# Ordinary Differential Equations Questions and Answers – Method of Undetermined Coefficients

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 = ex when solved using method of undetermined coefficients is _____
a) y = (c1 + c2)e2x + 2ex – 1
b) y = (c1 + c2 x)e2x + 4ex – 4
c) y = (c1 + c2 x)e2x + ex
d) y = (c1 + c2 x)ex + 4ex

Explanation: We have (D2 – 4D + 4)y = ex
A.E is m2 – 4m + 4 = 0 –> (m – 2)2 = 0 –> m = 2,2
thus yc = (c1 + c2 x) e2x, ∅(x) = ex and 1 is not a root of the A.E
we assume P.I in the form yp = aex…(1) to find a such that yp’’ – 4yp’ + 4yp = ex….(2)
yp’ = aex and yp’ = aex now (2) becomes aex – 4aex + 4aex = ex –> a = 1
substituting the value of a in (1) we get yp = ex
thus the solution is y = yc + yp –> y = (c1 + c2 x) e2x + ex.

2. Solution of the D.E y’’ + 3y’ + 2y = 12x2 when solved using the method of undetermined coefficients is ________
a) y = c1 ex + c2 e2x + 2 – 11x + x2
b) y = c1 e – x + c2 e – 2x + 18 + 21x + 3x2
c) y = c1 ex + c2 e – 2x + 11 + 18x + 2x2
d) y = c1 e – x + c2 e – 2x + 21 – 18x + 6x2

Explanation: We have (D2 + 3D + 2)y = 12x2
A.E is m2 + 3m + 2 = 0 –> (m + 1)(m + 2) = 0 –> m = – 1, – 2
yc = 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 yp = a + bx + cx2….(1)to find a,b & c such that
yp’’ + 3yp’ + 2yp = 12x2….(2), yp‘ = b + 2cx, yp” = 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
yp = 21 – 18x + 6x2 thus complete solution is
y = yc + yp –> 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) yp = 4 cos⁡x – 3 sin⁡x
b) yp = 2 sin⁡x – 4 cos⁡x
c) yp = – 3 cos⁡x + 4 sin⁡x
d) yp = 2 cos⁡x – 4 sin⁡x

Explanation: ∅(x) = 20 cos x,we assume P.I in the form yp = 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 yp” – 4yp‘ + 3yp = 20 cos⁡x……(2)
from (1) we get yp‘ = – a sin⁡x + b cos⁡x, yp” = – 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) yp = 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) = 3et + sin⁡t.
a) xp = 3et + $$\frac{t}{2}$$ (cos⁡t – 2 sin⁡t )
b) xp = 3tet + $$\frac{1}{2}$$ (cos⁡t + sin⁡t )
c) xp = 3tet + $$\frac{t}{3}$$ (4cos⁡t + 2sin⁡t )
d) xp = 3et + $$\frac{1}{2}$$ (cos⁡t – sin⁡t )

Explanation: We have (D3 – D2)x(t) = 3et + sin⁡t, where D = d/dt, A.E is m3 – m2 = 0
m2 (m – 1) = 0 –> m = 0, 0, 1 –> xc (t) = (c1 + c2 t) + c3 et
∅(t) = 3et + sin⁡t we assume for P.I in the form xp = atet + 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
xp’’’(t) – xp’’(t) = 3et + sin⁡t……(2)
from (1) we have xp‘ = a(tet + et) – b sin⁡t + c cos⁡t
xp” = a(tet + 2et) – b cos⁡t – c sin⁡t
xp”’ = a(tet + 3et) + b sin⁡t – c cos⁡t,now (2) becomes
atet + 3aet + b sin⁡t – c cos⁡t, – atet – 2aet + b cos⁡t + c sin⁡t = 3et + sin⁡t
aet + (b + c) sin⁡t + (b – c) cos⁡t = 3et + sin⁡t
– – > a = 3 and b + c = 1, b – c = 0 –> a = 3 and b = 1/2, c = 1/2
hence from (1) xp = 3tet + 1/2 (cos⁡t + sin⁡t) is the particular integral solution.

5. What is the solution of D.E (D2 – 2D)y = ex 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) $$y = c_1 + c_2 \,e^{2x} – \frac{e^x (sin x + x cos⁡x)}{4}$$

Explanation: A.E is m2 – 2m = 0 or m(m – 2) = 0 –> m = 0,2
yc = c1 + c2 e2x and ∅(x) = ex sin⁡x. we assume PI in the form
yp = ex (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 yp” – 2yp‘ = ex sin⁡x…..(2)
from (1) yp‘ = ex (- a sin x + b cos x) + ex (a cos x + b sin x)
yp” = ex (- a sin x + b cos x) + ex (a cos x + b sin x) + ex (- a cos x – b sin x) + ex (- a sin x + b cos x) = 2ex (- a sin⁡x + b cos⁡x) hence (2) becomes
2ex (- a sin⁡x + b cos⁡x) – 2ex (- a sin⁡x + b cos⁡x)
– 2ex (a cos⁡x + b sin⁡x) = ex sin⁡x i.e – 2aex cos⁡x – 2bex sin⁡x = ex sin⁡x
–> – 2a = 0, – 2b = 1 –> a = 0, b = – 1/2 hence (1) becomes $$y_p = \frac{-e^x sin x}{2}$$
y = yc + yp = c1 + c2 e2x – $$\frac{e^x sin x}{2}$$
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Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.

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