Ordinary Differential Equations Interview Questions and Answers for Freshers

This set of Ordinary Differential Equations Interview Questions and Answers for freshers focuses on “Method of Undetermined Coefficients”.

1. Solution of the D.E y’’ – 4y’ + 4y = e^x when solved using method of undetermined coefficients is _____
a) y = (c₁ + c₂)e^2x + 2e^x – 1
b) y = (c₁ + c₂ x)e^2x + 4e^x – 4
c) y = (c₁ + c₂ x)e^2x + e^x
d) y = (c₁ + c₂ x)e^x + 4e^x
View Answer

Answer: c
Explanation: We have (D² – 4D + 4)y = e^x
A.E is m² – 4m + 4 = 0 –> (m – 2)² = 0 –> m = 2,2
thus y_c = (c₁ + c₂ x) e^2x, ∅(x) = e^x and 1 is not a root of the A.E
we assume P.I in the form y_p = ae^x…(1) to find a such that y_p’’ – 4y_p’ + 4y_p = e^x….(2)
y_p’ = ae^x and y_p’ = ae^x now (2) becomes ae^x – 4ae^x + 4ae^x = e^x –> a = 1
substituting the value of a in (1) we get y_p = e^x
thus the solution is y = y_c + y_p –> y = (c₁ + c₂ x) e^2x + e^x.

2. Solution of the D.E y’’ + 3y’ + 2y = 12x² when solved using the method of undetermined coefficients is ________
a) y = c₁ e^x + c₂ e^2x + 2 – 11x + x²
b) y = c₁ e^{– x} + c₂ e^{– 2x} + 18 + 21x + 3x²
c) y = c₁ e^x + c₂ e^{– 2x} + 11 + 18x + 2x²
d) y = c₁ e^{– x} + c₂ e^{– 2x} + 21 – 18x + 6x²
View Answer

Answer: d
Explanation: We have (D² + 3D + 2)y = 12x²
A.E is m² + 3m + 2 = 0 –> (m + 1)(m + 2) = 0 –> m = – 1, – 2
y_c = c₁ e^{– x} + c₂ e^{– 2x} and ∅(x) = 12x² and 0 is not a root of the A.E,
we assume P.I in the form y_p = a + bx + cx²….(1)to find a,b & c such that
y_p’’ + 3y_p’ + 2y_p = 12x²….(2), y_p‘ = b + 2cx, y_p” = 2c now (2) becomes
2c + 3(b + 2cx) + 2(a + bx + cx²) = 12x²
(2a + 3b + 2c) + (2b + 6c)x + (2c)x² = 12x²
2a + 3b + 2c = 0, 2b + 6c = 0, 2c = 12 –> c = 6, b = – 18, a = 21 hence (1) becomes
y_p = 21 – 18x + 6x² thus complete solution is
y = y_c + y_p –> c₁ e^{– x} + c₂ e^{– 2x} + 21 – 18x + 6x².

3. Find the Particular integral solution of the D.E (D² – 4D + 3)y = 20 cos x by the method of undetermined coefficients.
a) y_p = 4 cos⁡x – 3 sin⁡x
b) y_p = 2 sin⁡x – 4 cos⁡x
c) y_p = – 3 cos⁡x + 4 sin⁡x
d) y_p = 2 cos⁡x – 4 sin⁡x
View Answer

Answer: d
Explanation: ∅(x) = 20 cos x,we assume P.I in the form y_p = a cos⁡x + b sin⁡x ….(1)
and since A.E has m = 1,3 as roots,∓i are not roots of A.E.we have to find a and b
such that y_p” – 4y_p‘ + 3y_p = 20 cos⁡x……(2)
from (1) we get y_p‘ = – a sin⁡x + b cos⁡x, y_p” = – a cos⁡x – b sin⁡x, (2) becomes
– a cos⁡x – b sin⁡x – 4( – a sin⁡x + b cos⁡x) + 3(a cos⁡x + b sin⁡x) = 20 cos x
(2a – b)cos x + (4a + 2b)sin x = 20 cos x –> 2a – b = 20 and 4a + 2b = 0, by solving we get
a = 2, b = – 4 from (1) y_p = 2 cos⁡x – 4 sin⁡x is the particular integral solution.

