## Thursday, January 16, 2014

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