2xy+x^2y^'=0

Lösung

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

y = \frac{c _{1}}{x ^{2}}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{c _{1}}{x ^{2}}

y = \frac{c _{1}}{x ^{2}}

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

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