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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 11 y^{^{'}} + 24 y = 280 e^{t}

y^{^{''}} + 9 y = 5 sec (3 x)

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

y^{^{''}} + 10000 y = 100 sin (50 x)

y^{^{''}} + 10 y^{^{'}} + 15 = 0

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