2y^'=y^3cos(x)

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

x \to 0 lim (x^{4} sin (\frac{2}{x}))

d er i v a t i v e g (u) = 7 u + 13 u

\frac{\partial}{\partial x} (x e^{3 x y})

\int \frac{cos ( 2 x )}{x} d x

\int 8 cos (θ) cos^{5} (sin (θ)) d θ