Mathematics Questions and Answers – Linear Second Order Differential Equations

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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 = aex + 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 = aex + be-x ……(1)
Differentiating (1) with respect to x, we get
x(dy/dx) + y = aex + be-x …..(2)
Differentiating (2) now, with respect to x, we get
x(d2y/ dx) + dy/dx + dy/dx = aex + be-x
From (1),
aex + be-x = xy, so that we get
x(d2y/ dx) + 2(dy/dx) = xy
which is the required differential equation.
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2. What is thedifferential equation whose solution represents the family y = ae3x + bex?
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 = ae3x + bex ……….(1)
Differentiating the above equation, we get
dy/dx = 3ae3x + bex ……….(2)
d2y/dx2 = 9ae3x + bex ……….(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 = 3ae3x + 3bex ……….(4)
And -4(dy/dx) = -12ae3x – 4 bex ……….(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
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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) Ae4x – Be-x
c) Ae4x + Be-x
d) Ae4x + Bex
View Answer

Answer: c
Explanation: d2y/dx2 – 3dy/dx + 4y = 0 …..(1)
Let, y = emx be a trial solution of (1), then,
=> dy/dx = memx and d2y/dx2 = m2emx
Clearly, y = emx will satisfy equation (1). Hence, we have,
m2emx – 3m * emx – 4emx = 0
=>m2 – 3m – 4 = 0 (as emx ≠ 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 = Ae4x + 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 = e2x(12 cos3x – 5 sin3x)? (here, A and B are integration constant)
a) y = ex sin3x + Ax + B
b) y = e2x sin3x + Ax + B
c) y = e2x sin3x + A
d) Data inadequate
View Answer

Answer: b
Explanation: Given, d2y/dx2 = e2x(12 cos3x – 5 sin3x) ………(1)
Integrating (1) we get,
dy/dx = 12∫ e2xcos3x dx – 5∫ e2x sin3x dx
= 12 * (e2x/22 + 32)[2cos3x + 3sin3x] – 5 * (e2x/22 + 32)[2sin3x – 3cos3x] + A (A is integrationconstant)
So dy/dx = e2x/13 [24 cos3x + 36 sin3x – 10 sin3x + 15 cos3x] + A
= e2x/13(39 cos3x + 26 sin3x) + A
=> dy/dx = e2x(3 cos3x + 2 sin3x) + A ……….(2)
Again integrating (2) we get,
y = 3*∫ e2x cos3xdx + 2∫ e2x sin3xdx + A ∫dx
y = 3*(e2x/22 + 32)[2cos3x + 3sin3x] + 2*(e2x/22 + 32)[2sin3x – 3cos3x] + Ax + B (B is integration constant)
y = e2x/13(6 cos3x + 9 sin3x + 4 sin3x – 6 cos3x) + Ax + B
or, y = e2x sin3x + Ax + B

7. What is the solution of the given equation (D + 1)2y = 0 given y = 2 loge 2 when x = loge 2 and y = (4/3) loge3 when x = loge3?
a) y = 4xe-x
b) y = 4xex
c) y = -4xe-x
d) y = -4xex
View Answer

Answer: a
Explanation:(D + 1)2y = 0
Or, (D2 + 2D+ 1)y = 0
=> d2y/dx2 + 2dy/dx + y = 0 ……….(1)
Let y = emx be a trial solution of equation (1). Then,
=> dy/dx = memx and d2y/dx2 = m2emx
Clearly, y = emx will satisfy equation (1). Hence, we have
=> m2.emx + 2m.emx + emx = 0
Or, m2 + 2m +1 = 0 (as, emx ≠ 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 loge 2 when x = loge 2
Therefore, from (3) we get,
2 loge 2 = (A + B loge2)e-x
Or, 1/2(A + B loge2) = 2 log e2
Or, A + B loge2 = 4 loge2 ……….(4)
Again y = (4/3) loge3 when x = loge3
So, from (3) we get,
4/3 loge3 = (A + Bloge3)
Or, A + Bloge3 = 4loge3 ……….(5)
Now, (5) – (4) gives,
B(loge3 – loge2) = 4(loge3 – loge2)
=> B = 4
Putting B = 4 in (4) we get, A = 0
Thus the required solution of (1) is y = 4xe-x
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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
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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 loge 2 when x = loge 2 and y = (4/3) loge3 when x = loge3?
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 = emx be a trial solution of equation (1). Then,
=> dy/dx = memx and d2y/dx2 = m2emx
Clearly, y = emx will satisfy equation (1). Hence, we have
=> m2.emx + 2m.emx + emx = 0
Or, m2 + 2m + 1 = 0 (as, emx ≠ 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 loge 2 when x = loge 2
Therefore, from (3) we get,
2 loge 2 = (A + B loge2)e-x
Or, 1/2(A + B loge2) = 2 log e2
Or, A + B loge2 = 4 loge2 ……….(4)
Again y = (4/3) loge3 when x = loge3
So, from (3) we get,
4/3 loge3 = (A + Bloge3)
Or, A + Bloge3 = 4loge3 ……….(5)
Now, (5) – (4) gives,
B(loge3 – loge2) = 4(loge3 – loge2)
=> 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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter