How do I solve for the initial value with the given equation:
$xy'+y = e^{\sin x}\cos x$ for y$\left(π\right)$ = 1
I have tried to isolate the $y'$ by multiplying the whole equation by $\frac 1x$ and I got:
$y' + \frac 1x$$y= \frac 1x$$e^{\sin x}\cos x$
I thus got my integrating factor as:
$I\left(x\right)$= $e^{∫ \frac 1x}$
= $e^{\log x}$
= x
I'm not really sure how do I continue from here though.
On
$$\begin{align} & Notice\ that\ x{y}'+y={{\left( xy \right)}^{\prime }} \\ & {{\left( xy \right)}^{\prime }}={{e}^{\sin \left( x \right)}}\cos \left( x \right)dx \\ & \int{{{\left( xy \right)}^{\prime }}dx}=\int{{{e}^{\sin \left( x \right)}}\cos \left( x \right)dx}=\int{{{e}^{u}}du}={{e}^{u}}+C={{e}^{\sin \left( x \right)}}+C\quad by\ setting\ u=\sin \left( x \right) \\ & hence \\ & xy={{e}^{\sin \left( x \right)}}+C\quad OR\quad y=\frac{1}{x}\left( {{e}^{\sin \left( x \right)}}+C \right) \\ \end{align}$$
For a differential equation of the form, $$y'+{\rm P}(x) \cdot y = {\rm Q}(x)$$ Once you've got Integrating Factor, the solution is given by $$y \cdot {\rm (IF)} = \int {\rm Q}(x) \cdot {\rm ( IF)} {\rm d}x + {\rm C}$$
Can you continue now?