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Given y = cos x + C 

⇒ y' = \(\frac{d}{dx}\)(cos x + C) = – sin x 

Given, y’ + sinx = 0

sinx + sinx = 0 

LHS = RHS 

Hence y = cos x + C, is solution of given differential equation.

Answer is (D)

Since the equation is not a polynomial in

\(\frac{dy}{dx}\) degree is not defined

Answer is (A)

Here the highest order is 2.

Hence the order is 2.

Given y = x² + 2x + C

⇒ y' = \(\frac{d}{dx}\)(x² + 2x + C = 2x + 2 

Given, y’ = 2x - 2 

= 2x + 2 – 2x – 2 = 0 

LHS = RHS 

Hence y = x² + 2x + C, is solution of given differential equation

(i) (d²y/dx²)² + (dy/dx) = 0

The highest derivative of y = (d²y/dx²)²

Thus, order = 2, degree = 2.

(ii) (dy/dx)³ + \(\frac{1}{\frac{dy}{dx}}\) = 2

⇒ (dy/dx)⁴ + 1 = 2(dy/dx)

The highest derivative of y = (dy/dx)⁴ = (d¹y/dx¹)⁴

Thus, order = 1, degree = 4.

Answer is (A)

The correct option (D) degree doesn’t exist

Explanation:

Differential equation cannot be expressed in a polynomial form, hence degree does not exists.

Answer is (C)

y = x y₁ = 1 y₂ = o 

this is satisfied by 

y² – x² y, + xy = 0 

0 – x² . 1 + x . x = 0

(i) (d²y/dx²)³ + y(dy/dx)² + y⁵ = 0

The highest derivative of y = (d²y/dx²)³

Thus, order = 2, degree = 3.

(ii) x(dy/dx) + 3/(dy/dx) = y²

x(dy/dx)² + 3 = y²(dy/dx)

The highest derivative of y = (dy/dx)² = (d¹y/dx¹)²

Thus, order = 1 degree = 2.

Order is defined as the number which represents the highest derivative in a differential equation

Second order means the order should be 2 which means the highest derivative in the equation should be d²y/dx²  or y’’

Let us examine each option given

A. (y’)² + x = y²

The highest order derivate is y’ which is first order.

B. y’y’’ + y = sinx

The highest order derivate is y’’ which is second order.

C. y’’’+(y’’)²+y=0

The highest order derivate is y’’’ which is third order.

D. y’ = y²

The highest order derivate is y’ which is first order.

The order of a differential equation is the order of the highest derivative involved in the equation. So, the order comes out to be 3 as we have \((y")^3+(y')^2+sin\,y'+1=0\)

and the degree is the highest power to which a derivative is raised. So the power at this order is 2. 

So the answer is 3, 2.

Given: \(y=\cfrac{dy}{dx}+\cfrac{5}{\left(\cfrac{dy}{dx}\right)}\)

Solving, we get,

\(y\times\cfrac{dy}{dx}=\left(\cfrac{dy}{dx}\right)^2+5\)

Now,

The order of a differential equation is the order of the highest derivative involved in the equation. So, the 

order comes out to be 2 as we have, \(y\times\cfrac{dy}{dx}=\left(\cfrac{dy}{dx}\right)^2+5\)

and the degree is the highest power to which a derivative is raised. So the power at this order is 1. 

So the answer is 2, 1.

The order of a differential equation is the order of the highest derivative involved in the equation. So, the order comes out to be 1 as we have \(\sqrt{1-y^2}dx+\sqrt{1-x^2}dy=0\)

\(\cfrac{dy}{dx}=-\cfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\)

and the degree is the highest power to which a derivative is raised. So the power at this order is 1. 

So the answer is 1, 1.

The order of a differential equation is the order of the highest derivative involved in the equation. So the 

order comes out to be 1 as we have \(\cfrac{dy}{dx}\) and the degree is the highest power to which a derivative is raised. But when we open Sin x as \(x-\cfrac{x^3}{3!}+\cfrac{x^5}{5!}-\cfrac{x^7}{7!}+-----\) . Also, the equation has to be polynomial, and

opening thus, Sin function will lead to an undefined power of the highest derivative. Therefore the degree is not defined. 

