What is the solution for y^{''}+9y=3xe^{2x} ?

The solution for y^{''}+9y=3xe^{2x} is y=c_{1}cos(3x)+c_{2}sin(3x)+(3e^{2x}x)/(13)-12/169 e^{2x}

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y^{''}+9y=3xe^{2x}

y^{^{''}} + 9 y = 3 x e^{2 x}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{3 e ^{2 x} x}{13} - \frac{12}{169} e^{2 x}

Solution steps

y^{''} (x) + 9 y = 3 x e^{2 x}

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{3 e ^{2 x} x}{13} - \frac{12}{169} e^{2 x}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{3 e ^{2 x} x}{13} - \frac{12}{169} e^{2 x}

t \to 0 lim (\frac{e ^{5 t} - 1}{t})

\int f^{- 1} (x) d x

4 cos (\frac{2 π}{3})

cos (a^{3} + x^{3})

\frac{d}{d x} (f (x) (5 x^{2} - 4))

What is the solution for y^{''}+9y=3xe^{2x} ?
The solution for y^{''}+9y=3xe^{2x} is y=c_{1}cos(3x)+c_{2}sin(3x)+(3e^{2x}x)/(13)-12/169 e^{2x}