What is the solution for y^{''}+y=-8e^{3t} ?

The solution for y^{''}+y=-8e^{3t} is y=c_{1}cos(t)+c_{2}sin(t)-4/5 e^{3t}

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y^{''}+y=-8e^{3t}

y^{^{''}} + y = - 8 e^{3 t}

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

Solution steps

y^{''} (t) + y = - 8 e^{3 t}

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} - 4 y = - 8 e^{t} (cos (2 t))

y^{^{''}} + y^{^{'}} - 12 y = - 32 t + 24 t^{2}

\frac{d ^{2} y}{d x ^{2}} + y = 3 + e^{- x} + 5 e^{2 x}

y^{^{''}} + 20 y^{^{'}} + 200 y = 320

y^{^{'''}} - 2 y^{^{'}} + 4 y = sin (2 x)

What is the solution for y^{''}+y=-8e^{3t} ?
The solution for y^{''}+y=-8e^{3t} is y=c_{1}cos(t)+c_{2}sin(t)-4/5 e^{3t}