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)/(16)+(e^4)/(32)

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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}{16} + \frac{e ^{4}}{32}

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}{16} + \frac{e ^{4}}{32}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{e ^{4} x}{16} + \frac{e ^{4}}{32}

y^{^{''}} - y = 12 x^{2} + e^{x}

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

y^{^{''}} + 4 y = t^{2} + 10 e^{t}

d (x) = \frac{1}{2} x^{2} - 10 x

y^{^{''}} - y = 1.7 sin (t)

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)/(16)+(e^4)/(32)