What is the solution for (1-x^2)y^{''}+2xy^'-2y=0 ?

The solution for (1-x^2)y^{''}+2xy^'-2y=0 is y=e^{c_{1}}x^2+e^{c_{1}}+c_{2}x

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(1-x^2)y^{''}+2xy^'-2y=0

(1 - x^{2}) y^{^{''}} + 2 x y^{^{'}} - 2 y = 0

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

Solution steps

(1 - x^{2}) y^{''} (x) + 2 x y^{'} (x) - 2 y = 0

Solve using substitution:

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

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

f (x) = ln (x^{3} - 5 x^{2} + 3 x - 1)

f (t) = 3 t^{3} - \frac{2}{t ^{2}}

x \to 0 + lim (\frac{1}{2 x ^{2}})

3 + 5 x

\frac{d}{d x} (x + y z)

What is the solution for (1-x^2)y^{''}+2xy^'-2y=0 ?
The solution for (1-x^2)y^{''}+2xy^'-2y=0 is y=e^{c_{1}}x^2+e^{c_{1}}+c_{2}x