y^'=-(1+y)^4

Lösung

y^{^{'}} = - (1 + y)^{4}

y = 3 \frac{1}{3 ( t + c _{1} )} - 1

Schritte zur Lösung

y^{'} (t) = - (1 + y)^{4}

Solve separable ODE:

y = 3 \frac{1}{3 ( t + c _{1} )} - 1

y = 3 \frac{1}{3 ( t + c _{1} )} - 1

\int t^{4} e^{- s t} d t

\frac{d}{d x} (\frac{5 x ln ( 3 x + 1 )}{1 - e ^{2 x}})

\int_{0}^{1} 7 cos (3 x) d x

\frac{d}{d θ} (θ)

\frac{\partial}{\partial x} (sin (7 x^{2} y - 3 x y^{3}))