(x+x^2)y^'y=y^2-2

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

2 x y + x^{2} y^{^{'}} = 0

P^{^{'}} = 0.6 \cdot P \cdot (1 - \frac{P}{840})

\frac{d y}{d x} = \frac{4 y cos ( x )}{( 3 + cos ^{2} ( x ) )}

\frac{d y}{d x} = - y + 2

u^{^{'}} cos (H (t)) = 8