- #1

bdw1386

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## Homework Statement

Differentiate the following expression with respect to t:

[tex]

exp[\int_{0}^{t}d\tau \lambda(\tau)]P(t)+R\int_0^tds(-\frac{dP(s)}{ds})exp[\int_0^sd\tau\lambda(\tau)]

[/tex]

## Homework Equations

N/A

## The Attempt at a Solution

Using the product rule and the FTC on both terms:

[tex]

exp[\int_{0}^{t}d\tau \lambda(\tau)]\frac{dP(t)}{dt}+P(t)exp[\int_{0}^{t}d\tau \lambda(\tau)]\lambda(t)

+

R\int_0^tds(-\frac{dP(s)}{ds})exp[\int_0^sd\tau\lambda(\tau)](0)

+

R(-\frac{dP(t)}{dt})exp[\int_0^sd\tau\lambda(\tau)]

[/tex]

The third term falls out, so we get:

[tex]

exp[\int_{0}^{t}d\tau \lambda(\tau)]\frac{dP(t)}{dt}+P(t)exp[\int_{0}^{t}d\tau \lambda(\tau)]\lambda(t)

+

R(-\frac{dP(t)}{dt})exp[\int_0^sd\tau\lambda(\tau)]

[/tex]

The textbook matches my solution exactly EXCEPT that the integral in the last term goes from 0 to t, rather than 0 to s. I couldn't figure out why the s changed to a t. The final exp[.] expression comes from the product rule and is simply copied from the original equation:

[tex]

x=R\int_0^tds(-\frac{dP(s)}{ds})

[/tex]

[tex]

y=exp[\int_0^sd\tau\lambda(\tau)]

[/tex]

[tex]

\frac{d}{dt}[xy] = x\frac{dy}{dt}+y\frac{dx}{dt} = y\frac{dx}{dt}

[/tex]

because [itex]\frac{dy}{dt}[/itex] = 0.

What am I missing?