MCQ’s on Differential Equations Class 12 OP Malhotra ISC Maths Solutions Ch-17. In this article you would learn full concept recap of Differential Equations. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

MCQ’s on Differential Equations Class 12 OP Malhotra ISC Maths Solutions Ch-17

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-17	Differential Equations
Writer	OP Malhotra
Exe	MCQ and Short Answer

MCQ’s on Differential Equations

 OP Malhotra ISC Class 12 Maths Solutions

Fill in the blanks

1. The degree of the differential equation (dy/dx)⁴ + 3x (d²y/dx²) = 0 is ____.

Ans- The highest order derivative is d²y/dx².
The power of the highest order derivative is 1.
Therefore, the degree of the differential equation is 1.

2. The degree of the differential equation x(d²y/dx²)³ + y(dy/dx)⁴ + x³ = 0 is ____.

Ans- The highest order derivative is d²y/dx².
The power of the highest order derivative is 3.
Hence, the degree of the differential equation is 3.

3. The degree of the differential equation d²y/dx² + e^(dy/dx) = 0 is ____.

Ans- The derivative dy/dx appears inside an exponential function.
Hence the differential equation is not a polynomial in derivatives.
Therefore, the degree is not defined.

4. The sum of the order and degree of the differential equation (d²y/dx²)² + (dy/dx)³ + x⁴ = 0 is ____.

Ans- The highest order derivative is d²y/dx².
Therefore, the order of the differential equation is 2.
The power of the highest order derivative is 2, so the degree is 2.

Sum of order and degree = 2 + 2.
Hence, the answer is 4.

5. The sum of the order and degree of the differential equation d/dx [(dy/dx)³] = 0 is ____.

Ans- d/dx[(dy/dx)³] = 3(dy/dx)²(d²y/dx²).
The highest order derivative is d²y/dx².
Therefore, order = 2.

The power of the highest order derivative is 1.
Hence, degree = 1.
Sum = 2 + 1 = 3.

6. The differential equation representing the family of non-horizontal lines y = mx + c in a plane is ____.

Ans- Given equation: y = mx + c.
Differentiating with respect to x:
dy/dx = m.

Differentiating again:
d²y/dx² = 0.
Hence, the required differential equation is d²y/dx² = 0.

7. The differential equation obtained by eliminating the arbitrary constant c in the equation representing the family of curves xy = c cos x is ____.

Ans- Given equation: xy = c cos x.

Differentiating with respect to x:
y + x(dy/dx) = -c sin x.

From the original equation:
c = xy / cos x.

Substituting this value of c:
y + x(dy/dx) = -xy tan x.

8. The solution of the differential equation dy/dx + 2x = e^3x is ____.

Ans- Given equation:
dy/dx + 2x = e^3x.

Rearranging:
dy/dx = e^3x − 2x.

Integrating both sides:
y = ∫e^3xdx − ∫2x dx.
y = (1/3)e^3x − x² + C.

9. The equation dy/dx + y/(x log x) = 1/x is a ____ differential equation.

Ans- The equation is of the form:
dy/dx + P(x)y = Q(x).
Hence, it is a linear differential equation.

10. The integrating factor of dy/dx = (x + 2y)/x is ____.

Ans- Rewrite the equation:
dy/dx − (2/x)y = 1.
Here P(x) = −2/x.

Integrating factor (I.F.) = e^∫P(x)dx.
I.F. = e^∫(-2/x)dx.
I.F. = e^{-2 ln x}.
I.F. = x^-2 = 1/x².

Multiple Choice Questions

1. The degree of the differential equation [1 + (dy/dx)²]^3/2 = d²y/dx² is
(a) 4
(b) 3/2
(c) not defined
(d) 2

Ans- (d) 2
Given equation:
[1 + (dy/dx)²]^3/2 = d²y/dx²

To remove the fractional power we square both sides.
[1 + (dy/dx)²]³ = (d²y/dx²)²

The highest order derivative is d²y/dx².
Its power is 2.

Therefore the degree is 2.

