Answer
The given equation
$$
\frac{d y}{d x}=-\frac{a x+b y}{b x+c y}
$$
is exact, and its solutions are
$$
a x^{2}+2bxy+ c y=d
$$
where $d$ is an arbitrary constant
Work Step by Step
$$
\frac{d y}{d x}=-\frac{a x+b y}{b x+c y}
\quad \quad (i)
$$
this equation can be written as
$$
\left(a x+b y\right) d x+\left(b x+c y\right) d y=0
$$
Comparing this Equation with the differential form:
$$
M(x,y) d x+N(x,y) d y=0
$$
we observe that
$$
\begin{aligned}M(x,y) &=\left(a x+b y\right)
\\ N(x,y) &=\left(b x+c y\right) \end{aligned}
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
M_{y}(x, y)=b=N_{x}(x, y)
$$
so the given equation is exact. Thus there is a $\psi (x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=M(x,y) =\left(a x+b y\right),
\\ \psi_{y}(x, y) &=N(x,y) =\left(b x+c y\right) \end{aligned} \quad \quad (ii)
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\psi(x, y)=\frac{1}{2}a x^{2}+bxy+h(y) \quad \quad (iii)
$$
where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (ii), we compute $\psi_{y}(x, y) $ from Eq. (iii) and set it equal to $N$, obtaining
$$
\psi_{y}(x, y)=bx+h^{\prime}(y)=\left(b x+c y\right)
$$
Thus $h^{\prime}(y)=c y $ and $h(y)=\frac{1}{2} c y^{2}$ The constant of integration can be omitted since any solution of the preceding differential equation is satisfactory.
Now substituting for h(y) in Eq. (iii) gives
$$
\psi(x, y)=\frac{1}{2}a x^{2}+bxy+\frac{1}{2} c y^{2}
$$
Hence solutions of Eq. (i) are given implicitly by
$$
\begin{split}
\psi(x, y) & = \frac{1}{2}a x^{2}+bxy+\frac{1}{2} c y^{2}=d_{1} \\
& = a x^{2}+2bxy+ c y^{2}=d
\end{split}
$$
where $ d_{1}, d=2d_{1} $ are arbitrary constants