What is the solution for y^{''}+2y^'-3y=sin(t) ?

The solution for y^{''}+2y^'-3y=sin(t) is y=c_{1}e^t+c_{2}e^{-3t}-1/5 sin(t)-1/10 cos(t)

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y^{''}+2y^'-3y=sin(t)

y^{^{''}} + 2 y^{^{'}} - 3 y = sin (t)

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{1}{5} sin (t) - \frac{1}{10} cos (t)

Solution steps

y^{''} (t) + 2 y^{'} (t) - 3 y = sin (t)

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{1}{5} sin (t) - \frac{1}{10} cos (t)

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{1}{5} sin (t) - \frac{1}{10} cos (t)

(d^{2} - 2 d + 2) y = x^{2} e^{3} x

y^{^{''}} + y^{^{'}} - 2 y = 8 sin (2 x)

(d^{2} + 2 d + 1) y = cos^{2} (x)

(d^{2} + 3 d + 2) y = e - x + cos (x)

(d^{2} + 5 d + 6) y = e^{- 2} x sin^{2} (x)

What is the solution for y^{''}+2y^'-3y=sin(t) ?
The solution for y^{''}+2y^'-3y=sin(t) is y=c_{1}e^t+c_{2}e^{-3t}-1/5 sin(t)-1/10 cos(t)