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:
\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 $(0,1)$.

Question of the week #7

(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 $(0,1)$.

Any thoughts or solutions would be highly appreciable !!

Enjoy!

Question of the week #5 - the answer

Let us come to consider a detailed look at the solution of last week's question:
For the reader's convenience we repeat the statement of the question:
Question of the week #5:
(a). Use integration by parts to show that:
$$
\int sinx \cdot cosx \cdot e^{-sinx} dx = -e^{-sinx} \cdot \big( 1 + sinx \big) + c
$$
Now consider the following differential equation:
$$
\frac{dy}{dx}-y \cdot cosx = sinx \cdot cosx
$$
   (b). Determine the integration factor and find the general solution $y=f(x)$
   (c). Find the special solution satisfying $f(0)=-2$

Solution: (a). Noticing (by applying the chain rule of differentiation) that $\big( e^{-sinx} \big)' = e^{-sinx}(-sinx)' = - cosx \cdot e^{-sinx}$ we can proceed with a straightforward integration by parts:
$$
\begin{array}{c}
\int sinx \cdot cosx \cdot e^{-sinx} dx = -\int sinx \frac{d\big( e^{-sinx} \big)}{dx} dx = \\
     \\
= -\int sinx \big( e^{-sinx} \big)' dx = -sinx \cdot e^{-sinx} + \int e^{-sinx} \cdot cosx dx = \\
    \\
= -sinx \cdot e^{-sinx} - \int \frac{d\big( e^{-sinx} \big)}{dx} dx = -sinx \cdot e^{-sinx} - \int \big( e^{-sinx} \big)' dx = \\
       \\
= -sinx \cdot e^{-sinx} -  e^{-sinx} + c = - e^{-sinx} \big( 1 + sinx \big) + c
\end{array}
$$
(b). The given DE is $y'-y \cdot cosx = sinx \cdot cosx$. So we compute for the integration factor:

• $-\int cosx dx = -sinx + d$, where $d \in \mathbb{R}$ is a constant of integration. Since we need only one (actually: anyone) of the indefinite integrals -in order to find an integration factor- we can pick $d=0$
• The integration factor thus reads: $\mu(x) = e^{-sinx}$ 

(c). Multiplying both sides of the DE with the integration factor $\mu$ determined in (b) and using the result of (a) we get:
$$
\begin{array}{c}
e^{-sinx} \cdot y'- e^{-sinx} \cdot y \cdot cosx = e^{-sinx} \cdot sinx \cdot cosx  \Leftrightarrow  \\
     \\
\Leftrightarrow \big( e^{-sinx} \cdot y \big) ' = e^{-sinx} \cdot sinx \cdot cosx   \Leftrightarrow \\
     \\
\Leftrightarrow e^{-sinx} \cdot y  = \int e^{-sinx} \cdot sinx \cdot cosx dx \Leftrightarrow \\
    \\
\Leftrightarrow e^{-sinx} \cdot y = -sinx \cdot e^{-sinx} -  e^{-sinx} + c \Leftrightarrow   \\
     \\
\Leftrightarrow y = -sinx + c \cdot e^{sinx} - 1
\end{array}
$$
Hence, we have determined the general solution of the given DE. It is a family of functions parameterized by $c \in \mathbb{R}$. In order to single out that special solution satisfying $x=0$, $y=2$ we have to simply substitute these values in the expression of the general solution and solve the resulting expression for $c$:
$$
-2 = sin0 + c \cdot e^{sin0} - 1 \Leftrightarrow -2 = c - 1 \Leftrightarrow c = -1
$$
thus, the special solution is
$$
y = -sinx - e^{sinx} - 1
$$ 

Question of the week #5

This week's post comes from differential equations:
   (a). Use integration by parts to show that:
$$
\int sinx \cdot cosx \cdot e^{-sinx} dx = -e^{-sinx} \cdot \big( 1 + sinx \big) + c
$$
Now consider the following differential equation:
$$
\frac{dy}{dx}-y \cdot cosx = sinx \cdot cosx
$$
   (b). Determine the integration factor and find the general solution $y=f(x)$
   (c). Find the special solution satisfying $f(0)=-2$