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

x^{^{''}} + 240 x^{^{'}} - 9.8 = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3} t, y (0) = 2, y^{^{'}} (0) = 17

y^{^{''}} + 6 y^{^{'}} + 13 y = e^{- 3 x} (cos (2 x))

y^{^{''}} + 36 y = sec^{2} (x)

y^{^{''}} + 8 y^{^{'}} + 12 = 1 - H (t - 3)

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