ty^2-y^'=0

Lösung

t y^{2} - y^{^{'}} = 0

y = - \frac{2}{t ^{2} + c _{1}}

Schritte zur Lösung

t y^{2} - y^{'} (t) = 0

Solve separable ODE:

y = - \frac{2}{t ^{2} + c _{1}}

y = - \frac{2}{t ^{2} + c _{1}}

\frac{d y}{d x} = 6 - \frac{y}{100}

y^{^{'}} + y (\frac{1}{x} - \frac{3}{4}) = 0

d v = e^{2 x} d x

(1 + y^{2}) d x - d y = 0

\frac{1}{2} y y^{^{'}} = t^{3}, y (0) = 2