What is the solution for y^{''}-4y^'=2x-1 ?

The solution for y^{''}-4y^'=2x-1 is y=c_{1}e^{4x}+c_{2}-(x^2)/4+x/8

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y^{''}-4y^'=2x-1

y^{^{''}} - 4 y^{^{'}} = 2 x - 1

y = c_{1} e^{4 x} + c_{2} - \frac{x ^{2}}{4} + \frac{x}{8}

Solution steps

y^{''} (x) - 4 y^{'} (x) = 2 x - 1

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{x ^{2}}{4} + \frac{x}{8}

y = c_{1} e^{4 x} + c_{2} - \frac{x ^{2}}{4} + \frac{x}{8}

y^{^{'''}} + 2 y^{^{''}} - 9 y^{^{'}} - 18 = 10 + 20 sin (5 t)

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

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

y^{^{''}} + 12 y^{^{'}} + 100 y = 6

y^{^{''}} - 2 y^{^{'}} - 7 y = 3 x^{2}

What is the solution for y^{''}-4y^'=2x-1 ?
The solution for y^{''}-4y^'=2x-1 is y=c_{1}e^{4x}+c_{2}-(x^2)/4+x/8