Linear Second Order Differential Equations - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 9 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(d²y/ dx) + 2(dy/dx) = xy
b) x(d²y/ dx) – 2(dy/dx) = xy
c) 3x(d²y/ dx) + 2(dy/dx) = xy
d) x(d²y/ dx) + 2(dy/dx) = 2xy
View Answer

Answer: a
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(d²y/ dx) + dy/dx + dy/dx = ae^x + be^-x
From (1),
ae^x + be^-x = xy, so that we get
x(d²y/ 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) d²y/dx² – 3dy/dx + 4y = 0
b) d²y/dx² – 4dy/dx + 3y = 0
c) d²y/dx² + 4dy/dx + 3y = 0
d) d²y/dx² – 4dy/dx – 3y = 0
View Answer

Answer: b
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)
d²y/dx² = 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,
d²y/dx² – 4dy/dx + 3y = 0

3. What is the differential equation of all parabolas whose directrices are parallel to the x-axis?
a) d³x/dy³ = 0
b) d³y/(dx³ + d²y/dx²) = 0
c) d³y/dx³ = 0
d) d²y/dx² = 0
View Answer

Answer: c
Explanation: The equation of family of parabolas is Ax² + 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, d³y/dx³ = 0

4. If y = t(x) be a differentiable function ᵾ x € R, then which of the following is always true?
a) d²y/dx² – (dy/dx)³ = 0
b) d²y/dx² + (dy/dx)³ d²x/dy² = 0
c) d²y/dx² – (dx/dy)³ = 0
d) d²y/dx² + (dy/dx)³ = 0
View Answer

Answer: b
Explanation: (dy/dx)= (dx/dy)^-1
So, d²y/dx² = – (dx/dy)^-2 d/dx(dx/dy)
= – (dy/dx)²(d²x/dy²)(dy/dx)
=> d²y/dx² + (dy/dx)³ d²x/dy² = 0

5. What will be the required solution of d²y/dx² – 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

Answer: c
Explanation: d²y/dx² – 3dy/dx + 4y = 0 …..(1)
Let, y = e^mx be a trial solution of (1), then,
=> dy/dx = me^mx and d²y/dx² = m²e^mx
Clearly, y = e^mx will satisfy equation (1). Hence, we have,
m²e^mx – 3m * e^mx – 4e^mx = 0
=>m² – 3m – 4 = 0 (as e^mx ≠ 0) …….(2)
=> m² – 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.

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6. What will be the general solution of the differential equation d²y/dx² = 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

Answer: b
Explanation: Given, d²y/dx² = e^2x(12 cos3x – 5 sin3x) ………(1)
Integrating (1) we get,
dy/dx = 12∫ e^2xcos3x dx – 5∫ e^2x sin3x dx
= 12 * (e^2x/2² + 3²)[2cos3x + 3sin3x] – 5 * (e^2x/2² + 3²)[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/2² + 3²)[2cos3x + 3sin3x] + 2*(e^2x/2² + 3²)[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)²y = 0 given y = 2 log_e 2 when x = log_e 2 and y = (4/3) log_e3 when x = log_e3?
a) y = 4xe^-x
b) y = 4xe^x
c) y = -4xe^-x
d) y = -4xe^x
View Answer

Answer: a
Explanation:(D + 1)²y = 0
Or, (D² + 2D+ 1)y = 0
=> d²y/dx² + 2dy/dx + y = 0 ……….(1)
Let y = e^mx be a trial solution of equation (1). Then,
=> dy/dx = me^mx and d²y/dx² = m²e^mx
Clearly, y = e^mx will satisfy equation (1). Hence, we have
=> m².e^mx + 2m.e^mx + e^mx = 0
Or, m² + 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_e2)e^-x
Or, 1/2(A + B log_e2) = 2 log _e2
Or, A + B log_e2 = 4 log_e2 ……….(4)
Again y = (4/3) log_e3 when x = log_e3
So, from (3) we get,
4/3 log_e3 = (A + Blog_e3)
Or, A + Blog_e3 = 4log_e3 ……….(5)
Now, (5) – (4) gives,
B(log_e3 – log_e2) = 4(log_e3 – log_e2)
=> 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) d²x/dt² – αx = 0
b) d²x/dt² + αx = 0
c) d²x/dt² – α²x = 0
d) d²x/dt² + α²x = 0
View Answer

Answer: d
Explanation: Since, x = a cos(αt + β)
Therefore, dx/dt = a cos(αt + β)
And, d²x/dt² = -a α² cos(αt + β)
= -α² a cos(αt + β)
Or, d²x/dt² = -α² [as a cos(αt + β) = x]
So, d²x/dt² + α²x = 0

9. If, A and B are arbitrary constants then what will be the differential equation of y = Ax + B/x?
a) x² d² y/dx² – xdy/dx + y = 0
b) x² d² y/dx² + xdy/dx + y = 0
c) x² d² y/dx² + xdy/dx – y = 0
d) x² d² y/dx² – xdy/dx – y = 0
View Answer

Answer: c
Explanation: Given, y = Ax + B/x
=> xy = Ax² + B ……….(1)
Differentiating (1) with respect to x, we get,
d(xy)/dx = d/dx(Ax² + B)
or, xdy/dx + y = A * 2x ……….(2)
Differentiating again with respect to x, we get,
x*d²y/dx² + dy/dx + dy/dx = A*2 ……….(3)
Eliminating A from (2) and (3) we get,
x² d² y/dx² + 2xdy/dx = 2Ax [multiplying (3) by x]
or, x² d² y/dx² + 2xdy/dx = xdy/dx + y [using (2)]
or, x² d² y/dx² + 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)²y = 0 given y = 2 log_e 2 when x = log_e 2 and y = (4/3) log_e3 when x = log_e3?
a) 2
b) -2
c) -4
d) 4
View Answer

Answer: c
Explanation: (D + 1)²y = 0
Or, (D² + 2D+ 1)y = 0
=> d²y/dx² + 2dy/dx + y = 0 ……….(1)
Let y = e^mx be a trial solution of equation (1). Then,
=> dy/dx = me^mx and d²y/dx² = m²e^mx
Clearly, y = e^mx will satisfy equation (1). Hence, we have
=> m².e^mx + 2m.e^mx + e^mx = 0
Or, m² + 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_e2)e^-x
Or, 1/2(A + B log_e2) = 2 log _e2
Or, A + B log_e2 = 4 log_e2 ……….(4)
Again y = (4/3) log_e3 when x = log_e3
So, from (3) we get,
4/3 log_e3 = (A + Blog_e3)
Or, A + Blog_e3 = 4log_e3 ……….(5)
Now, (5) – (4) gives,
B(log_e3 – log_e2) = 4(log_e3 – log_e2)
=> B = 4
Putting B = 4 in (4) we get, A = 0
Thus the required solution of (1) is y = 4xe^-x
So, C = 4

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