What is the solution for y^{''}-2y^'+5y=2cos(2t)e^t ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = \frac{2 x}{3 - cot ( x )}

(e)^{^{'}}

\int_{0}^{\infty} \frac{1}{x} d x

x + e^{2 x}

(0, 0), (5, 1)

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