What is the solution for y^{''}-4y^'+4y=-3+2x+2x^2 ?

The solution for y^{''}-4y^'+4y=-3+2x+2x^2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^2)/2+(3x)/2+1/2

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y^{''}-4y^'+4y=-3+2x+2x^2

y^{^{''}} - 4 y^{^{'}} + 4 y = - 3 + 2 x + 2 x^{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{2} + \frac{3 x}{2} + \frac{1}{2}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = - 3 + 2 x + 2 x^{2}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{2} + \frac{3 x}{2} + \frac{1}{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{2} + \frac{3 x}{2} + \frac{1}{2}

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

\frac{d ^{2} y}{d x ^{2}} + y = 32 x^{3}

y^{^{''}} - 0.16 y = 32 cosh (0.4 x)

y^{^{''}} + 9 y = e^{4 x} + 3 x^{2} - 5

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

What is the solution for y^{''}-4y^'+4y=-3+2x+2x^2 ?
The solution for y^{''}-4y^'+4y=-3+2x+2x^2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^2)/2+(3x)/2+1/2