e^yy^'=1

Lösung

e^{y} y^{^{'}} = 1

y = ln (t + c_{1})

Schritte zur Lösung

e^{y} y^{'} (t) = 1

Solve separable ODE:

y = ln (t + c_{1})

y = ln (t + c_{1})

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

4 \cdot y^{^{'}} + y = 0

\frac{d y}{d t} = - \frac{y}{15 + t}

\frac{x ^{2}}{y} - 4 x e^{y^{2}} y^{^{'}} = 0

\frac{d P}{d t} = 1.4 P (1 - \frac{P}{4600})