Differential Q[1]

Let $u=x+y$

$8 \frac{dy}{dx}=u$ $[1]$

$\frac{du}{dx}=1+\frac{dy}{dx}$ $[2]$

$8\frac{dy}{dx} dx =u dx$

$8 dy =u dx$ $[3]$

$\frac{dy}{dx}=\frac{u}{8}$

From[2] and [3]

$\frac{du}{dx}=1+\frac{dy}{dx}=1+\frac{u}{8}$

$dx=\frac{du}{1+\frac{u}{8}}$

Substituting into [3]

$8 dy =u \frac{du}{1+\frac{u}{8}}$

$1+\frac{u}{8}=\frac{8+u}{8}$

$8 dy =8 u \frac{du}{8+u}$

$dy = u \frac{du}{8+u}$

$\displaystyle \int dy = \int u \frac{ du}{8+u}$

$\displaystyle \int dy = \int [\frac{-8}{8+u} +1] du$

$y = -8\ ln( 8+u) +u +C$

$y = -8 \ln(8+x+y) +x+y +C$

$-8\ ln( 8+x+y)= x +C$

Substituting x=-3 and y=-4

$-8 \ln( 8-3-4)= -3 +C$

$-8\ ln( 1)= -3 +C$

$-3 +C= 0$

$C= 3$