This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Linear Second Order Differential Equations”.

1. Which one of the following is correct if we differentiate the equation xy = ae^{x} + be^{-x} two times?

a) x(d^{2}y/ dx) + 2(dy/dx) = xy

b) x(d^{2}y/ dx) – 2(dy/dx) = xy

c) 3x(d^{2}y/ dx) + 2(dy/dx) = xy

d) x(d^{2}y/ dx) + 2(dy/dx) = 2xy

View Answer

Explanation: We have xy = ae

^{x}+ be

^{-x}……(1)

Differentiating (1) with respect to x, we get

x(dy/dx) + y = ae

^{x}+ be

^{-x}…..(2)

Differentiating (2) now, with respect to x, we get

x(d

^{2}y/ dx) + dy/dx + dy/dx = ae

^{x}+ be

^{-x}

From (1),

ae

^{x}+ be

^{-x}= xy, so that we get

x(d

^{2}y/ dx) + 2(dy/dx) = xy

which is the required differential equation.

2. What is thedifferential equation whose solution represents the family y = ae^{3x} + be^{x}?

a) d^{2}y/dx^{2} – 3dy/dx + 4y = 0

b) d^{2}y/dx^{2} – 4dy/dx + 3y = 0

c) d^{2}y/dx^{2} + 4dy/dx + 3y = 0

d) d^{2}y/dx^{2} – 4dy/dx – 3y = 0

View Answer

Explanation: The original given equation is,

y = ae

^{3x}+ be

^{x}……….(1)

Differentiating the above equation, we get

dy/dx = 3ae

^{3x}+ be

^{x}……….(2)

d

^{2}y/dx

^{2}= 9ae

^{3x}+ be

^{x}……….(3)

Now, to obtain the final equation we have to make the RHS = 0,

So, to get RHS = 0, we multiply, (1) with 3 and (2) with -4,

Thus , 3y = 3ae

^{3x}+ 3be

^{x}……….(4)

And -4(dy/dx) = -12ae

^{3x}– 4 be

^{x}……….(5)

Now, adding the above three equations,(i.e. (3) + (4) + (5)) we get,

d

^{2}y/dx

^{2}– 4dy/dx + 3y = 0

3. What is the differential equation of all parabolas whose directrices are parallel to the x-axis?

a) d^{3}x/dy^{3} = 0

b) d^{3}y/(dx^{3} + d^{2}y/dx^{2}) = 0

c) d^{3}y/dx^{3} = 0

d) d^{2}y/dx^{2} = 0

View Answer

Explanation: The equation of family of parabolas is Ax

^{2}+ Bx + C = 0 where, A, B, C are arbitrary constant.

By differentiating the equation with respect to x till all the constants get eliminated,

Hence, d

^{3}y/dx

^{3}= 0

4. If y = t(x) be a differentiable function ᵾ x € R, then which of the following is always true?

a) d^{2}y/dx^{2} – (dy/dx)^{3} = 0

b) d^{2}y/dx^{2} + (dy/dx)^{3} d^{2}x/dy^{2} = 0

c) d^{2}y/dx^{2} – (dx/dy)^{3} = 0

d) d^{2}y/dx^{2} + (dy/dx)^{3} = 0

View Answer

Explanation: (dy/dx)= (dx/dy)

^{-1}

So, d

^{2}y/dx

^{2}= – (dx/dy)

^{-2}d/dx(dx/dy)

= – (dy/dx)

^{2}(d

^{2}x/dy

^{2})(dy/dx)

=> d

^{2}y/dx

^{2}+ (dy/dx)

^{3}d

^{2}x/dy

^{2}= 0

5. What will be the required solution of d^{2}y/dx^{2} – 3dy/dx + 4y = 0?

a) Ae^{-4x} + Be^{-x}

b) Ae^{4x} – Be^{-x}

c) Ae^{4x} + Be^{-x}

d) Ae^{4x} + Be^{x}

View Answer

Explanation: d

^{2}y/dx

^{2}– 3dy/dx + 4y = 0 …..(1)

Let, y = e

^{mx}be a trial solution of (1), then,

=> dy/dx = me

^{mx}and d

^{2}y/dx

^{2}= m

^{2}e

^{mx}

Clearly, y = e

^{mx}will satisfy equation (1). Hence, we have,

m

^{2}e

^{mx}– 3m * e

^{mx}– 4e

^{mx}= 0

=>m

^{2}– 3m – 4 = 0 (as e

^{mx}≠ 0) …….(2)

=> m

^{2}– 4m + m – 4 = 0

=> m(m – 4) + 1(m – 4) = 0

Or, (m – 4)(m + 1) = 0

Thus, m = 4 or m = -1

Clearly, the roots of the auxiliary equation (2) are real and unequal.

