What is the solution for 2y^{''}+3y^'+y=x^2,y(0)=0,y^'(0)=1 ?

The solution for 2y^{''}+3y^'+y=x^2,y(0)=0,y^'(0)=1 is y=x^2-6x-14e^{-x/2}+14

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2y^{''}+3y^'+y=x^2,y(0)=0,y^'(0)=1

2 y^{^{''}} + 3 y^{^{'}} + y = x^{2}, y (0) = 0, y^{^{'}} (0) = 1

y = x^{2} - 6 x - 14 e^{- \frac{x}{2}} + 14

Solution steps

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

Solve linear ODE:

y = x^{2} - 6 x - 14 e^{- \frac{x}{2}} + 14

y = x^{2} - 6 x - 14 e^{- \frac{x}{2}} + 14

(d^{2} - 1) y = sin^{2} (x)

41 H (x) - 8 u^{^{'}} + u^{^{''}} = 0

y^{^{''}} - 3 y = 6 x

y^{^{''}} + y = e^{t} sin (t), y (0) = 0, y^{^{'}} (0) = 0

\frac{1}{2} y^{^{''}} + 5 y^{^{'}} + 12 y = e^{- 4 x} (4 x - 2)

What is the solution for 2y^{''}+3y^'+y=x^2,y(0)=0,y^'(0)=1 ?
The solution for 2y^{''}+3y^'+y=x^2,y(0)=0,y^'(0)=1 is y=x^2-6x-14e^{-x/2}+14