What is the solution for y^{''}+2y^'+y=e^{3x},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+2y^'+y=e^{3x},y(0)=0,y^'(0)=0 is y=-1/16 e^{-x}-(e^{-x}x)/4+1/16 e^{3x}

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y^{''}+2y^'+y=e^{3x},y(0)=0,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + y = e^{3 x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{16} e^{- x} - \frac{e ^{- x} x}{4} + \frac{1}{16} e^{3 x}

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = e^{3 x}

Solve linear ODE:

y = - \frac{1}{16} e^{- x} - \frac{e ^{- x} x}{4} + \frac{1}{16} e^{3 x}

y = - \frac{1}{16} e^{- x} - \frac{e ^{- x} x}{4} + \frac{1}{16} e^{3 x}

What is the solution for y^{''}+2y^'+y=e^{3x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+2y^'+y=e^{3x},y(0)=0,y^'(0)=0 is y=-1/16 e^{-x}-(e^{-x}x)/4+1/16 e^{3x}