2. The curve y = (cos x + 1)^1/2 satisfies the differential equation
(a) (2y − 1)d²y/dx² + 2(dy/dx)² + cos x = 0
(b) d²y/dx² − 2y(dy/dx)² + cos x = 0
(c) (2y − 1)d²y/dx² − 2(dy/dx)² + cos x = 0
(d) (2y − 1)d²y/dx² − (dy/dx)² + cos x = 0

Ans- (a) 2y(d²y/dx²) + 2(dy/dx)² + cos x = 0
Given:
y = √(cos x + 1)

Squaring both sides:
y² = cos x + 1

Differentiating:
2y (dy/dx) = −sin x

Differentiating again:
2(dy/dx)² + 2y(d²y/dx²) = −cos x

Rearranging:
2y(d²y/dx²) + 2(dy/dx)² + cos x = 0

3. If xy = A sin x + B cos x is the solution of the differential equation x d²y/dx² − 5a dy/dx + xy = 0, then the value of a is
(a) 2/5
(b) 5/2
(c) −2/5
(d) −5/2

Ans- (c) 2/5
Given solution:
xy = A sin x + B cos x

Differentiating with respect to x and substituting into the differential equation.
After simplification the coefficient comparison gives:
a = 2/5.

4. The solution of the differential equation xdy − ydx = 0 represents
(a) a rectangular hyperbola
(b) parabola whose vertex is at origin
(c) straight line passing through origin
(d) a circle whose centre is at origin

Ans- (c) straight line passing through origin.
Given equation:
xdy − ydx = 0

Dividing by x²:
(xdy − ydx)/x² = 0d(y/x) = 0

Integrating:
y/x = C
y = Cx

This represents a straight line passing through origin.

5. The integrating factor of x (dy/dx) − y = x⁴ − 3x is
(a) x
(b) log x
(c) 1/x
(d) −x

Ans- (c) 1/x
Rewrite the equation:
x dy/dx − y = x⁴ − 3x

Divide by x:dy/dx − (1/x) y = x³ − 3
Here P(x) = −1/x.

Integrating factor:
I.F = e^∫(-1/x)dx
I.F = e^{-ln x}
I.F = 1/x.

6. The solution of dy/dx + y tan x = sec x, y(0) = 0 is
(a) y sec x = tan x
(b) y tan x = sec x
(c) tan x = y tan x
(d) x sec x = tan y

Ans- (a) y sec x = tan x
Given differential equation:
dy/dx + y tan x = sec x

This is a linear differential equation of the form:
dy/dx + P(x)y = Q(x)
Here P(x) = tan x.

Integrating Factor (I.F):
I.F = e^{∫tan x dx}
I.F = e^{−ln cos x}
I.F = sec x

Multiplying the equation by sec x:
sec x dy/dx + y sec x tan x = sec² x

The left side becomes:
d(y sec x)/dx = sec² x

Integrating both sides:
y sec x = tan x + C

Using y(0) = 0:
C = 0
Hence:
y sec x = tan x

7. The particular solution of the differential equation x dy + 2y dx = 0 when x = 2, y = 1 is
(a) xy = 4
(b) x²y = 4
(c) xy² = 4
(d) x²y² = 4

Ans- (b) x²y = 4
Given equation:
x dy + 2y dx = 0

Divide by dx:
x (dy/dx) + 2y = 0
dy/dx = −2y/x

Separate variables:
dy/y = −2 dx/x

Integrating:
ln y = −2 ln x + C
ln y = ln (C/x²)
y = C/x²
x²y = C

Using x = 2, y = 1:
4 = C

Hence:
x²y = 4

8. The general solution of the differential equation (1 + y²) dx + (1 + x²) dy = 0 is
(a) x − y = c (1 − xy)
(b) x − y = c (1 + xy)
(c) x + y = c (1 − xy)
(d) x + y = c (1 + xy)

Ans- (c) x + y = C (1 − xy)
Given equation:
(1 + y²) dx + (1 + x²) dy = 0

Divide by (1 + x²)(1 + y²):
dx/(1 + x²) + dy/(1 + y²) = 0

Integrating:
tan⁻¹ x + tan⁻¹ y = C

Using identity of tangent:
tan(A + B) = (x + y)/(1 − xy)

Therefore:
x + y = C (1 − xy)

9. The solution of the differential equation dy/dx = x/y + y/x is
(a) log (y/x) = x² + C
(b) 2 log (y/x) = x² + C
(c) (y/x)² = log x + C
(d) (y/x)² = 2 log x + C

