What is the solution for y^{''}+9y=te^{3t} ?

The solution for y^{''}+9y=te^{3t} is y=c_{1}cos(3t)+c_{2}sin(3t)+(e^{3t}t)/(18)-1/54 e^{3t}

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y^{''}+9y=te^{3t}

y^{^{''}} + 9 y = t e^{3 t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{3 t} t}{18} - \frac{1}{54} e^{3 t}

Solution steps

y^{''} (t) + 9 y = t e^{3 t}

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{3 t} t}{18} - \frac{1}{54} e^{3 t}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{3 t} t}{18} - \frac{1}{54} e^{3 t}

y^{^{'}} - y = 8 e^{x} + 18 e^{4 x}

\int 4 sec^{2} (x) tan (x) d x

f (x) = x^{3} - 2

\frac{d}{d x} (lo g_{4} (x^{2} - 7 x))

What is the solution for y^{''}+9y=te^{3t} ?
The solution for y^{''}+9y=te^{3t} is y=c_{1}cos(3t)+c_{2}sin(3t)+(e^{3t}t)/(18)-1/54 e^{3t}