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.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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