Reducible to homogeneous differential equation general solution

Question

Reduce the equation $\frac{dy}{dx}=\frac{2y+6}{x+y+1}$ to homogeneous by a suitable change of variables $X=x-x_0$, $Y=y-y_0$. Hence find the general solution.

I've solved this question but don't have the correct answer to it. Would someone please be able to check it for me? The answer I got is: $$\ln \frac{y+3}{x-2} - 2\ln \left( 1-\frac{y+3}{2-x}\right)=\ln (x-2)-C$$

To solve the separable equation: $$\int \frac{1}x \, dx = \int \frac{1}v + \frac{2}{1-v} +C$$

$$ln(x) = ln(v) - 2ln(1-v) + C$$

$$v = \frac{y}{x} = \frac{y+3}{x-2}$$

$$ln(x-2) = ln(\frac{y+3}{x-2}) - 2ln(1-\frac{y+3}{x-2}) + C$$

Answer 1

Hint

Rewrite $$\frac{dy}{dx}=\frac{2y+6}{x+y+1}=\frac{2(y+3)}{(x-2)+(y+3)}$$ So, set $Y=y+3$,$X=x-2$ and the equation becomes $$\frac{dY}{dX}=\frac{2Y}{X+Y}$$ Rewrite now the reverse so $$\frac{dX}{dY}=\frac{X+Y}{2Y}=\frac{X}{2Y}+\frac{1}{2}$$ which is now separable.

I am sure that you can take from here.

Answer 2

The idea is to choose suitable constants $x_0$ and $y_0$ such that the constant terms (in this case, $6$ and $1$) disappear.

We have $\frac{dy}{dx}=\frac{dY}{dX}=\frac{2Y}{X+Y}=\frac{2\left( \frac{Y}{X}\right)}{1+\left(\frac{Y}{X}\right)}$.

Now let $v=\frac{Y}{X} \iff Y=vX \implies \frac{dY}{dX}=v+X\frac{dv}{dX}$

Then we have $$v+X\frac{dv}{dX}=\frac{2v}{1+v} \tag{1}.$$

Re-arranging $(1)$ yields $$X\frac{dv}{dX}=\frac{v(1-v)}{v+1}$$ which is separable.

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You need to solve $\begin{cases} 2y_0+6=0 \\ x_0+y_0+1=0\end{cases}$ in order to find $x_0$ and $y_0$

You should find that $x_0=2$ and $y_0=-3.$