91Ó°ÊÓ

A first-order ordinary differential equation$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions$\mathbf{g}\left(x\right)$ that is a function of x alone and$\mathbf{h}\left(y\right)$ that is a function of y alone.

A separable differential equation can be expressed as $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{g}\left(x\right)\text{ }\mathbf{h}\left(y\right)$. By separating the variables, the equation follows. Then, on direct integration of both sides, the solution of the differential equation is determined.

The given equation is

$\mathrm{x}\text{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-4{\mathrm{v}}^2}{3\mathrm{v}}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)$

After separating the variables, equation (1) can be written as

$\frac{3\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(2\right)$

Integrate both sides of equation (2). It results,

$\begin{array}{c}∫\frac{3\mathrm{v}}{1-4{\mathrm{v}}^2}\mathrm{dv}=∫\frac{dx}{x}\\ -\frac{3}{8}∫\frac{-8\mathrm{v}\text{ }\mathrm{dv}}{1-4{\mathrm{v}}^2}=∫\frac{dx}{x}\\ -\frac{3}{8}∫\frac{\mathrm{d}\left(1-4{\mathrm{v}}^2\right)}{1-4{\mathrm{v}}^2}=∫\frac{dx}{x}\\ -\frac{3}{8}\ln \left(1-4{\mathrm{v}}^2\right)=\ln \text{ }\mathrm{x}+\ln \text{ }\mathrm{k}\left[\ln \text{ }\mathrm{k}=\mathrm{IntegratingConstant}\right]\end{array}$

$\begin{array}{c}\ln \left(1-4{\mathrm{v}}^2\right)=-\frac{8}{3}\ln \text{ }\mathrm{x}+\left(-\frac{8}{3}\right)\ln \text{ }\mathrm{k}\\ \ln \left(1-4{\mathrm{v}}^2\right)=\ln \text{ }{\mathrm{x}}^{-\frac{8}{3}}+\ln \text{ }\mathrm{C}\left[\ln \text{ }\mathrm{C}=-\frac{8}{3}\ln \text{ }\mathrm{k}=\mathrm{Constant}\right]\\ \ln \left(1-4{\mathrm{v}}^2\right)-\ln \text{ }\mathrm{C}=\ln \text{ }{\mathrm{x}}^{-\frac{8}{3}}\\ \ln \left(\frac{1-4{\mathrm{v}}^2}{\mathrm{C}}\right)=\ln \text{ }{\mathrm{x}}^{-\frac{8}{3}}\end{array}$

$\begin{array}{c}\frac{1-4{\mathrm{v}}^2}{C}={\mathrm{x}}^{-\frac{8}{3}}\\ 4{\mathrm{v}}^2=1-\mathrm{C}\text{ }{\mathrm{x}}^{-\frac{8}{3}}\\ \mathrm{v}=Â\pm \frac{\sqrt{1-{\mathrm{Cx}}^{-\frac{8}{3}}}}{2}\end{array}$

Therefore, the solution of the given equation is$\mathbf{v}\mathbf{=}\mathbf{Â\pm }\frac{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{Cx}}^{\mathbf{-}\frac{8}{3}}}}{2}$.