I need help justifying the jump from the expression on the left side of the equal sign, to the right.
\begin{array},\int_0^L \mathrm{d}x \ \frac{\partial y}{\partial t} \frac{\partial}{\partial x} (T(x) \frac{\partial y}{\partial t})\, = \int_0^L \!\mathrm{d}x \ T(x) \frac{\partial y}{\partial t} \frac{\partial}{\partial x} (\frac{\partial y}{\partial t})\, \end{array}
The preceding passage - in the book from which the problem hails - claims to use integration by parts to do so, but I can't seem to cancel out the right terms after employing the product rule... \begin{array}1 \int_0^L \! \mathrm{d}x \ \frac{\partial}{\partial x} [ \frac{\partial T}{\partial x} \frac{\partial y}{\partial x} + T \frac{\partial ^2 y}{\partial x^2}] \end{array} Any and all help is appreciated!
Thanks.
Integration by parts allows to switch a derivative from one factor to another...
$$ \int_a^b u'(x) v(x) dx = \left[u(x)v(x)\right]_a^b - \int_a^b u(x) v'(x) dx $$
Are you sure about the formula you posted? There should be a minus sign in the RHS. Also, since there are no boundary terms on the RHS, there must be some assumption on $T$ or $y$ that makes them go away.