# Differential equation intermezzo!

Posted on January 25, 2011 by phimuemue

Hi all,

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

$$\frac{d}{dt}x(t)=-\frac{x(t)}{t}+sin(t)$$,

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.