Therefore, the required general solution of (1) is

y = Ae

^{4x}+ Be

^{-x}where A and B are constants.

6. What will be the general solution of the differential equation d^{2}y/dx^{2} = e^{2x}(12 cos3x – 5 sin3x)? (here, A and B are integration constant)

a) y = e^{x} sin3x + Ax + B

b) y = e^{2x} sin3x + Ax + B

c) y = e^{2x} sin3x + A

d) Data inadequate

View Answer

Explanation: Given, d

^{2}y/dx

^{2}= e

^{2x}(12 cos3x – 5 sin3x) ………(1)

Integrating (1) we get,

dy/dx = 12∫ e

^{2x}cos3x dx – 5∫ e

^{2x}sin3x dx

= 12 * (e

^{2x}/2

^{2}+ 3

^{2})[2cos3x + 3sin3x] – 5 * (e

^{2x}/2

^{2}+ 3

^{2})[2sin3x – 3cos3x] + A (A is integrationconstant)

So dy/dx = e

^{2x}/13 [24 cos3x + 36 sin3x – 10 sin3x + 15 cos3x] + A

= e

^{2x}/13(39 cos3x + 26 sin3x) + A

=> dy/dx = e

^{2x}(3 cos3x + 2 sin3x) + A ……….(2)

Again integrating (2) we get,

y = 3*∫ e

^{2x}cos3xdx + 2∫ e

^{2x}sin3xdx + A ∫dx

y = 3*(e

^{2x}/2

^{2}+ 3

^{2})[2cos3x + 3sin3x] + 2*(e

^{2x}/2

^{2}+ 3

^{2})[2sin3x – 3cos3x] + Ax + B (B is integration constant)

y = e

^{2x}/13(6 cos3x + 9 sin3x + 4 sin3x – 6 cos3x) + Ax + B

or, y = e

^{2x}sin3x + Ax + B

7. What is the solution of the given equation (D + 1)^{2}y = 0 given y = 2 log_{e} 2 when x = log_{e} 2 and y = (4/3) log_{e}3 when x = log_{e}3?

a) y = 4xe^{-x}

b) y = 4xe^{x}

c) y = -4xe^{-x}

d) y = -4xe^{x}

View Answer

Explanation:(D + 1)

^{2}y = 0

Or, (D

^{2}+ 2D+ 1)y = 0

=> d

^{2}y/dx

^{2}+ 2dy/dx + y = 0 ……….(1)

Let y = e

^{mx}be a trial solution of equation (1). Then,

=> dy/dx = me

^{mx}and d

^{2}y/dx

^{2}= m

^{2}e

^{mx}

Clearly, y = e

^{mx}will satisfy equation (1). Hence, we have

=> m

^{2}.e

^{mx}+ 2m.e

^{mx}+ e

^{mx}= 0

Or, m

^{2}+ 2m +1 = 0 (as, e

^{mx}≠ 0) ………..(2)

Or, (m + 1)

^{2}= 0

=> m = -1, -1

So, the roots of the auxiliary equation (2) are real and equal. Therefore, the general solution of equation (1) is

y = (A + Bx)e

^{-x}where A and B are two independent arbitrary constants ……….(3)

Given, y = 2 log

_{e}2 when x = log

_{e}2

Therefore, from (3) we get,

2 log

_{e}2 = (A + B log

_{e}2)e

^{-x}

Or, 1/2(A + B log

_{e}2) = 2 log

_{e}2

Or, A + B log

_{e}2 = 4 log

_{e}2 ……….(4)

Again y = (4/3) log

_{e}3 when x = log

_{e}3

So, from (3) we get,

4/3 log

_{e}3 = (A + Blog

_{e}3)

Or, A + Blog

_{e}3 = 4log

_{e}3 ……….(5)

Now, (5) – (4) gives,

B(log

_{e}3 – log

_{e}2) = 4(log

_{e}3 – log

_{e}2)

