What is the solution for -8t^2y^{''}-4t(t-4)y^'+4(t-4)y=0 ?

The solution for -8t^2y^{''}-4t(t-4)y^'+4(t-4)y=0 is y=c_{1}t+c_{2}e^{-t/2}t

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-8t^2y^{''}-4t(t-4)y^'+4(t-4)y=0

- 8 t^{2} y^{^{''}} - 4 t (t - 4) y^{^{'}} + 4 (t - 4) y = 0

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

Solution steps

- 8 t^{2} y^{''} (t) - 4 t (t - 4) y^{'} (t) + 4 (t - 4) y = 0

Solve using substitution:

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

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

y^{2} = x, x - 2 y = 3

\frac{d}{d x} (3 x (x^{2} - x + 1) (5 x - 3))

\frac{d y}{d t} = - 2 y, y (0) = \frac{1}{10}

x^{2} - 8

\frac{d}{d x} (sin (2 x) cos (2 x))

What is the solution for -8t^2y^{''}-4t(t-4)y^'+4(t-4)y=0 ?
The solution for -8t^2y^{''}-4t(t-4)y^'+4(t-4)y=0 is y=c_{1}t+c_{2}e^{-t/2}t