$\bullet$ Last week's exercise was the following:

**Exercise**

**(a).** Find the general solution of the following differential equation:

$$\frac{dy}{dx}lnx+\frac{y}{x}= cotx$$

**(b). **Find also a special solution coming through $(0,1)$.

$\bullet$ In today's post let us see how this could have been worked out:

**Solution:**

**(a). **Let us begin by noticing that $(lnx)'=\frac{1}{x}$. Thus we have that:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}lnx+\frac{y}{x}= cotx \ \ \ \Leftrightarrow \ \ \ y' lnx + y (lnx)' = cotx \Leftrightarrow$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Leftrightarrow (y lnx)' = \frac{cosx}{sinx} \ \ \ \Leftrightarrow \ \ \ (y lnx)' = \frac{(sinx)'}{sinx} \Leftrightarrow $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Leftrightarrow (y lnx)' = \big( ln(sinx) \big)' \ \ \ \Leftrightarrow \ \ \ y lnx = ln(sinx) + c$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Leftrightarrow lnx^{y} = ln(sinx)+c \ \ \ \Leftrightarrow \ \ \ x^{y} = d sinx$

where $x > 0$ and both $c$ and $d=e^{c}$ are integration constants.

So the general solution consists of all the functions defined (in implicit form) by the parametric family of equations:

\begin{equation} \label{implgensol}

x^{y} = d sinx

\end{equation}

where $x > 0$.

**(b). **In order to determine the special solution coming through $(0,1)$ we have to substitute $x=0$, $y=1$ in the general solution \eqref{implgensol}, getting: $ \ \ \ 0=d sin0$.

Thus, all solutions \eqref{implgensol}, for any $d \in \mathbb{R}$ are coming through $(0,1)$.