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

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

y^{^{'}} = y (\frac{a}{x ^{2}} + b x)

\frac{d y}{d t} = 0.48 - \frac{y}{240}

f (x) = f^{^{'}} (x) (4 x + 9)^{5}

\frac{d y}{d x} = - 7 (- 4 - y)

\frac{d h}{d t} = - \frac{1}{72} h