e^yy^'=((ln(t))^2)/t

Lösung

e^{y} y^{^{'}} = \frac{( ln ( t ) ) ^{2}}{t}

y = ln (\frac{ln ^{3} ( t )}{3} + c_{1})

Schritte zur Lösung

e^{y} y^{'} (t) = \frac{( ln ( t ) ) ^{2}}{t}

Solve separable ODE:

y = ln (\frac{ln ^{3} ( t )}{3} + c_{1})

y = ln (\frac{ln ^{3} ( t )}{3} + c_{1})

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

x \frac{d v}{d x} = \frac{- v}{1 + v}

\frac{d y}{d x} x^{2} - y^{2} = 16

\frac{d y}{d x} + 2 x y^{3} = 0

\frac{d y}{d x} = 10 y - 2 y^{2}