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

The solution for y^{''}+4y=te^t is y=c_{1}cos(2t)+c_{2}sin(2t)+(e^tt)/5-2/25 e^t

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{e ^{t} t}{5} - \frac{2}{25} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{e ^{t} t}{5} - \frac{2}{25} e^{t}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{e ^{t} t}{5} - \frac{2}{25} e^{t}

x \to 2 - lim (\frac{4 x + 1}{x - 2})

\int x^{2} arccos (x) d x

\int 6 sin^{6} (x) cos^{3} (x) d x

x \to 4 lim (\frac{4}{x - 4})

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What is the solution for y^{''}+4y=te^t ?
The solution for y^{''}+4y=te^t is y=c_{1}cos(2t)+c_{2}sin(2t)+(e^tt)/5-2/25 e^t