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

The solution for (1-x^2)y^{''}+2xy^'=0 is y=e^{c_{1}}x-(e^{c_{1}}x^3)/3+c_{2}

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

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

y = e^{c_{1}} x - \frac{e ^{c_{1}} x ^{3}}{3} + c_{2}

Solution steps

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

Solve using substitution:

y = e^{c_{1}} x - \frac{e ^{c_{1}} x ^{3}}{3} + c_{2}

y = e^{c_{1}} x - \frac{e ^{c_{1}} x ^{3}}{3} + c_{2}

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What is the solution for (1-x^2)y^{''}+2xy^'=0 ?
The solution for (1-x^2)y^{''}+2xy^'=0 is y=e^{c_{1}}x-(e^{c_{1}}x^3)/3+c_{2}