y^'=1+y+y^2

Lösung

y^{^{'}} = 1 + y + y^{2}

y = \frac{tan ( \frac{3 ( t + c _{1} )}{2} ) 3}{2} - \frac{1}{2}

Schritte zur Lösung

y^{'} (t) = 1 + y + y^{2}

Solve separable ODE:

y = \frac{tan ( \frac{3 ( t + c _{1} )}{2} ) 3}{2} - \frac{1}{2}

y = \frac{tan ( \frac{3 ( t + c _{1} )}{2} ) 3}{2} - \frac{1}{2}

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

\frac{d y}{d x} = 2 x^{2} y^{2}

(2 + x) d y - y d x = 0

(y + 1) d x + (6 x - 1) d y = 0

y^{^{'}} = y^{2} + 2