What is the solution for y^{''}+9y=5e^x-162x ?

The solution for y^{''}+9y=5e^x-162x is y=c_{1}cos(3x)+c_{2}sin(3x)+1/2 e^x-18x

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y^{''}+9y=5e^x-162x

y^{^{''}} + 9 y = 5 e^{x} - 162 x

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

Solution steps

y^{''} (x) + 9 y = 5 e^{x} - 162 x

Solve linear ODE:

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

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

\int - \frac{x ^{3}}{3} - x^{2} + \frac{4}{3} d x

\frac{d}{d x} (2 arcsin (\frac{x}{2}))

\frac{d y}{d x} = 7 x 5 y + 11, y (0) = 9

\frac{d y}{d x} - x y = 0

What is the solution for y^{''}+9y=5e^x-162x ?
The solution for y^{''}+9y=5e^x-162x is y=c_{1}cos(3x)+c_{2}sin(3x)+1/2 e^x-18x