# 1st order ODE problem

http://img440.imageshack.us/img440/7352/11unled.jpg

Having some trouble. Im not sure what d_t(u) and :u(with the . above it) means. Would appreciate some help. Thanks

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The "overdot" notation for the derivative with respect to time is due to Newton. The notation now has limited popularity. –  André Nicolas Oct 8 '11 at 22:39

The first several things listed are just different notations for the derivative of u with respect to t: $du/dt$, $d_t(u)$, and $\dot{u}$ are all the same thing.

You're asked to solve $u'-g(t)u=h(t)$ with initial condition $u(0)=u_0$.

The trick to solving such an equation is to multiply by some function so that the left hand side is the result of differentiating using the product rule: Say multiply by $X$ and get $Xu'-Xg(t)u = Xh(t)$. We want...

$$Xu'+X'u=\frac{d}{dt}\left[Xu\right] = Xu'-Xg(t)u$$

Thus we need $X'u=-Xg(t)u$ that is $X'=-Xg(t)$ so that $X=exp(-\int_0^t g(x)\,dx)$

Hopefully this will get you started. By the way this kind of equation is a "first order linear ODE". http://en.wikipedia.org/wiki/Linear_differential_equation (see first order)

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$\mathrm d_t u$ and $\dot u$ are both just alternative notations for $\frac{\mathrm du}{\mathrm dt}$.

In fact, that is what the $=:$ symbol means: "The symbol to the right of $=:$ is is hereby defined to be a name for the thing to the left".

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