What is the solution for y^{''}-4y^'+4y=10x+3 ?

The solution for y^{''}-4y^'+4y=10x+3 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(5x)/2+13/4

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y^{''}-4y^'+4y=10x+3

y^{^{''}} - 4 y^{^{'}} + 4 y = 10 x + 3

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 x}{2} + \frac{13}{4}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 10 x + 3

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 x}{2} + \frac{13}{4}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 x}{2} + \frac{13}{4}

\frac{d q}{20 - q} = \frac{1}{16} d T

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

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{2 x} + 1

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

\frac{d ^{2} y}{d x ^{2}} + y = 3 sin (2 x) + x cos (2 x)

What is the solution for y^{''}-4y^'+4y=10x+3 ?
The solution for y^{''}-4y^'+4y=10x+3 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(5x)/2+13/4