y^'+4e^{x+y}=0

Lösung

y^{^{'}} + 4 e^{x + y} = 0

y = - ln (4 e^{x} - c_{1})

Schritte zur Lösung

y^{'} (x) + 4 e^{x + y} = 0

Solve separable ODE:

y = - ln (4 e^{x} - c_{1})

y = - ln (4 e^{x} - c_{1})

x \to 0 lim (\frac{9 x ^{2} - 2 x}{x})

\frac{d}{d x} (arccos (- x))

\int \frac{1}{t ^{2} 441 - t ^{2}} d t

\int (4 x^{5} + 1) x^{4} d x

\frac{\partial}{\partial x} (cos^{3} (x^{7} y^{3}))