y^'=1+x+y^2+xy^2,y(0)=1

Lösung

y^{^{'}} = 1 + x + y^{2} + x y^{2}, y (0) = 1

y = tan (\frac{x ^{2}}{2} + x + \frac{π}{4})

Schritte zur Lösung

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

Solve separable ODE:

y = tan (\frac{x ^{2}}{2} + x + \frac{π}{4})

y = tan (\frac{x ^{2}}{2} + x + \frac{π}{4})

\frac{d x}{d t} = k (a - x)

y^{^{'}} = \frac{tan ( y )}{x}

(x^{2} + 1) \cdot \frac{d y}{d x} = 6 x - 3 x y

y^{^{'}} = 5 - \frac{y}{200}

\frac{d x}{d t} = - x + 2