y^'=e^{2y-x},y(0)=0

Lösung

y^{^{'}} = e^{2 y - x}, y (0) = 0

y = - \frac{ln ( 2 e ^{- x} - 1 )}{2}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{ln ( 2 e ^{- x} - 1 )}{2}

y = - \frac{ln ( 2 e ^{- x} - 1 )}{2}

(y - 1) y^{^{'}} = 1

\frac{d N}{d t} = 0.005 N

\frac{d y}{d x} = y^{2} x cos (x^{2})

y^{^{'}} = \frac{e ^{y}}{( x ^{2} - 1 )}

2 y^{^{'}} + y = 5, y (3) = 0