What is the solution for y^{''}+9y=sin(3x)+e^{-x} ?

The solution for y^{''}+9y=sin(3x)+e^{-x} is y=c_{1}cos(3x)+c_{2}sin(3x)-(xcos(3x))/6+1/10 e^{-x}

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

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

y = c_{1} cos (3 x) + c_{2} sin (3 x) - \frac{x cos ( 3 x )}{6} + \frac{1}{10} e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) - \frac{x cos ( 3 x )}{6} + \frac{1}{10} e^{- x}

y = c_{1} cos (3 x) + c_{2} sin (3 x) - \frac{x cos ( 3 x )}{6} + \frac{1}{10} e^{- x}

y^{^{''}} - y^{^{'}} - 2 y = cosh (2 x)

(D^{2} - 2 D + 1) y = x^{2} e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 x + 3

(d^{2} - 1) y = x sin (x) + (1 + x^{2}) e^{x}

y^{^{''}} + 4 y = sec^{3} (2 x)

What is the solution for y^{''}+9y=sin(3x)+e^{-x} ?
The solution for y^{''}+9y=sin(3x)+e^{-x} is y=c_{1}cos(3x)+c_{2}sin(3x)-(xcos(3x))/6+1/10 e^{-x}