What is the solution for y^{''}-4y^'+4y=-7x+4x^2,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-4y^'+4y=-7x+4x^2,y(0)=0,y^'(0)=1 is y= 1/4 e^{2x}+(e^{2x}x)/4+x^2+x/4-1/4

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y^{''}-4y^'+4y=-7x+4x^2,y(0)=0,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 4 y = - 7 x + 4 x^{2}, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{4} e^{2 x} + \frac{e ^{2 x} x}{4} + x^{2} + \frac{x}{4} - \frac{1}{4}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = - 7 x + 4 x^{2}

Solve linear ODE:

y = \frac{1}{4} e^{2 x} + \frac{e ^{2 x} x}{4} + x^{2} + \frac{x}{4} - \frac{1}{4}

y = \frac{1}{4} e^{2 x} + \frac{e ^{2 x} x}{4} + x^{2} + \frac{x}{4} - \frac{1}{4}

What is the solution for y^{''}-4y^'+4y=-7x+4x^2,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-4y^'+4y=-7x+4x^2,y(0)=0,y^'(0)=1 is y= 1/4 e^{2x}+(e^{2x}x)/4+x^2+x/4-1/4