e^yy^'=1,y(0)=ln(x+c)

Lösung

e^{y} y^{^{'}} = 1, y (0) = ln (x + c)

y = ln (x + c_{1})

Schritte zur Lösung

e^{y} y^{'} (x) = 1

Solve separable ODE:

y = ln (x + c_{1})

y = ln (x + c_{1})

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

x^{2} y^{^{'}} - 2 y = 0

f^{^{'}} (x) = - 2 \cdot f (x)

\frac{d y}{d x} = 7 y - 1

(y^{2} sec (2 x) tan (2 x)) d x + y (sec (2 x) + 2) d y = 0