y^'=0.1y^2

Lösung

y^{^{'}} = 0.1 y^{2}

y = - \frac{10}{t + c _{1}}

Schritte zur Lösung

y^{'} (t) = 0.1 y^{2}

Solve separable ODE:

y = - \frac{10}{t + c _{1}}

y = - \frac{10}{t + c _{1}}

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

x y d x - 3 (y - 2) d y = 0

y^{^{'}} - y tan (x) = 0

y^{^{'}} = x y - x y^{2}, y (0) = 3

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