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
$$
y = -sinx - e^{sinx} - 1
$$