What is the solution for y^{''}-4y^'+4y=10x+2 ?

The solution for y^{''}-4y^'+4y=10x+2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(5x)/2+3

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y^{''}-4y^'+4y=10x+2

y^{^{''}} - 4 y^{^{'}} + 4 y = 10 x + 2

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 x}{2} + 3

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 10 x + 2

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 x}{2} + 3

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 x}{2} + 3

2 y^{^{''}} + y = 9 e^{2 x}

y^{^{''}} - 3 y^{^{'}} = 27 x^{2}

4 y^{^{''}} + 324 y = 8 sin (t)

y^{^{'''}} - y^{^{'}} = 5 cos (2 x) + 7 e^{x} + 2 x^{2}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 36 y = e^{10 x}

What is the solution for y^{''}-4y^'+4y=10x+2 ?
The solution for y^{''}-4y^'+4y=10x+2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(5x)/2+3