I'm really stuck on how to go about solving the following first order ODE; I've got little idea on how to approach it, and I'd really appreciate if someone could give me some hints and/or working for a solution so I can have a reference point on how to approach these sorts of problems.
The following is one of many ODE's I've gotten off a problem set I found in a textbook at a library:
$$y' = xe^{-\sin(x)} - y\cos(x)$$
Can anyone help?
On
The particular solution is $$y=\frac{1}{2}x^2 e^{-\sin x}$$
Then solve for the homogeneous solution.
Can you take it from here?
On
This kind of ODE should be solved as follows:
In your case it is $y'+y\cos(x)=0$ which has solution $y=c\cdot e^{-\sin(x)}$.
So, we have $y(x)=c(x)\cdot e^{-\sin(x)}$ and should substitute it into $y'+y\cos(x) = xe^{-\sin(x)}$.
This leads us to general solution in the form of $y(x) = \frac{1}{2}x^2e^{-\sin(x)}+ce^{-\sin(x)}$.
I always like to think of these type of ODE's in terms of the product rule. \begin{equation}x=y'e^{\sin(x)}+y\cos(x)e^{\sin(x)}=\left(ye^{\sin(x)}\right)' \end{equation} So integrating both sides and dividing by $e^{\sin(x)}$ yields\begin{equation}y=e^{-\sin(x)}\left(\frac{1}{2}x^2+c\right). \end{equation}