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

解

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

y = - \frac{1}{2} ln (1 - x^{2}) + c_{1} (\frac{ln ( x + 1 )}{2} - \frac{ln ( x - 1 )}{2}) + c_{2}

解答ステップ

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

置換を使用して解く:

y = - \frac{1}{2} ln (1 - x^{2}) + c_{1} (\frac{ln ( x + 1 )}{2} - \frac{ln ( x - 1 )}{2}) + c_{2}

y = - \frac{1}{2} ln (1 - x^{2}) + c_{1} (\frac{ln ( x + 1 )}{2} - \frac{ln ( x - 1 )}{2}) + c_{2}