= Solution
Assume small transverse slopes, a flexible string, constant <tension> $T$ and uniform <linear mass density> $\rho$. For an element of equilibrium length $dx$, the vertical <force> is the difference of the vertical tension components at its ends. To first order in slope,
$$
F_y=T\big[y_x(x+dx,t)-y_x(x,t)\big]=T y_{xx}\,dx.
$$
Its <mass> is $\rho\,dx$ and its transverse <acceleration> is $y_{tt}$. <Newton's second law> therefore gives $\rho\,dx\,y_{tt}=T\,dx\,y_{xx}$, or
$$
\boxed{y_{xx}=\frac1{c^2}y_{tt},\qquad c=\sqrt{\frac{T}{\rho}}.}
$$
This is the one-dimensional <wave equation> for small-amplitude <transverse waves>. Changes in string length and horizontal <acceleration> are second order in slope and are neglected in this linear approximation.
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