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

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

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

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

y = - cos (x) + sin (x) + e^{- x}

Solution steps

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

Solve linear ODE:

y = - cos (x) + sin (x) + e^{- x}

y = - cos (x) + sin (x) + e^{- x}

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5} x

\frac{d ^{2} x}{d t ^{2}} + 5 \frac{d x}{d t} + 4 x = e^{6 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 27 e^{6 x}

y^{^{'''}} + 4 y^{^{''}} + 3 y^{^{'}} = 2 + e^{2} t

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