=> B = 4

Putting B = 4 in (4) we get, A = 0

Thus the required solution of (1) is y = 4xe

^{-x}

8. Which of the following is the valid differential equation x = a cos(αt + β)?

a) d^{2}x/dt^{2} – αx = 0

b) d^{2}x/dt^{2} + αx = 0

c) d^{2}x/dt^{2} – α^{2}x = 0

d) d^{2}x/dt^{2} + α^{2}x = 0

View Answer

Explanation: Since, x = a cos(αt + β)

Therefore, dx/dt = a cos(αt + β)

And, d

^{2}x/dt

^{2}= -a α

^{2}cos(αt + β)

= -α

^{2}a cos(αt + β)

Or, d

^{2}x/dt

^{2}= -α

^{2}[as a cos(αt + β) = x]

So, d

^{2}x/dt

^{2}+ α

^{2}x = 0

9. If, A and B are arbitrary constants then what will be the differential equation of y = Ax + B/x?

a) x^{2} d^{2} y/dx^{2} – xdy/dx + y = 0

b) x^{2} d^{2} y/dx^{2} + xdy/dx + y = 0

c) x^{2} d^{2} y/dx^{2} + xdy/dx – y = 0

d) x^{2} d^{2} y/dx^{2} – xdy/dx – y = 0

View Answer

Explanation: Given, y = Ax + B/x

=> xy = Ax

^{2}+ B ……….(1)

Differentiating (1) with respect to x, we get,

d(xy)/dx = d/dx(Ax

^{2}+ B)

or, xdy/dx + y = A * 2x ……….(2)

Differentiating again with respect to x, we get,

x*d

^{2}y/dx

^{2}+ dy/dx + dy/dx = A*2 ……….(3)

Eliminating A from (2) and (3) we get,

x

^{2}d

^{2}y/dx

^{2}+ 2xdy/dx = 2Ax [multiplying (3) by x]

or, x

^{2}d

^{2}y/dx

^{2}+ 2xdy/dx = xdy/dx + y [using (2)]

or, x

^{2}d

^{2}y/dx

^{2}+ xdy/dx – y = 0

10. What will be the value of C if C the constant of the coefficient of the solution of the given equation (D + 1)^{2}y = 0 given y = 2 log_{e} 2 when x = log_{e} 2 and y = (4/3) log_{e}3 when x = log_{e}3?

a) 2

b) -2

c) -4

d) 4

View Answer

Explanation: (D + 1)

^{2}y = 0

Or, (D

^{2}+ 2D+ 1)y = 0

=> d

^{2}y/dx

^{2}+ 2dy/dx + y = 0 ……….(1)

Let y = e

^{mx}be a trial solution of equation (1). Then,

=> dy/dx = me

^{mx}and d

^{2}y/dx

^{2}= m

^{2}e

^{mx}

Clearly, y = e

^{mx}will satisfy equation (1). Hence, we have

=> m

^{2}.e

^{mx}+ 2m.e

^{mx}+ e

^{mx}= 0

Or, m

^{2}+ 2m + 1 = 0 (as, e

^{mx}≠ 0) ………..(2)

Or, (m + 1)

^{2}= 0

=> m = -1, -1

So, the roots of the auxiliary equation (2) are real and equal. Therefore, the general solution of equation (1) is

y = (A + Bx)e

^{-x}where A and B are two independent arbitrary constants ……….(3)

Given, y = 2 log

_{e}2 when x = log

_{e}2

Therefore, from (3) we get,

2 log

_{e}2 = (A + B log

_{e}2)e

^{-x}

Or, 1/2(A + B log

_{e}2) = 2 log

_{e}2

Or, A + B log

_{e}2 = 4 log

_{e}2 ……….(4)

Again y = (4/3) log

_{e}3 when x = log

_{e}3

So, from (3) we get,

4/3 log

_{e}3 = (A + Blog

_{e}3)

Or, A + Blog

_{e}3 = 4log

_{e}3 ……….(5)

Now, (5) – (4) gives,

B(log

_{e}3 – log

_{e}2) = 4(log

_{e}3 – log

_{e}2)

=> B = 4

Putting B = 4 in (4) we get, A = 0

Thus the required solution of (1) is y = 4xe

^{-x}

So, C = 4

**Sanfoundry Global Education & Learning Series – Mathematics – Class 12**.

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