What is the solution for y^{''}+2y^'+y=9-e^{2x},y(0)=0,y^'(0)=1 ?

The solution for y^{''}+2y^'+y=9-e^{2x},y(0)=0,y^'(0)=1 is y=-80/9 e^{-x}-(23e^{-x}x)/3+9-1/9 e^{2x}

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y^{''}+2y^'+y=9-e^{2x},y(0)=0,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + y = 9 - e^{2 x}, y (0) = 0, y^{^{'}} (0) = 1

y = - \frac{80}{9} e^{- x} - \frac{23 e ^{- x} x}{3} + 9 - \frac{1}{9} e^{2 x}

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = 9 - e^{2 x}

Solve linear ODE:

y = - \frac{80}{9} e^{- x} - \frac{23 e ^{- x} x}{3} + 9 - \frac{1}{9} e^{2 x}

y = - \frac{80}{9} e^{- x} - \frac{23 e ^{- x} x}{3} + 9 - \frac{1}{9} e^{2 x}

(3 d^{2} + 3) y = 3 x

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 + 5 x

y^{^{''}} + 3 y^{^{'}} - 10 y = 20 + 5 e^{3 x}

x^{^{''}} + 4 x^{^{'}} + 14 x = 9 cos (3 t)

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 sin (6 t)

What is the solution for y^{''}+2y^'+y=9-e^{2x},y(0)=0,y^'(0)=1 ?
The solution for y^{''}+2y^'+y=9-e^{2x},y(0)=0,y^'(0)=1 is y=-80/9 e^{-x}-(23e^{-x}x)/3+9-1/9 e^{2x}