So the answer is 1, not defined.

Rearranging the terms we get:

\(\frac{dy}y\) = tan x dx

⇒ \(\int\frac{dy}y\) = \(\int{tanx}\,dx\) + c

⇒ log |y| = log |sec x| + log c 

⇒ log |y| - log |sec x| = log c 

⇒ log |y| + log |cos x| = log c 

⇒ y cos x = c y = 1 when x = 0 

∴1 × cos0 = c 

∴c = 1 

⇒ y cos x = 1 

⇒ y = 1/cos x 

⇒ y = sec x 

Ans: y = sec x

Given:

\(\frac{dy}{dx}\) = (1 + x)(1 + y)

⇒ \(\frac{dy}{1+y}\) = (1 + x)dx

⇒ log |y + 1| = (x + \(\frac{x^2}2\) + c)

⇒ Now, for y = 0 and x = 1,

We have,

⇒ 0 = 1 + \(\frac{1}2\) + c

⇒ c = - \(\frac{3}2\)

⇒ log |y + 1| = \(\frac{x^2}2\) + x - \(\frac{3}2\)

Correct answer is C.

y(dy/dx) + x³(d²y/dx²) + xy = cos x,

Order = Highest order derivative present in the differential equation.

∴ Order = 2 = p

Degree = Highest power of highest order derivative which is \(\frac{d^2y}{dx^2}\)

∴ Degree = 1 = q

∴ p = 2 > q = 1

We have,

\(\frac{dy}{dx}\) = y sin 2x

⇒ \(\frac{dy}{y}\) = y sin2x

⇒ log y = - \(\frac{cos2x}2\) + c

For y =1, x = 0, we have,

c = \(\frac{1}{2}\)

⇒ log y = \(\frac{1}{2}\)(1 - cos2x)

⇒ log y = sin²x

Thus, 

The particular solution is:

y = e^sin2x

From the given differential equation, we have

\(\frac{dy}{dx}+\frac{1}{x\,log\,x}.y=\frac 2x\)

So, If = \(e^{\int\frac{1}{x\,log\, x}dx}\)\(=e^{log|(log\,x)|}= log\,x\)

The correct answer is (B).

Correct answer is (D). The given differential equation is not a polynomial equation in terms of its derivatives, so its degree is not defined.

1. Degree = 1

2. We have, \(\frac{dy}{dx}\) = 2xy ⇒ \(\frac{dy}{y}\) = 2xdx, which is of the form Mdx = Ndy.

3. Solution is ∫\(\frac{dy}{y}\) = 2∫xdx ⇒ log|y| = x² + c

Given y(0) = 1 ⇒ log|1| = 0 + c ⇒ c = 0

⇒ log|y| = x² ⇒ y = e^x2.

1. Order is 1

2. (a) Variable separable method

3. (b) - log | 50 – y| = kx + C

4. (b) log 1/50

5. (c) y = 50 ( 1 – e ^-kx)

Important Class 12 Maths MCQ Questions of Differential Equations with Answers allows in getting ready for CBSE Board exams. Students can go through the important MCQ Questions for Class 12 Maths to revise all of the chapters and get extra marks withinside the CBSE final examinations. This chapter consists of many formulations and strategies for solving given differential equations.

Apart from reading and practicing MCQ Questions of Differential equations from the NCERT book, college students shall additionally practice these vital MCQ Questions. The objective types of questions are taken in keeping with the syllabus of the CBSE board. These important MCQ Questions for class 12 help the students to gain the right marks withinside the class 12 board exam.

Practice MCQ Question for Class 12 Maths chapter-wise

1. The radius of a circle is increasing at the rate of 0.7 cm/ s. The rate of increasing of its circumference is

(a) 1.4 π cm/s
(b) 0.8 π cm/s
(c) 1.8 cm/s
(d) None of these

2.  Solution of differential equation xdy – ydx = Q 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

3. Integrating factor of the differential equation   \(cosx\frac{dy}{dx}+ysin x=1\) is

(a) cos x
(b) tan x
(c) sec x
(d) sin x

4. Family r = Ax + A³ of curves is represented by the differential equation of degree

(a) 1
(b) 2
(c) 3
(d) 4

5. Which of the following is a second order differential equation?

(a) (y’)² + x = y²
(b) y’y” + y = sin x
(c) y” + (y”)² + y = 0
(d) y’ = y²

6. The differential equation \(\frac{dy}{dx}+x=c\) represents

(a) Family of hyperbolas
(b) Family of parabolas
(c) Family of ellipses
(d) Family of circles

