What is the solution for y^{''}+4y=8sin(2t)+5e^t ?

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

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y^{''}+4y=8sin(2t)+5e^t

y^{^{''}} + 4 y = 8 sin (2 t) + 5 e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 2 t cos (2 t) + e^{t}

Solution steps

y^{''} (t) + 4 y = 8 sin (2 t) + 5 e^{t}

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 2 t cos (2 t) + e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 2 t cos (2 t) + e^{t}

y^{^{''}} - 8 y^{^{'}} + 16 y = 45 e^{7 t}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = 6 x^{2} + 2 x - 2

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 e^{4 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{x} cos (x)

y^{^{''}} + 4 y^{^{'}} + 13 y = 5 cos (2 t)

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