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 form
y_p = e^x (a cos x + b sin x)?.(1) since 1�i are not roots of 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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\(m \frac{d^2 y}{dt^2} + ky = A \,sin \,?t\, \,or\, \frac{d^2 y}{dt^2} + \frac{k}{m} y = \frac{A}{m} sin ?t \rightarrow \frac{d^2 y}{dt^2} + ?^2 y = ? sin ?t\)
A.E is m² + ?² = 0 –> m = �?i –> y = c₁ cos??t + c₂ sin??t at t=0 y=y₀
–> c₁ = y₀ refore y_c = y₀ cos??t + c₂ sin??t
now differentiate and substituting y?(0)=y₁
y’ = -y₀ ?sin??t + ?c₂ cos??t at t = 0 y₁/? = c₂ and y_c = y₀ cos??t + (y₁/?)sin??t
to find particular solution \(y_p = \frac{? sin ?t}{D^2+?^2}\) assuming ? as constant \(y_p = \frac{? sin ?t}{-?^2+?^2}\)
?y=y₀ cos??t + (y₁/?)sin??t + \(\frac{? sin ?t}{-?^2+?^2}\).

?(x) = 20 cos x,we assume P.I in 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 particular integral solution.

x” (t) + 25x(t) = 21 cos?t comparing it with \( \frac{d^2 y}{dt^2} + \frac{k}{m} y= \frac{A}{m} sin ?t\)
\( \frac{d^2 y}{dt^2} + ?^2 y = ? sin \,?t\) where ?²=k/m, ?=A/m and its solution is
y=y₀ cos??t + (y₁/?)sin??t + \( \frac{? sin ?t}{-?^2+?^2}\) where y(0)=y₀ and y?(0)=y₁.
we got ?² = 25 –> ?=5 and ?=21, ?=1. Its solution is
x = x₀ cos?5t + (x₁/5)sin?5t + \(\frac{21 sin t}{-1^2+25}\) particle is at rest at t = 0 –> x(0) = 0 = x₀
and x?(0) = 0 = x₁ refore \(x = \frac{21 sin t}{24} = \frac{7 sin t}{8}.\)

The governing D.E is given by \(m \frac{d^2 y}{dt^2} + c \frac{dy}{dt} + ky=0;\) c>0
–> \( \frac{d^2 y}{dt^2} + \frac{c}{m} \frac{dy}{dt} + \frac{k}{m}y = 0\) given c/m=2?, k/m=?² thus D.E takes form
\( \frac{d^2 y}{dt^2} + 2? d\frac{dy}{dt} + ?^2 y = 0\) it?s A.E is m² + 2?m+ ?² = 0
\(m=\frac{-2?�\sqrt{4?^2-4?^2}}{2} = – ?�vi\)?. given \(v = \sqrt{?^2-?^2}\)
y_c = y = e^{– ?t} (c₁ cos?vt + c₂ sin?vt ) to find constants we use initial condition
at t=0 y = y₀ –> c₁ = y₀ refore y = e^-?t (y₀ cos?vt + c₂ sin?vt)
now differentiate and substituting y?(0)=y₁
y? = – ?e^{– ?t} (y₀ cos?vt + c₂ sin?vt) + e^{– ?t} (-y₀ vsin?vt + vc₂ cos?vt)
at t=0, y₁ = – ?y₀ + vc₂ –> c₂ = \(\left(\frac{y_1+ ?y_0}{v}\right)\) thus y=e^{– ?t} (y₀ cos?vt+\(\left(\frac{y_1+ ?y_0}{v}\right)\) sin?vt).

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 A.E,
we assume P.I in 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.

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 form x_p = ate^t + b cos?t + c sin?t ?(1)
since 1 is a root and ?i are not a roots of 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 particular integral solution.