y^'=x^2+x^2y^2

Lösung

y^{^{'}} = x^{2} + x^{2} y^{2}

y = tan (\frac{x ^{3}}{3} + c_{1})

Schritte zur Lösung

y^{'} (x) = x^{2} + x^{2} y^{2}

Solve separable ODE:

y = tan (\frac{x ^{3}}{3} + c_{1})

y = tan (\frac{x ^{3}}{3} + c_{1})

(y + 1) (t^{2} + 1) = t y^{^{'}}

x y^{^{'}} - 5 y = 0

\frac{d y}{d x} = x^{6} (y - 4), y (0) = 6

y^{^{'}} = \frac{( 1 - x )}{y}, y (1) = 1

(x^{2} + 9) y^{^{'}} = x y