250+ TOP MCQs on Method of Undetermined Coefficients and Answers

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
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
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
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 )
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})
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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