What is the solution for y^{''}-4y^'=2e^{4x} ?

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

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

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

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

Solution steps

y^{''} (x) - 4 y^{'} (x) = 2 e^{4 x}

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 16, y (0) = 1, y^{^{'}} (0) = - 4

θ^{^{''}} (t) - θ (t) = t sin (t)

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

y^{^{''}} + \frac{5}{6} y^{^{'}} + \frac{2}{15} y = \frac{4}{3}

y^{^{''}} - y = x + sin (x), y (0) = 2, y^{^{'}} (0) = 3

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