Solve dy\dx = {2y+3}/{4x+5} by separation of variables

(1) $\begin{equation*} \frac{dy}{dx} = \left(\frac{2y+3}{4x+5}\right)^2 \end{equation*}$

To solve this differential equation, we can separate the variables by putting all the y terms on one side and all the x terms on the other side:

(2) $\begin{equation*}\frac{1}{(2y+3)^2} \frac{dy}{dx} = \frac{1}{(4x+5)^2}\end{equation*}$

Now we can integrate both sides with respect to their respective variables:

(3) $\begin{equation*}\int \frac{1}{(2y+3)^2} \frac{dy}{dx} dx = \int \frac{1}{(4x+5)^2} dx\end{equation*}$

On the left side, we can use the substitution $u = 2y+3$ , $du/dx = 2(dy/dx)$ , so that we have:

(4) $\begin{equation*}\int \frac{1}{u^2} \frac{du}{2} = -\frac{1}{2u} + C_1\end{equation*}$

On the right side, we can use the substitution $v = 4x+5$ , $dv/dx = 4$ , so that we have:

(5) $\begin{equation*}\int \frac{1}{v^2} dv = -\frac{1}{v} + C_2\end{equation*}$

Putting these together, we have:

(6) $\begin{equation*}-\frac{1}{2(2y+3)} + C_1 = -\frac{1}{4x+5} + C_2\end{equation*}$

We can simplify this expression as

(7) $\begin{equation*}-\frac{1}{2(2y+3)} = -\frac{1}{4x+5} + C \end{equation*}$