y^'-e^{x+y}=0

Lösung

y^{^{'}} - e^{x + y} = 0

y = - ln (- e^{x} - c_{1})

Schritte zur Lösung

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

Solve separable ODE:

y = - ln (- e^{x} - c_{1})

y = - ln (- e^{x} - c_{1})

y^{^{'}} = - y + 9

y d x + (x - 6) d y = 0, y (7) = 9

(3 y 2 - t 2 y 5) \frac{d y}{d t} + t 2 y 4 = 0

y^{^{'}} = tan (x) y

y^{^{'}} + 4 y = 12, y (0) = 0