What is the solution for y^{''}+y=9e^{-4x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=9e^{-4x},y(0)=0,y^'(0)=0 is y=-9/17 cos(x)+36/17 sin(x)+9/17 e^{-4x}

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y^{''}+y=9e^{-4x},y(0)=0,y^'(0)=0

y^{^{''}} + y = 9 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{9}{17} cos (x) + \frac{36}{17} sin (x) + \frac{9}{17} e^{- 4 x}

Solution steps

y^{''} (x) + y = 9 e^{- 4 x}

Solve linear ODE:

y = - \frac{9}{17} cos (x) + \frac{36}{17} sin (x) + \frac{9}{17} e^{- 4 x}

y = - \frac{9}{17} cos (x) + \frac{36}{17} sin (x) + \frac{9}{17} e^{- 4 x}

\frac{d}{d x} (c e^{- 2 x^{2}})

\frac{d}{d x} (sin (x) os (x))

\int csc (3 x) d x

y^{^{'}} = y - y^{2}, y (0) = 9

\frac{2}{( s + 2 ) ( s + 3 ) ^{2}}

What is the solution for y^{''}+y=9e^{-4x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=9e^{-4x},y(0)=0,y^'(0)=0 is y=-9/17 cos(x)+36/17 sin(x)+9/17 e^{-4x}