What is the solution for y^{''}-8y+16y=t^{-3}e^{4t} ?

The solution for y^{''}-8y+16y=t^{-3}e^{4t} is

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y^{''}-8y+16y=t^{-3}e^{4t}

y^{^{''}} - 8 y + 16 y = t^{- 3} e^{4 t}

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t) + cos (22 t) \cdot \int - \frac{2 e ^{4 t} sin ( 2 2 t )}{4 t ^{3}} d t + sin (22 t) \cdot \int \frac{2 e ^{4 t} cos ( 2 2 t )}{4 t ^{3}} d t

Solution steps

y^{''} (t) - 8 y + 16 y = t^{- 3} e^{4 t}

Solve linear ODE:

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t) + cos (22 t) \cdot \int - \frac{2 e ^{4 t} sin ( 2 2 t )}{4 t ^{3}} d t + sin (22 t) \cdot \int \frac{2 e ^{4 t} cos ( 2 2 t )}{4 t ^{3}} d t

y = c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t) + cos (22 t) \cdot \int - \frac{2 e ^{4 t} sin ( 2 2 t )}{4 t ^{3}} d t + sin (22 t) \cdot \int \frac{2 e ^{4 t} cos ( 2 2 t )}{4 t ^{3}} d t

What is the solution for y^{''}-8y+16y=t^{-3}e^{4t} ?
The solution for y^{''}-8y+16y=t^{-3}e^{4t} is