given differential equation is: (ex+ e-x) dy - (ex- e-x) dx = 0
- Rearranging the equation gives: dy/dx = (ex- e-x) / (ex+ e-x)

Integrating the equation:
- To solve this differential equation, separate the variables and integrate both sides.
- Integrate the left side with respect to y and the right side with respect to x.
- Integrate (ex+ e-x) dy = Integrate (ex- e-x) dx

Applying Integration:
- The integral of (ex+ e-x) dy is y = log|ex+e-x| + C1, where C1 is the constant of integration.
- The integral of (ex- e-x) dx is x = log|ex-e-x| + C2, where C2 is the constant of integration.

General Solution:
- Combining the results from the integrations, the general solution of the given differential equation is:
- y = log(ex+e-x) + C1
- This is the general solution for the given differential equation.

The order of the differential equation: 
  • a)
    4
  • b)
    2
  • c)
    3
  • d)
    1
Correct answer is option 'C'. Can you explain this answer?

Anuj Seth answered
Order is seen from highest efficiency of differentiability I.e. in hindi , kisi bhi equation me jisme sabse jyada differential power hai wo jeeta

Differential equation of the family of ellipses having foci on y-axis and centre at origin is
  • a)
    xy′′ + x(y′)2  yy′ = 0
  • b)
    xyy′′ + x(y′)2  yy′ = 0
  • c)
    yy′′ + x(y′)2  yy′ = 0
  • d)
    xyy′′ −x(y′)2 +yy′ = 0
Correct answer is option 'B'. Can you explain this answer?

Palak Banerjee answered
The differential equation of the family of ellipses having foci on the y-axis and center at the origin can be found as follows:

Let's consider an ellipse with foci (0, c) and (0, -c), where c is a positive constant. The distance between the foci is 2c.

The general equation of an ellipse centered at the origin is given by:

x^2/a^2 + y^2/b^2 = 1,

where a and b are positive constants representing the semi-major and semi-minor axes, respectively.

Since the foci lie on the y-axis, the equation of the ellipse becomes:

x^2/a^2 + (y - c)^2/b^2 = 1.

We know that the distance between the foci is 2c, so we have:

2c = 2b^2/a.

Simplifying this equation, we get:

b^2 = ac.

Now, differentiating both sides of the equation with respect to x, we have:

2b(b') = a'c + ac',

where b' and a' represent the derivatives of b and a with respect to x, respectively.

Since the derivatives of a and b are unknown, we cannot solve for the differential equation in terms of x and y directly. However, we can eliminate a' and b' by using the relationship b^2 = ac, which gives:

2b(b') = 2b(c'/b) + ac'.

Simplifying this equation, we find:

b' = c'/b + ac'/(2b^2).

Rearranging terms, we get:

b'(2b^2) = c' + ac'.

Substituting b^2 = ac, we obtain:

b'(2b^2) = c' + b^2c'.

Finally, dividing both sides by 2b^2, we get the differential equation:

b' = (c' + b^2c')/(2b^2).

Therefore, the differential equation of the family of ellipses having foci on the y-axis and center at the origin is:

b' = (c' + b^2c')/(2b^2).

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