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

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

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

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

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

Solution steps

y^{''} (t) + 4 y = e^{t}

Solve linear ODE:

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

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

x^{^{''}} + 4 x^{^{'}} + 4 x = 4 t^{2} + 6 e^{t}

y^{^{''}} - y = 2 x^{2} - x - 3

\frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} + 16 y = 16 (x - 1)

y^{^{''}} + 9 y = 9 sec^{2} (3 x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 8 e^{- t} cos (2 t)

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