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

The solution for y^{''}+9y=xe^x is y=c_{1}cos(3x)+c_{2}sin(3x)+(e^xx)/(10)-1/50 e^x

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

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

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{e ^{x} x}{10} - \frac{1}{50} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{e ^{x} x}{10} - \frac{1}{50} e^{x}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{e ^{x} x}{10} - \frac{1}{50} e^{x}

\int_{1200}^{1201} 0.1

\frac{d}{d x} (- 5 ln (cos (x)))

\int_{- 5}^{5} \frac{1}{v} d v

\int e^{3 x} \cdot e^{3 x} - 4 d x

t e^{- t} cos (5 t)

What is the solution for y^{''}+9y=xe^x ?
The solution for y^{''}+9y=xe^x is y=c_{1}cos(3x)+c_{2}sin(3x)+(e^xx)/(10)-1/50 e^x