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}

x^{^{''}} + 2 x^{^{'}} + 5 x = 10 cos (3 t)

y^{^{''}} - y^{^{'}} = 1, y (0) = 19, y^{^{'}} (0) = 1

d, d (x) = \frac{5}{4} x^{2} + 56 x^{4} + \frac{8}{x} - 9 x

y^{^{''}} + 8 y^{^{''}} - 9 y = - 10 e^{- x} - 425 cos (2 x)

y^{^{''}} + 16 y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

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}