(y-xy)y^'-y^2+4=0

Lösung

(y - x y) y^{^{'}} - y^{2} + 4 = 0

y = \frac{c _{1} + 4 ( x ^{2} - 2 x + 1 )}{x ^{2} - 2 x + 1}, y = - \frac{c _{1} + 4 ( x ^{2} - 2 x + 1 )}{x ^{2} - 2 x + 1}

Schritte zur Lösung

(y - x y) y^{'} (x) - y^{2} + 4 = 0

Solve separable ODE:

y = \frac{c _{1} + 4 ( x ^{2} - 2 x + 1 )}{x ^{2} - 2 x + 1}, y = - \frac{c _{1} + 4 ( x ^{2} - 2 x + 1 )}{x ^{2} - 2 x + 1}

y = \frac{c _{1} + 4 ( x ^{2} - 2 x + 1 )}{x ^{2} - 2 x + 1}, y = - \frac{c _{1} + 4 ( x ^{2} - 2 x + 1 )}{x ^{2} - 2 x + 1}

\frac{d y}{d x} + 6 x^{2} y = 9 x^{2}

\frac{d P}{d t} = 8 Pt, P (1) = 3

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

y^{^{'}} = - 10 x^{3} e^{y}

\frac{d y}{d x} + 9 x y = 4 x