What is the solution for y^{''}+9y=te^{2t} ?

The solution for y^{''}+9y=te^{2t} is y=c_{1}cos(3t)+c_{2}sin(3t)+(e^{2t}t)/(13)-4/169 e^{2t}

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y^{''}+9y=te^{2t}

y^{^{''}} + 9 y = t e^{2 t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{2 t} t}{13} - \frac{4}{169} e^{2 t}

Solution steps

y^{''} (t) + 9 y = t e^{2 t}

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{2 t} t}{13} - \frac{4}{169} e^{2 t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{2 t} t}{13} - \frac{4}{169} e^{2 t}

\frac{d}{d x} (x^{9} ln (4 x))

\frac{d}{d t} (3)

f (x) = ln (4)

x d x + (x^{2} y + 4 y) d y = 0, y (4) = 0

What is the solution for y^{''}+9y=te^{2t} ?
The solution for y^{''}+9y=te^{2t} is y=c_{1}cos(3t)+c_{2}sin(3t)+(e^{2t}t)/(13)-4/169 e^{2t}