What is the solution for t^2y^{''}-2ty^'-4y=0 ?

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

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t^2y^{''}-2ty^'-4y=0

t^{2} y^{^{''}} - 2 t y^{^{'}} - 4 y = 0

y = c_{1} t^{4} + \frac{c _{2}}{t}

Solution steps

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

Solve linear ODE:

y = c_{1} t^{4} + \frac{c _{2}}{t}

y = c_{1} t^{4} + \frac{c _{2}}{t}

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

\frac{d}{d x} (x e^{y}) = \frac{d}{d x} (x - y)

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