y^'=(x-4)e^{-2y},y(4)=ln(4)

Lösung

y^{^{'}} = (x - 4) e^{- 2 y}, y (4) = ln (4)

y = \frac{ln ( x ^{2} - 8 x + 32 )}{2}

Schritte zur Lösung

y^{'} (x) = (x - 4) e^{- 2 y}

Solve separable ODE:

y = \frac{ln ( x ^{2} - 8 x + 32 )}{2}

y = \frac{ln ( x ^{2} - 8 x + 32 )}{2}

\frac{d}{d x} (2 x^{- \frac{14}{5}})

x \to 3 lim (\frac{6 x}{x ^{2} - 9})

\int - t^{2} d t

\int_{1}^{5} \frac{1}{3 x + 1} d x

\frac{d}{d x} (\frac{4 x ^{2} + 2 x + 2}{x})