What is the solution for y^{''}-5y^'+4y=x ?

The solution for y^{''}-5y^'+4y=x is y=c_{1}e^{4x}+c_{2}e^x+x/4+5/16

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y^{''}-5y^'+4y=x

y^{^{''}} - 5 y^{^{'}} + 4 y = x

y = c_{1} e^{4 x} + c_{2} e^{x} + \frac{x}{4} + \frac{5}{16}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = x

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} + \frac{x}{4} + \frac{5}{16}

y = c_{1} e^{4 x} + c_{2} e^{x} + \frac{x}{4} + \frac{5}{16}

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

y^{^{''}} + 4 y = 3 sin (x) + 5 cos (3 x)

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 4, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 5 y^{^{'}} + 6 y = 20 sin (2 t)

What is the solution for y^{''}-5y^'+4y=x ?
The solution for y^{''}-5y^'+4y=x is y=c_{1}e^{4x}+c_{2}e^x+x/4+5/16