What is the solution for y^{''}-8y^'+16y=e^{4x} ?

The solution for y^{''}-8y^'+16y=e^{4x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+(e^{4x}x^2)/2

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y^{''}-8y^'+16y=e^{4x}

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x}

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

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = e^{4 x}

Solve linear ODE:

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

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

\frac{d}{d x} (4 x^{2} + 6 x)

\frac{\partial}{\partial s} (s^{6} + t^{6})

\frac{df}{d x} + 2 f = f x^{3}

n = 0 \sum \infty (\frac{1}{3})^{n}

a re a 2 x^{2} + 3, x, (- 1, 2)

What is the solution for y^{''}-8y^'+16y=e^{4x} ?
The solution for y^{''}-8y^'+16y=e^{4x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+(e^{4x}x^2)/2