What is the solution for y^{''}+2ay^'+b^2y=sin(ct) ?

The solution for y^{''}+2ay^'+b^2y=sin(ct) is

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y^{''}+2ay^'+b^2y=sin(ct)

y^{^{''}} + 2 a y^{^{'}} + b^{2} y = sin (c t)

y = c_{1} e^{(- a + a^{2} - b^{2}) t} + c_{2} e^{(- a - a^{2} - b^{2}) t} + \frac{sin ( c t ) ( - c ^{2} + b ^{2} )}{4 c ^{2} a ^{2} + c ^{4} - 2 c ^{2} b ^{2} + b ^{4}} - \frac{2 c a cos ( c t )}{4 c ^{2} a ^{2} + c ^{4} - 2 c ^{2} b ^{2} + b ^{4}}

Solution steps

y^{''} (t) + 2 a y^{'} (t) + b^{2} y = sin (c t)

Solve linear ODE:

y = c_{1} e^{(- a + a^{2} - b^{2}) t} + c_{2} e^{(- a - a^{2} - b^{2}) t} + \frac{sin ( c t ) ( - c ^{2} + b ^{2} )}{4 c ^{2} a ^{2} + c ^{4} - 2 c ^{2} b ^{2} + b ^{4}} - \frac{2 c a cos ( c t )}{4 c ^{2} a ^{2} + c ^{4} - 2 c ^{2} b ^{2} + b ^{4}}

y = c_{1} e^{(- a + a^{2} - b^{2}) t} + c_{2} e^{(- a - a^{2} - b^{2}) t} + \frac{sin ( c t ) ( - c ^{2} + b ^{2} )}{4 c ^{2} a ^{2} + c ^{4} - 2 c ^{2} b ^{2} + b ^{4}} - \frac{2 c a cos ( c t )}{4 c ^{2} a ^{2} + c ^{4} - 2 c ^{2} b ^{2} + b ^{4}}

x^{^{''}} + 10 x^{^{'}} + 16 x = 2581 cos (5 t)

(D^{4} + 10 D^{2} + 9) y = sin (3 x) + cos (3 x)

y^{^{''}} + 36 y = 2 e^{6 x}

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 e^{4 t}

d (x) = \frac{3}{2 x}

What is the solution for y^{''}+2ay^'+b^2y=sin(ct) ?
The solution for y^{''}+2ay^'+b^2y=sin(ct) is