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I was solving a PDE using a change of variables. The result is

$$V(\xi,\eta) = e^{-\frac{1}{2}\xi}\int A(\eta)\;d\eta$$

where $A(\eta)$ is an arbitrary function of $\eta$. I had originally used the change of variables

\begin{align} \xi&=x+at\\\eta&=x+bt \end{align}

so that $u(x(\xi,\eta),t(\xi,\eta)) = V(\xi,\eta)$. Now I want the answer in terms of $x$ and $t$. When I try to make the change of variables back, I get

$$u(x,t)=e^{-\frac{1}{2}(x+at)}\int A(x+bt)\;d(x+bt)$$

But now I have a problem! How can I write $d(x+bt)$ in terms of $dx$ and/or $dt$?

Here would be my guess: $d(x+bt) = dx+b\,dt$, in which case

$$\int A(x+bt)\;d(x+bt)=\int A(x+bt)\;dx + \int A(x+bt)\;dt.$$

This seems sketchy to me. Is this correct?

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    $\begingroup$ If $b$ is constant then your good. Only one thing, when you integrated you forgot the $b$ out in front. $\endgroup$ – Ahmed S. Attaalla Jun 13 '16 at 23:27
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If $b$ and $t$ are independent of $x$

$$\frac{d(x+bt)}{dx}=1$$

So:

$$d(x+bt)=dx$$

If it is not independent then,

$$\frac{d(x+bt)}{dx}=1+\frac{d(bt)}{dx}$$

Thus,

$$d(x+bt)=dx+d(bt)$$

If $b$ is constant with a change in $x$ then,

$$d(x+bt)=dx+bdt$$

If $b$ is not constant then through the product rule we have,

$$d(x+bt)=dx+bdt+tdb$$

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