Differential equation intermezzo!

Posted on January 25, 2011 by phimuemue

Hi all,

I've encountered (and solved) the following differential equation:


which can be easily transformed into:

\(x'(t) \cdot t - t\cdot \sin(t)=-x(t)\)

Integrating both sides yields:

\(\int_0^t x'(u)\cdot u \ du - \int_0^t t \sin(u) \ du = \int_0^t -x(u)\ du\)

We can transform the above to:

\(\int_0^t x'(u)\cdot u \ du + \int_0^t x(u)\ du = \int_0^t t \sin(u) \ du\)

The left side of the above can be evaluated to \([x(u)\cdot u]_0^t\) (via partial integration) and the right side evaluates to \(\sin(t) - t\cdot\cos(t)\) (also via partial integration). Thus, we obtain, that

\(x(t)\cdot t=\sin(t) - t\cdot\cos(t)\), from which we conclude that \(x(t)=\frac{\sin(t) - t\cdot\cos(t)}{t}\). That's it.