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

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

Solution steps

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

Solve using substitution:

y = - c_{1} - c_{1} x arctan (x) + c_{2} x

y = - c_{1} - c_{1} x arctan (x) + c_{2} x

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