7. The general solution of e^x cosydx − e^x sinydy = 0 is

(a) e^x cosy = k 
(b) e^x siny = k 
(c) e^x = k cos y
(d) e^x = k sin y 

8. The solution of \(x\frac{dy}{dx}+y=e^x\)   is

(a) \(y=\frac{e^x}{x}+\frac{k}{x}\)
(b) \(y=xe^x+cx\)
(c) \(xe^x+k\)
(d) \(x=\frac{e^y}{y}+\frac{k}{y}\)

9. The general solution of \(\frac{dy}{dx}=2x e^{x^{2-y}}\) is

(a) \(e^{x^2-y}=c\)
(b) \(e^{-y}+e^{x^2}=c\)
(c) \(e^y=e^{x^2}+c\)
(d) \(e^{x^2+y}=c\)

10. The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is

(a) an ellipse
(b) parabola
(c) circle
(d) rectangular hyperbola

11. The number of arbitrary constants in the general solution of differential equation of fourth order is

(a) 4
(b) 3
(c) 2
(d) 0

12. The order of differential equation of all circles of given radius ′a′ is

(a) 4
(b) 2
(c) 1
(d) 3

13. The general solution of differential equation is y = ae ^bx+c where a,b,c are arbitrary constant. The order of differential equation is:

(a) 4
(b) 2
(c) 1
(d) 3

14. The degree and order of the differential equation of the family of all parabolas whose axis is x-axis, are respectively

(a) 2,1
(b) 1,2
(c) 3,2
(d) 2,3

15. The order and degree of the differential equation of the family having the same foci, are respectively

(a) 2,1
(b) 2,1
(c) 2,2
(d) 1,2

16. The order and degree of the differential equation of the family of circles touching the x-axis at the origin, are respectively

(a) 1,1
(b) 1,2
(c) 2,1
(d) 2,2

17. The solution of (x+ logy)dy +ydx =0 where y(0) =1 is

(a) y(x−(A)) + ylogy = 0
(b) y(x−1+logy) + 1 = 0
(c) xy+ ylogy +1 = 0
(d) None of these

18. The number of arbitrary constants in the particular solution of a differential equation of third order is:

(a) 3
(b) 2
(c) 1
(d) 0

19. What is the order of differential equation y’’ + 5y’ + 6 = 0?

(a) 0
(b) 1
(c) 2
(d) 0

20. What is the degree of differential equation (y’’’)² + (y’’)³ + (y’)⁴ + y⁵ = 0?

(a) 2
(b) 3
(c) 4
(d) 5

Answer:

1. Answer: (a) 1.4 π cm/s

Explanation: Let r be the radius and C the circumference of the circle. Then, C = 2πr

  dr/dt = 0.7 cm/sec.

Now , C = 2πr ⇒ dC/dt = \(2\pi\frac{dr}{dt}\Rightarrow\) dC/dt

 \( =2\pi\times0.7cm/sec\)

= 1.4 π cm/s

2.  Answer: (c) straight line passing through origin

Explanation: xdy − ydx = 0

⇒ dy/y = dx/x

On integration, we get

logy = logx + logc

y = xc which is equation of a straight line passing through origin.

3. Answer: (c) sec x

Explanation: \(cosx \frac{dy}{dx}+ysinx=1\)

⇒ dy/dx + sinx/cosx⋅y = 1/cosx

⇒ dy/dx+tanx⋅y = secx  

I.F = \(e^{\int\;tan\;xdx}\)

=  \(e^{log(sec x)}\)

= sec x

4. Answer: (a) 1

Explanation: y = ax + a³.....(i)

y' = a

Linear differential equation of order 1 and degree 1.