Ans- (d) (y/x)² = 2 log x + C
Given:
dy/dx = x/y + y/x

Let v = y/x
Then y = vx
dy/dx = v + x dv/dx

Substituting:
v + x dv/dx = 1/v + v
x dv/dx = 1/v
v dv = dx/x

Integrating:
v²/2 = ln x + C
(y/x)² = 2 ln x + C

Hence:
(y/x)² = 2 log x + C

10. The differential equation y (dy/dx) + x = a (a constant) represents
(a) a set of circles having centre on the y-axis
(b) a set of circles having centre on the x-axis
(c) a set of ellipses
(d) None of these

Ans- (b) a set of circles having centre on the x-axis
Given equation:
y dy/dx + x = a

Multiply by dx:
y dy + x dx = a dx

Integrating:
y²/2 + x²/2 = ax + C

Rearranging:
x² + y² − 2ax = C

This represents a circle whose centre lies on the x-axis.
Hence the answer is:
a set of circles having centre on the x-axis.

11. If p and q are the order and degree respectively of the differential equation y (dy/dx) + x³ (d²y/dx²)³ + xy = cos x, then
(a) p < q
(b) p = q
(c) p > q
(d) None of these

Ans- (a) p < q
The highest order derivative present is d²y/dx².
Therefore order p = 2.

The power of the highest order derivative is 3.
Therefore degree q = 3.
Thus p < q.

12. The degree and order of the differential equation [1 + (dy/dx)³]^(7/3) = (7 d²y/dx²) respectively ar
(a) 3 and 7
(b) 3 and 2
(c) 7 and 3
(d) 2 and 3

Ans- (b) 3 and 2
Given equation:
[1 + (dy/dx)³]^(7/3) = 7 d²y/dx²

To remove fractional power we cube both sides.
[1 + (dy/dx)³]⁷ = 343 (d²y/dx²)³

The highest order derivative is d²y/dx²
.Order = 2.

Power of the highest order derivative = 3.
Degree = 3.

Hence the answer is 3 and 2.

13. The order and degree of the differential equation y = x (dy/dx) + 2/(dy/dx) are
(a) 1, 2
(b) 1, 3
(c) 2, 1
(d) 1, 1

Ans- (a) 1,2
Given equation:
y = x(dy/dx) + 2/(dy/dx)

Multiply both sides by dy/dx.
y(dy/dx) = x(dy/dx)² + 2

The highest derivative present is dy/dx.
Order = 1.

The power of dy/dx is 2.
Degree = 2.

14. The degree and order of the differential equation y = x (dy/dx)² + (dx/dy)² are respectively
(a) 1, 1
(b) 2, 1
(c) 4, 1
(d) 1, 4

Ans- (c) 4,1
Given equation:
y = x (dy/dx)² + (dx/dy)²

Since dx/dy = 1/(dy/dx)

Substitute:
y = x(dy/dx)² + 1/(dy/dx)²

Multiply by (dy/dx)².
y(dy/dx)² = x(dy/dx)⁴ + 1

Highest derivative is dy/dx.
Order = 1.
Highest power = 4.
Degree = 4.

15. The degree of the differential equation d²y/dx² = (5y + dy/dx) / √(d²y/dx²) is
(a) 2
(b) 3
(c) 4
(d) 5/2

Ans- (b) 3
Given equation:
d²y/dx² = (5y + dy/dx) / √(d²y/dx²)

Multiply both sides by √(d²y/dx²).
(d²y/dx²)^(3/2) = 5y + dy/dx

To remove fractional power square both sides.
(d²y/dx²)³ = (5y + dy/dx)²

Highest derivative = d²y/dx².
Power = 3.
Therefore degree = 3.

16. The order and degree of the differential equation d²y/dx² + y + (dy/dx + d³y/dx³)^(5/2) = 0 respectively are
(a) 3, 2
(b) 2, 3
(c) 3, 1
(d) 3, 5

Ans- (d) 3,5
The highest derivative present is d³y/dx³.
Therefore order = 3.

Removing fractional power gives the highest power as 5.
Hence degree = 5.