4. Using the method of undetermined coefficients find the P.I for the D.E x’’’(t) – x’’(t) = 3e^t + sin⁡t.
a) x_p = 3e^t + \(\frac{t}{2}\) (cos⁡t – 2 sin⁡t )
b) x_p = 3te^t + \(\frac{1}{2}\) (cos⁡t + sin⁡t )
c) x_p = 3te^t + \(\frac{t}{3}\) (4cos⁡t + 2sin⁡t )
d) x_p = 3e^t + \(\frac{1}{2}\) (cos⁡t – sin⁡t )
View Answer

Answer: b
Explanation: We have (D³ – D²)x(t) = 3e^t + sin⁡t, where D = d/dt, A.E is m³ – m² = 0
m² (m – 1) = 0 –> m = 0, 0, 1 –> x_c (t) = (c₁ + c₂ t) + c₃ e^t
∅(t) = 3e^t + sin⁡t we assume for P.I in the form x_p = ate^t + b cos⁡t + c sin⁡t …(1)
since 1 is a root and ∓i are not a roots of the A.E. To find a, b& c such that
x_p’’’(t) – x_p’’(t) = 3e^t + sin⁡t……(2)
from (1) we have x_p‘ = a(te^t + e^t) – b sin⁡t + c cos⁡t
x_p” = a(te^t + 2e^t) – b cos⁡t – c sin⁡t
x_p”’ = a(te^t + 3e^t) + b sin⁡t – c cos⁡t,now (2) becomes
ate^t + 3ae^t + b sin⁡t – c cos⁡t, – ate^t – 2ae^t + b cos⁡t + c sin⁡t = 3e^t + sin⁡t
ae^t + (b + c) sin⁡t + (b – c) cos⁡t = 3e^t + sin⁡t
– – > a = 3 and b + c = 1, b – c = 0 –> a = 3 and b = 1/2, c = 1/2
hence from (1) x_p = 3te^t + 1/2 (cos⁡t + sin⁡t) is the particular integral solution.

5. What is the solution of D.E (D² – 2D)y = e^x sin⁡x when solved using the method of undetermined coefficients?
a) \( y = c_1 + c_2 \,e^{2x} – \frac{e^x (xsin x + cos⁡x)}{2}\)
b) \(y = c_1 + c_2 \,e^{2x} – \frac{e^x sin x}{2}\)
c) \(y = c_1 + c_2 \,e^{2x} – \frac{e^x cos x}{2}\)
d) \(y = c_1 + c_2 \,e^{2x} – \frac{e^x (sin x + x cos⁡x)}{4}\)
View Answer

Answer: b
Explanation: A.E is m² – 2m = 0 or m(m – 2) = 0 –> m = 0,2
y_c = c₁ + c₂ e^2x and ∅(x) = e^x sin⁡x. we assume PI in the form
y_p = e^x (a cos x + b sin x)….(1) since 1±i are not roots of the A.E.
we have to find a, b such that y_p” – 2y_p‘ = e^x sin⁡x…..(2)
from (1) y_p‘ = e^x (- a sin x + b cos x) + e^x (a cos x + b sin x)
y_p” = e^x (- a sin x + b cos x) + e^x (a cos x + b sin x) + e^x (- a cos x – b sin x) + e^x (- a sin x + b cos x) = 2e^x (- a sin⁡x + b cos⁡x) hence (2) becomes
2e^x (- a sin⁡x + b cos⁡x) – 2e^x (- a sin⁡x + b cos⁡x)
– 2e^x (a cos⁡x + b sin⁡x) = e^x sin⁡x i.e – 2ae^x cos⁡x – 2be^x sin⁡x = e^x sin⁡x
–> – 2a = 0, – 2b = 1 –> a = 0, b = – 1/2 hence (1) becomes \(y_p = \frac{-e^x sin x}{2}\)
y = y_c + y_p = c₁ + c₂ e^2x – \(\frac{e^x sin x}{2}\)
.

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