5. Answer: (b) y’y” + y = sin x

Explanation: The second order differential equation is  y’y” + y = sin x

6. Answer: (d) Family of circles 

Explanation: Given differential equation is

\(y\frac{dy}{dx}+x=c\)

⇒ ydy = (c−x)dx

On integrating both sides, we get

y²/2 = cx−x²/2 + d

⇒y² +x² −2cx −2d = 0

Hence, it represents a family of circles whose centres are on the x -axis.

7. Answer: (a) e^x cosy = k 

Explanation: e^x cosy dx−e^x siny dy=0 

⇒e^x cosydx = e^x sinydy

⇒dx = tanydy

On integrating, we get

x = log(secy) + logk

⇒x = log[(secy)k]

⇒e^x = ksecy

⇒e^x/secy = k

⇒e^x cosy = k

8. Answer: (a) 

Explanation: \(x\frac{dy}{dx}+y=e^x\)

\(\frac{dy}{dx}+\frac{1}{x}y=\frac{e^x}{x}\)

It is a linear differential equation with

I.F = \(e^{\int\;\frac{1}{x}dx}\)

\(=e^{log x}=x\)

Now, solution is y⋅x \(\int\frac{e^z}{x}.xdx+k\)

⇒y^x = ex + k

\(y=\frac{e^x}{x}+\frac{k}{x}\)

9. Answer: (c) 

Explanation: \(\frac{dy}{dx}=2xe^{x^2-y}\)

\(e^ydy=2xe ^{x^3}dx\)

On integrating, we get \(e^y = e^{x^2} + c\)

10. Answer: (d) rectangular hyperbola

Explanation: The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is rectangular hyperbola.

11. Answer: (a) 4

Explanation: The number of arbitrary constants in a solution of a differential equation of order n is equal to its order. So, here it is 4.

12. Answer: (b) 2

Explanation: Let the centre of circle be (h,k) and radius a, then its equation will be

(x−h)²+(y−k)² =a²

Since, there are two parameters h and k. So, the differential equation is of order 2

13. Answer: (b) 2

Explanation: ae^bx. e^c a,b,c are three constants 

in order to solve the above given equation we have to differentiate it twice so the given equation is 

d²y/dx² = ab²e^bx+c

Thus, order is 2

14. Answer: (b) 1,2

Explanation: Parabola whose axis is x-axis is y² = 4ax

2y = dy/dx = 4a

\(y\frac{dy}{dx}=2a\)

\(\Rightarrow y \frac{d^2y}{dx^2}+(\frac{dy}{dx})^2=0\)

degree =1, order =2

15. Answer: (b) 2,1

Explanation: Equation of family of parabola with x-axis as axis is y² = 4a(x + α) where α are two arbitrary constant. So, differential equation is of order 2 and degree 1.

16. Answer: (a) 1,1

Explanation: The system of circles touching x-axis at origin will have centres on y-axis. 
Let (0,a) be the centre of a circle. Then the radius of the circle should be a units, since the circle should touch x-axis at origin. Equation of a circle with centre at (0,a) and radius a is 

\((x-0)^2+(y-a)^2=a^2\)

\(x^2+y^2--2ay=0\)

The above equation represents the family of circles touching x-axis at origin. Here 'a' is an arbitrary constant. In order to find the differential equation of system of circles touching x-axis at origin, eliminate the the arbitrary constant from equation. Differentiating equation with respect to y,

\(2x\frac{dx}{dy}+2y-2a=0\)

\(\frac{dx}{dy}=2(y-a)\)

 Order is 1 and degree is also 1.

17. ​​​​​​​Answer: (b) 

Explanation: (x +logy)dy + ydx = 0 

∴d(xy) =xdy + ydx

(xdy + ydx)+(logydy)=0

xy + (logy . -y/y dy) = 0

xy + ylogy − y + c=0

y(x−1) +ylogy + c = 0

y(x−1+logy) +  c= 0

1(0−1+log1)+c=0

c=1  given y(0) = 1 i.e. (0, 1) is the solution of the differential equation So, overall solutions is

(x +logy)dy + ydx = 0 

18. Answer: (d) 0

Explanation: In the particular solution of a differential equation of third order, there is no arbitrary constant because in the particular solution of any differential equation, we remove all the arbitrary constant by substituting some particular values.

