We have,
$(3 y-7 x+7) d x+(7 y-3 x+3) d y=0$
$\Rightarrow(7 x-3 y-7) d x+(3 x-7 y-3) d y=0$ $\ldots(\mathrm{i})$
Given equation is non-homogeneous and
$a_1 b_2-a_2 b_1=-40 \neq 0$
On solving $7 x-3 y-7=0$ and $3 x-7 y-3=0$
we get $x=1, y=0$
Now, substitude $x=1+u$ and $y=0+v=v$
$\therefore d x=d u$ and $d y=d v \ldots$ in Eq. (i),
we get $(7 u-3 v) d u+(3 u-7 v) d v=0$ $\ldots(\mathrm{ii})$
This homogeneous equation in $u$ and $v$.
Put, $u=t v \Rightarrow d u=t d v+v d t$ in Eq. (ii),
we get $(7 t v-3 v)(t d v+v d t)+(3 t v-7 v) d v=0$
$(7 t-3)(t d v+v d t)+(3 t-7) d v=0$
$\Rightarrow \quad\left(7 t^2-7\right) d v+v(7 t-3) d t=0$
$\Rightarrow \quad \int \frac{d v}{v}+\int \frac{7 t-3}{7 t^2-7} d t=0$
$\Rightarrow \log |v|+\int \frac{t-3 / 7}{t^2-1} d t=C_1$
$\Rightarrow \log |v|+\left[\int \frac{2}{7(t-1)} d t+\int \frac{5 / 7}{(t+1)} d t\right]=\log C_1$
$\Rightarrow \log |v|+\frac{2}{7} \log |t-1|+\frac{5}{7} \log |t+1|=\log C_1$
$\Rightarrow \log |v|+\frac{2}{7} \log \left|\frac{u}{v}-1\right|+\frac{5}{7} \log \left|\frac{u}{v}+1\right|=\log C_1$
$\Rightarrow \log |v|+\frac{2}{7} \log \left(\frac{u-v}{v}\right)+\frac{5}{7} \log \left(\frac{u+v}{v}\right)=\log C_1$
$\Rightarrow \log |v|+\log \left(\frac{u-v}{v}\right)^{2 / 7}+\log \left(\frac{u+v}{v}\right)^{5 / 7}=\log C_1$
$\Rightarrow \log \left[|v| \cdot \frac{(u-v)^{2 / 7}}{(v)^{2 / 7}} \cdot \frac{(u+v)^{5 / 7}}{(v)^{5 / 7}}\right]=\log C_1$
$\Rightarrow \quad \log \left[(u-v)^{2 / 7}(u+v)^{5 / 7}\right]=\log C_1$
$\Rightarrow \log \left[(x-y-1)^{2 / 7}(x+y-1)^{5 / 7}\right]=\log C_1$
$\Rightarrow \quad(x-y-1)^{2 / 7}(x+y-1)^{5 / 7}=C_1$
$\Rightarrow \quad(x-y-1)^2(x+y-1)^5=C$