What is the solution for y^{''}+y=6e^{-2x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=6e^{-2x},y(0)=0,y^'(0)=0 is y=-6/5 cos(x)+12/5 sin(x)+6/5 e^{-2x}

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y^{''}+y=6e^{-2x},y(0)=0,y^'(0)=0

y^{^{''}} + y = 6 e^{- 2 x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{6}{5} cos (x) + \frac{12}{5} sin (x) + \frac{6}{5} e^{- 2 x}

Solution steps

y^{''} (x) + y = 6 e^{- 2 x}

Solve linear ODE:

y = - \frac{6}{5} cos (x) + \frac{12}{5} sin (x) + \frac{6}{5} e^{- 2 x}

y = - \frac{6}{5} cos (x) + \frac{12}{5} sin (x) + \frac{6}{5} e^{- 2 x}

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

y^{^{''}} + 16 y = g (t), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + y = 3 x^{\frac{3}{2}} e^{x}

y^{^{''}} + 16 y = g (t), y (0) = 0, y^{^{'}} (0) = 5

2 y^{^{''}} + 5 y^{^{'}} + 2 y = x e^{- x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+y=6e^{-2x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=6e^{-2x},y(0)=0,y^'(0)=0 is y=-6/5 cos(x)+12/5 sin(x)+6/5 e^{-2x}