4xy^2dy=2y^3+2ydx

Lösung

4 x y^{2} d y = 2 y^{3} + 2 y d x

y = c_{1} x - 1, y = - c_{1} x - 1

Schritte zur Lösung

4 x y^{2} d y = 2 y^{3} + 2 y d x

Solve separable ODE:

y = c_{1} x - 1, y = - c_{1} x - 1

y = c_{1} x - 1, y = - c_{1} x - 1

y^{2} d y = cos (5 x) d x

\frac{d y}{d x} = \frac{x ^{3}}{( y + 1 )}

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

y^{^{'}} = 3 x (1 - y)

\frac{d z}{d t} = 6 t e^{5 z}, t (0) = 0