What is the solution for y^{''}-5y^'+4y=8e^x ?

The solution for y^{''}-5y^'+4y=8e^x is y=c_{1}e^{4x}+c_{2}e^x-(8e^xx)/3

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y^{''}-5y^'+4y=8e^x

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{x}

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{8 e ^{x} x}{3}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 4 y = 8 e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{8 e ^{x} x}{3}

y = c_{1} e^{4 x} + c_{2} e^{x} - \frac{8 e ^{x} x}{3}

What is the solution for y^{''}-5y^'+4y=8e^x ?
The solution for y^{''}-5y^'+4y=8e^x is y=c_{1}e^{4x}+c_{2}e^x-(8e^xx)/3