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

The solution for y^{''}+2y^'+y=x^2,y(0)=1,y^'(0)=2 is y=-5e^{-x}+e^{-x}x+x^2-4x+6

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

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

y = - 5 e^{- x} + e^{- x} x + x^{2} - 4 x + 6

Solution steps

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

Solve linear ODE:

y = - 5 e^{- x} + e^{- x} x + x^{2} - 4 x + 6

y = - 5 e^{- x} + e^{- x} x + x^{2} - 4 x + 6

y^{^{''}} - y = 0, y (0) = 3, y^{^{'}} (0) = - 1

\frac{\partial}{\partial θ} (r (θ) θ)

x^{2}, - 1, 1, - 2 x^{4}

x ln (x)

\int \frac{1}{( x - 3 ) ( 2 x + 1 ) ( x ^{2} + x + 4 )} d x

What is the solution for y^{''}+2y^'+y=x^2,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+2y^'+y=x^2,y(0)=1,y^'(0)=2 is y=-5e^{-x}+e^{-x}x+x^2-4x+6