y^'=e^{(t+y)},y(0)=0

Lösung

y^{^{'}} = e^{(t + y)}, y (0) = 0

y = - ln (- e^{t} + 2)

Schritte zur Lösung

y^{'} (t) = e^{(t + y)}

Solve separable ODE:

y = - ln (- e^{t} + 2)

y = - ln (- e^{t} + 2)

x^{2} (\frac{d y}{d x}) + 4 x y - y^{3} = 0

x \to 2 lim ((3 x - 1)^{x^{2} - 3 x + 4})

\frac{\partial}{\partial y} (ln (\frac{x}{y}))

x \to 0 lim (\frac{2}{x} - \frac{2}{x})

a re a \frac{2}{x}, \frac{2}{x ^{2}}, x = 4