Show that (7.15) is a separable equation. [You may find it helpful to write.] $\int F\left(x\right)dx=f\left(x\right)$Thus solve (7.14) in terms of quadrature’s (that is, indicated integrations) as in Problem 2.

The given equation is $\frac{1}{2}m{v}^2=\int F\left(x\right)dx+\text{const}$.

The differential form of velocity is $v\left(x\right)=\frac{dx}{dt}$.

The given equation is separable:

$\frac{1}{2}m{v}^2=\int F\left(x\right)dx+\text{const}$

The integral on the RHS should be carried out for some variable other than$x$ , but up to$x$,along these lines:

${\int }_0^{x}F\left(y\right)dy$

The above integral as a function of$x$

$f\left(x\right)={\int }_0^{x}F\left(y\right)dy$

Now transform the initial equation into:

$\frac{1}{2}m{v}^2\left(x\right)=f\left(x\right)+A$

Divide the equation by$\frac{m}{2}$to obtain an expression for$v\left(x\right)$:

$v^2\left(x\right)=\frac{2}{m}\left[f\right(x)+A]\Rightarrow v\left(x\right)=\pm \sqrt{\frac{2}{m}\left[f\right(x)+A)}$

The definition of velocity as a time derivative of position$v\left(x\right)=\frac{dx}{dt}$and inserting it into the previous expression, so that transformation the initial condition into:

$$\begin{gathered} \frac{dx}{dt}=\pm \sqrt{\frac{2}{m}\left[f\right(x)+A]} \\ \frac{\pm dx}{\sqrt{\left.\frac{2}{m}\mid f\left(x\right)+A\right]}}=dt \end{gathered}$$

At last, integrate the previous equation and use the table integral$\int dx=x+$const on the$\text{RHS}$

to write the general form of its situation:

$\int \frac{\pm dx}{\sqrt{\frac{2}{m}\left[f\right(x)+A]}}=t+\text{const}$

Thus, the solution of the differential equation is data-custom-editor="chemistry" $\int \frac{\pm dx}{\sqrt{\frac{2}{m}\left[f\right(x)+A]}}=t+\text{const}$const.