17. A solution of the differential equation (dy/dx)² − x (dy/dx) + y = 0 is
(a) y = 2x
(b) y = −2x
(c) y = 2x − 4
(d) y = 2x + 4

Ans- (c) y = 2x − 4
Substitute y = 2x.
Then dy/dx = 2.

LHS = (2)² − x(2) + 2x
= 4 − 2x + 2x
= 4 ≠ 0.

Testing y = 2x + 4.
dy/dx = 2.

Substitute in equation:
4 − 2x + (2x + 4) = 0
8 ≠ 0.

After checking options the valid solution is y = 2x − 4.

18. The integrating factor of cos x (dy/dx) + y sin x = 1 is
(a) cos x
(b) tan x
(c) sec x
(d) sin x

Ans- (c) sec x
Given equation:
cos x dy/dx + y sin x = 1

Divide by cos x.
dy/dx + y tan x = sec x
Here P(x) = tan x.

Integrating Factor:
I.F = e^(∫ tan x dx)
I.F = e^(−ln cos x)
I.F = sec x.

Very Short Answer Type Questions

1. Find the degree of the differential equation 1 + (dy/dx)² = x

Ans- 2
Given equation:
1 + (dy/dx)² = x

Rearranging:
(dy/dx)² = x − 1

Highest order derivative is dy/dx and its power is 2.
Therefore degree = 2.

2. Find the order and degree of the differential equation
x² (d²y/dx²) = [1 + (dy/dx)²]⁴

Ans- Order = 2, Degree = 1

Highest order derivative present is d²y/dx².
Hence order = 2.

The power of the highest order derivative is 1.
Therefore degree = 1.

3. Find the order and degree of the differential equation
x³(d²y/dx²)² + x(dy/dx)⁴ = 0

Ans- Order = 2, Degree = 2

Highest order derivative is d²y/dx².
Therefore order = 2.

Power of highest order derivative = 2.
Therefore degree = 2.

4. Form the differential equation representing the family of curves:

(i) y = A/x + 5

Ans- x(dy/dx) + y − 5 = 0

Given:
y = A/x + 5

Differentiate with respect to x:
dy/dx = −A/x²

Multiply by x:
x(dy/dx) = −A/x

But A/x = y − 5

So,
x(dy/dx) = −(y − 5)

x(dy/dx) + y − 5 = 0

(ii) y = A sin x

Ans- dy/dx = y cot x

Given:
y = A sin x

Differentiate:
dy/dx = A cos x

Divide both equations:
(dy/dx)/y = (A cos x)/(A sin x)

dy/dx = y cot x

5. Write the general solution of differential equation dy/dx = e^x+y

Ans- e^-y = C − e^x

dy/dx = e^x+y
dy/dx = e^x e^y

Separate variables:
e^-y dy = e^x dx

Integrate both sides:
∫ e^-y dy = ∫ e^x dx

−e^-y = e^x + C
e^-y = C − e^x

6. Show that y = ax + 2a² is a solution of differential equation
2(dy/dx)² + x(dy/dx) − y = 0

Ans- Verified
Given y = ax + 2a²
dy/dx = a

Substitute in equation:
2(a)² + x(a) − (ax + 2a²)
= 2a² + ax − ax − 2a²
= 0
Hence LHS = RHS, therefore verified.

7. What form of differential equation is
dy/dx + y/(x log x) = 1/x ?

Ans- Linear Differential Equation
Standard linear form is:
dy/dx + P(x)y = Q(x)

Given equation:
dy/dx + y/(x log x) = 1/x

This matches the linear form.
Therefore it is a Linear Differential Equation.

8. Write the integrating factor of the differential equation
(i) dy/dx + 2y = x²

Ans- I.F. = e^2x
Standard form:
dy/dx + Py = Q

Here P = 2

Integrating Factor = e^{∫P dx}
I.F. = e^{∫2 dx}
I.F. = e^2x

(ii) (tan⁻¹y − x) dy = (1 + y²) dx

Ans- I.F. = 1/(1 + y²)

(iii) (e − 2√x)/√x − y/√x (dx/dy) = 1

Ans- I.F. = 1/x

–: End of MCQ’s on Differential Equations Class 12 OP Malhotra ISC Maths Solutions Ch-17 :–

Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions
Please share with your friends
Thanks