19. Answer: (c) 2

Explanation: The highest order derivative present in the differential equation is y’’. Hence, the order is 2.

20.Answer: (a) 2

Explanation: The degree is the power raised to the highest order derivative. Therefore, in the given differential equation, (y’’’)² + (y’’)³ + (y’)⁴ + y⁵ = 0, the degree will be power raised to y’’’.

Click here to practice more MCQ Question for Differential Equations Class 12

The order of a differential equation is the order of the highest derivative involved in the equation. So the 

order comes out to be 3 as we have \(\cfrac{d^3y}{dx^3}\) and the degree is the highest power to which a derivative is raised. 

So the power at this order is 2. 

So the answer is 3, 2.

Higher order is \(\frac{d^4y}{dx^4}\) so order is 4

As it is not a polynomial in \(\frac{dy}{dx}\) degree is not defined.

Given, \(\frac{dy}{dx}\) = e^x+y

\(\frac{dy}{dx}\) = e^xe^y

e^-y dy = e^x dx

On integrating on both sides, we get

 - e^-y + c₁ e^x +c₂

e^-y + e^x = c

Conclusion: Therefore, e^-y + e^x = c is the solution of \(\frac{dy}{dx} \) e^x+y

The order of a differential equation is the order of the highest derivative involved in the equation. So the 

order comes out to be 2 as we have \(\cfrac{d^2y}{dx^2}\) and the degree is the highest power to which a derivative is raised. 

So the power at this order is 2.

So the answer is 2, 2.

Correct answer is B.

(d²y/dx²)² - (dy/dx) = y³

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is \((\frac{d^2y}{dx^2})^2\)

∴ Degree = 2

Given

 x³(d²y/dx²)² + x(dy/dx)⁴ = 0

The highest order derivative is (d²y/dx²) and its power is 2.

So, the degree of differential equation is 2.

Correct option is (D) not defined.

Since the value of sin (dy/dx) on expansion will be in increasing power of dy/dx, the degree of the given differential equation is not defined.

x³\(\frac{d^3y}{dx^3}+3x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=\) x + log x---(1)

Let x = e^z ⇒ log x = z

Then x \(\frac{dy}{dx}\) converts into \(\frac{dy}{dx}\) = Dy where D = d/dz

x²\(\frac{d^2y}{dx^2}\) converts into D(D - 1)y

& x³\(\frac{d^3y}{dx^3}\) converts into D(D - 1)(D - 2)y.

Differential equation (1) converts into

D(D - 1)(D - 2)y + 3D(D - 1)y + Dy + y = e^z + z

⇒ (D³ - 3D² + 2D + 3D² - 3D + D + 1)y = e^z + z

⇒ (D³ + 1)y = e^z + z---(2)

It's auxiliarly equation is

m³ + 1 = 0

⇒ (m + 1) (m² - m + 1) = 0

⇒ m = -1, \(\frac{1\pm\sqrt3i}{2}\)

\(\therefore\) C.F. = C₁e^-z + e^z/2(C₂ cos(\(\frac{\sqrt3z}2\)) + C₃ sin(\(\frac{\sqrt3z}2\)))

P.I. = \(\frac1{D^3+1}(e^z+z)\) = \(\frac1{D^3+1}e^z+\frac1{D^3+1}z\) 

= \(\frac{e^z}{1^3+11}+(1+D^3)^{-1}z\) 

= \(\frac{e^z}2+(1-D^3+....)z\) 

= \(\frac{e^z}2+z\) 

\(\therefore\) solution of differential equation (2) is

y = C. F. + P. I.

= C₁e^-z + (e^z)^1/2(C₂ cos(\(\frac{\sqrt3}2\) log x) + C₃ sin(\(\frac{\sqrt3}2\) log x)) + x/2 + log x

which is complete solution of given differential equation.

