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

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

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