What is the solution for t^2y^{''}+4ty^'-10y=0 ?

The solution for t^2y^{''}+4ty^'-10y=0 is y=c_{1}t^2+(c_{2})/(t^5)

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t^2y^{''}+4ty^'-10y=0

t^{2} y^{^{''}} + 4 t y^{^{'}} - 10 y = 0

y = c_{1} t^{2} + \frac{c _{2}}{t ^{5}}

Solution steps

t^{2} y^{''} (t) + 4 t y^{'} (t) - 10 y = 0

Solve linear ODE:

y = c_{1} t^{2} + \frac{c _{2}}{t ^{5}}

y = c_{1} t^{2} + \frac{c _{2}}{t ^{5}}

What is the solution for t^2y^{''}+4ty^'-10y=0 ?
The solution for t^2y^{''}+4ty^'-10y=0 is y=c_{1}t^2+(c_{2})/(t^5)