What is the solution for y^{''}-8y^'+16y=t^{-4}e^{4t} ?

The solution for y^{''}-8y^'+16y=t^{-4}e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t})/(6t^2)

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y^{''}-8y^'+16y=t^{-4}e^{4t}

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 4} e^{4 t}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{6 t ^{2}}

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = t^{- 4} e^{4 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{6 t ^{2}}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} + \frac{e ^{4 t}}{6 t ^{2}}

\int_{1}^{2} 5 x^{2} ln (x) d x

f (x) = 6 x^{\frac{1}{2}}

y = 12 x^{2} - 7 x - 2 x^{- 2}

\int \frac{cos ^{3} ( x )}{sin ( x )} d x

\int cos (5 x - 1) d x

What is the solution for y^{''}-8y^'+16y=t^{-4}e^{4t} ?
The solution for y^{''}-8y^'+16y=t^{-4}e^{4t} is y=c_{1}e^{4t}+c_{2}te^{4t}+(e^{4t})/(6t^2)