What is the solution for y^{''}+4y^'+5y=15t ?

The solution for y^{''}+4y^'+5y=15t is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+3t-12/5

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y^{''}+4y^'+5y=15t

y^{^{''}} + 4 y^{^{'}} + 5 y = 15 t

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 3 t - \frac{12}{5}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 15 t

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 3 t - \frac{12}{5}

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + 3 t - \frac{12}{5}

y^{^{''}} - y^{^{'}} = 5 sin (2 t), y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} + 9 y = sin (3 x) sec^{2} (3 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = sin (t) - cos (2 t)

y^{^{''}} + 7 y^{^{'}} + 6 y = e^{2 t}

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

What is the solution for y^{''}+4y^'+5y=15t ?
The solution for y^{''}+4y^'+5y=15t is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+3t-12/5