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

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

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(1-t^2)y^{''}+2ty^'-2y=0

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

y = e^{c_{1}} t^{2} + e^{c_{1}} + c_{2} t

Solution steps

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

Solve using substitution:

y = e^{c_{1}} t^{2} + e^{c_{1}} + c_{2} t

y = e^{c_{1}} t^{2} + e^{c_{1}} + c_{2} t

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