## Friday, July 4, 2014

### Question of the week #8

Given that
$$siny+cosy = x$$
find the first $\frac{dy}{dx}$ and the second derivative $\frac{d^{2}y}{dx^{2}}$ as functions of x.

Hint: It deserves to be persistent (and to use some trigonometric identities as well) to get rid of the $y$ !

## Thursday, July 3, 2014

### Question of the week #7: the answer

$\bullet$ 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:
\label{implgensol}
x^{y} = d sinx

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 $(0,1)$.