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

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

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

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

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{e ^{x} x}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{e ^{x} x}{3}

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{e ^{x} x}{3}

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{3 x} - sin (x)

16 y^{^{''}} - 8 y^{^{'}} + y = 33 - 16 x + x^{2}

y^{^{''}} - 8 y^{^{'}} + 7 y = 4 e^{3} x

y^{^{''}} - 9 y^{^{'}} = \frac{9 x}{e ^{3 x}}

y^{^{''}} - 4 y^{^{'}} = 13 e^{12 x}

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