What is the solution for y^{''}-y^'-2y=sin(t) ?

The solution for y^{''}-y^'-2y=sin(t) is y=c_{1}e^{2t}+c_{2}e^{-t}-3/10 sin(t)+1/10 cos(t)

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y^{''}-y^'-2y=sin(t)

y^{^{''}} - y^{^{'}} - 2 y = sin (t)

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{3}{10} sin (t) + \frac{1}{10} cos (t)

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = sin (t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{3}{10} sin (t) + \frac{1}{10} cos (t)

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{3}{10} sin (t) + \frac{1}{10} cos (t)

x^{^{''}} + 14 x^{^{'}} + 98 x = 1960

y^{^{''}} + 4 y = sin (x), y (0) = 1, y^{^{'}} (0) = 2

x^{^{''}} + 4 x = 289 + e^{t} \cdot sin (2 t)

y^{^{''}} - 10 y^{^{'}} + 125 y = - 4 sin (3 t)

\frac{d ^{2} y}{d x ^{2}} + y = 2 x sin (x)

What is the solution for y^{''}-y^'-2y=sin(t) ?
The solution for y^{''}-y^'-2y=sin(t) is y=c_{1}e^{2t}+c_{2}e^{-t}-3/10 sin(t)+1/10 cos(t)