(1+x^2)y^'+xy=0

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

\frac{d p}{d x} = p - p^{2}

\frac{d y}{d x} = x + 2 x y, y (0) = 3

e^{x} d x - y d y = 0

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

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