(dy)/(dt)=e^{t+y}

解答

\frac{d y}{d t} = e^{t + y}

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

求解步骤

\frac{d y}{d t} = e^{t + y}

求解可分离方程:

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

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

x y^{^{'}} + y = 1

3 t \frac{d y}{d t} = y cos (t)

y^{^{'}} = 4 x y^{2}

\frac{d y}{d x} - 3 x^{2} y = 0

\frac{d y}{d x} = x e^{y - x^{2}}