# Thread: IVP using Leibniz formula

1. ## IVP using Leibniz formula

How does one formulate this integral equation as an initial value problem?

u(t)=1 + integral from 0 to t (s ln (s/t) u(s) ds)

we're supposed to use Leibniz formula but I'm not sure how to apply it. Thanks!

2. So your integral equation is

$\displaystyle u(t)=1+\int_{0}^{t}s\ln\left(\frac{s}{t}\right)u(s )\,ds,$

and you need to convert to an IVP using the Leibniz formula, which is, in this case,

$\displaystyle \frac{d}{dt}\int_{a(t)}^{b(t)}f(s,t)\,ds=\frac{db( t)}{dt}\,f(b(t),t)-\frac{da(t)}{dt}\,f(a(t),t)+\int_{a(t)}^{b(t)} \frac{\partial }{\partial t}\,f(s,t)\,ds.$

So, why not just differentiate the original integral equation with respect to t, and use the Leibniz formula on the integral. What do you get?

3. Originally Posted by Ackbeet

$\displaystyle u(t)=1+\int_{0}^{t}s\ln\left(\frac{s}{t}\right)u(s )\,ds,$

and you need to convert to an IVP using the Leibniz formula, which is, in this case,

$\displaystyle \frac{d}{dt}\int_{a(t)}^{b(t)}f(s,t)\,ds=\frac{db( t)}{dt}\,f(b(t),t)-\frac{da(t)}{dt}\,f(a(t),t)+\int_{a(t)}^{b(t)} \frac{\partial }{\partial t}\,f(s,t)\,ds.$

So, why not just differentiate the original integral equation with respect to t, and use the Leibniz formula on the integral. What do you get?
Why not just note that $\displaystyle \displaystyle \int _0^t s\log\left(\frac{s}{t}\right)u(s)\text{ }ds=\int_0^t s\log(s)u(s)\text{ }ds-\log(t)\int_0^t su(s)\text{ }ds$ from which all you need is the product rule and the FTC?

4. Originally Posted by Drexel28
Why not just note that $\displaystyle \displaystyle \int _0^t s\log\left(\frac{s}{t}\right)u(s)\text{ }ds=\int_0^t s\log(s)u(s)\text{ }ds-\log(t)\int_0^t su(s)\text{ }ds$ from which all you need is the product rule and the FTC?
True enough, but if the OP'er must use the Leibniz formula because of class restraints, then we're back to square one.