(x^2+1)dx-3ydy=0

Lösung

(x^{2} + 1) d x - 3 y d y = 0

y = \frac{2 x ^{3} + 6 x - c _{1}}{3}, y = - \frac{2 x ^{3} + 6 x - c _{1}}{3}

Schritte zur Lösung

(x^{2} + 1) d x - 3 y d y = 0

Solve separable ODE:

y = \frac{2 x ^{3} + 6 x - c _{1}}{3}, y = - \frac{2 x ^{3} + 6 x - c _{1}}{3}

y = \frac{2 x ^{3} + 6 x - c _{1}}{3}, y = - \frac{2 x ^{3} + 6 x - c _{1}}{3}

e^{x + y} \frac{d x}{d y} = x

\frac{d q}{d t} = k q (t)^{2}

y^{^{'}} = \frac{e ^{- x}}{sin ( y )}, y (1) = 0

y^{^{'}} + 12 x y = 0

\frac{d y}{d x} = 3 y + y^{2} - 2