What is the solution for y^{''}+5y^'+4y=sin(t) ?

The solution for y^{''}+5y^'+4y=sin(t) is y=c_{1}e^{-t}+c_{2}e^{-4t}+3/34 sin(t)-5/34 cos(t)

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y^{''}+5y^'+4y=sin(t)

y^{^{''}} + 5 y^{^{'}} + 4 y = sin (t)

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{3}{34} sin (t) - \frac{5}{34} cos (t)

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = sin (t)

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{3}{34} sin (t) - \frac{5}{34} cos (t)

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{3}{34} sin (t) - \frac{5}{34} cos (t)

\frac{d}{d x} ((\frac{x}{1 + x})^{5})

- 3 t (2 t^{4} - 1)^{7}

f (x) = ln (8 - 9 x)

7 x^{2}, \frac{1}{7} x

\int \frac{3 x}{x ^{2} + 4} d x

What is the solution for y^{''}+5y^'+4y=sin(t) ?
The solution for y^{''}+5y^'+4y=sin(t) is y=c_{1}e^{-t}+c_{2}e^{-4t}+3/34 sin(t)-5/34 cos(t)