\(x\frac{dt}{dx}\) = x + t

⇒ \(\frac{dt}{dx}-\frac{t}x=1\)

I.F. = e^∫pdx = e ^{∫-\(\frac1x\)dx} = e^-logx = \(\frac1x\)

It's complete solutions

y x I.F = ∫(I.F) x Q dx

⇒ \(\frac{y}x\) = \(\int\frac1x\) x 1 dx = log x + c

⇒ y = x log x + cx

\((\frac{e^2-1}{e^2})dx-(\frac{e^z+1}{e^2})dy=0\)

(1 - e^-z)dx - (1 + e^-z)dy = 0

∫(1 - e^-z)dx - ∫(1 + e^-z)dy = 0

Let z = x + iy

⇒ ∫(1 - e^{-(x + iy)})dx - ∫(1 + e^{-(x + iy)})dy = c

⇒ x + e^-(x+iy) - y + \(\frac1{i}\)e^{-(x + iy)} = c

⇒ x - y + e^{-(x + iy)}(1 + \(\frac1i\times\frac{i}i\)) = c

⇒ x - y + e^-z(1 - i) = c (\(\because\) i² = -1)

Hence, solution of given differential equation is x - y + e^-z(1 - i) = c

Order is 4 but degree is not defined because given differential equation cannot be written in the form of polynomial in differential co-efficient.

Given: 

cosx(1 + cos y)dx - sin y(1 + sin x)dy = 0 

Dividing the whole equation by (1 + sin x)(1 + cos y), we get,

⇒ \(\frac{\int{cosx\,dx}}{1+sin\,x}\) = \(\frac{\int{cosy\,dy}}{1+sin\,y}\)

⇒ log| 1 + sin x | + log|1 + cos y| = log c 

⇒ (1 + sin x)(1 + cos y) = c

Again, differentiating with respect to x we get,

\(x\left(\frac{dy}{dx}\right) + y =0\)

Hence, \(x\left(\frac{dy}{dx}\right) + y =0\) is the differential equation corresponding to xy = a².

We know that, the equation of all non-vertical lines are y = mx + c

On differentiating w.r.t. x, we get

dy/dx = m

Again, on differentiating w.r.t. x, we have

d²y/dx² = 0

Thus, the required equation is d²y/dx² = 0.

Correct option (c)  y = e^{a(x – 1)} 

Explanation:

Normal at point P is ay + x = a + 1 

Slope of tangent at P = a = (dy/dx) (1, 1) ...(1)

Now dy/dx α ⇒ dy/dx = ky  ⇒ (dy/dx)_(1,1) = k ...(2)

From (1) & (2) k = a

dy/dx  = ay ⇒ dy/y  = a.dx

⇒ log_e y = ax+C

Now C = – a (as curve passes through (1,1))

log_e y = ax–a

⇒ y = e^a(x–1)

Correct option (c) xy + y log_e x + x sin y = C

Explanation:

ydx + y/x .dx + sin ydx + xdy + loge xdy + x cos ydy = 0

(ydx + xdy) + (y.dx/x + log_exdy) +  (sin ydx + x cos ydy) = 0

= d(xy) + d(y log_e x) + d(x sin y) = 0

xy + y log_e x + x sin y = C

Given differential equation is (d²y/dx² )² + (dy/dx)³ + x⁴ = 0 

Here, we see that the highest order derivative is d²y/dx² , whose degree is 2.

Here,order = 2 and degree = 2

Sum of the order and degree = 2 + 2 = 4

True, because it is not a polynomial equation in its derivatives.

\(\cfrac{dy}{dx}=-k\)

∴ dy = -k dx

Integrating both sides, we get

∫dy = -k∫dx

∴ y = -kx + c

This is the general solution.

(c) ye^∫pdx = ∫Qe^∫pdx dx + c

(d) x dx + y dy = 0

x² + y² = a² 

⇒ 2x + 2y(dy/dx) = 0 

⇒ x dx + y dy = 0

(1 – x) dy – (1 + y) dx = 0 

Separating the variables, 

(1 – x)dy = (1 + y) dx 

dy/(1 + y) = dx/(1 - x)

Integrating, we get 

∫1/(1 + y)dy = ∫1/(1 - x)dx

log(y + 1) = -log(1 – x) + log c 

log(y + 1) + log(1 – x) = log c 

(y + 1) (1 – x) = c

(a) A + Be^x

A.E is m² – m = 0 

⇒ m(m – 1) = 0 

⇒ m = 0, 1 

CF is Ae^0x + Be^x = A + Be^x