What is the solution for y^{''}+y=5e^{-3x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=5e^{-3x},y(0)=0,y^'(0)=0 is y=-1/2 cos(x)+3/2 sin(x)+1/2 e^{-3x}

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y^{''}+y=5e^{-3x},y(0)=0,y^'(0)=0

y^{^{''}} + y = 5 e^{- 3 x}, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (x) + y = 5 e^{- 3 x}

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 3 y = x^{2}

y^{^{''}} + 4 y = sin (2 x) + 1

x^{^{''}} + 8 x^{^{'}} + 16 x = 6 sin (4 t)

y^{^{'''}} + y^{^{'}} = - 2 sin (x)

x^{^{''}} + 16 x = 2 sin (t)

What is the solution for y^{''}+y=5e^{-3x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=5e^{-3x},y(0)=0,y^'(0)=0 is y=-1/2 cos(x)+3/2 sin(x)+1/2 e^{-3x}