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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 9 x + 4

x^{^{''}} - 25 x = t^{2} + t

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

4 y^{^{''}} - 4 y^{^{'}} + y = 48 e^{\frac{t}{2}}

y^{^{''}} + 3 y^{^{'}} + 5 y = sin (t), y (0) = 1, y^{^{'}} (0) = 0

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