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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = sin (t), < y (0) = 0, y^{^{'}} (0) = 1

5 y^{^{''}} + y^{^{'}} = - 2 x, y (0) = 0, y^{^{'}} (0) = - 20

4 y^{^{''}} - 4 y^{^{'}} + y = e^{\frac{x}{2}}

2 y^{^{''}} - y^{^{'}} - y = x e^{x}

y^{^{''}} - 4 y^{^{'}} + 13 (t) = 0

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