4y^'=y^3cos(x)

Lösung

4 y^{^{'}} = y^{3} cos (x)

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

Schritte zur Lösung

4 y^{'} (x) = y^{3} cos (x)

Solve separable ODE:

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

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

y^{^{'}} = x - x y - y + 1

y (y^{2} - 1)^{\frac{1}{2}} d x - (1 - x^{2})^{\frac{1}{2}} d y = 0

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

y^{^{'}} = 6 x^{2} y

(2 x y + 4 x) \frac{d y}{d x} = y^{2} - 4