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Expert-verified Found in: Page 46 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # In problem 7-16, solve the equation. ${\mathbf{x}}\frac{\mathrm{dv}}{\mathrm{dx}}{\mathbf{=}}\frac{\mathbf{1}\mathbf{-}\mathbf{4}{\mathbf{v}}^{2}}{3v}$

The solution of the given differential equation is ${\mathbf{v}}{\mathbf{=}}{\mathbf{±}}\frac{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{Cx}}^{\mathbf{-}\frac{8}{3}}}}{2}$.

See the step by step solution

## Step 1: Concept of Separable Differential Equation

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.

## Step 2: Solution of the Equation

The given equation is

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

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

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

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

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

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

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

Therefore, the solution of the given equation is ${\mathbf{v}}{\mathbf{=}}{\mathbf{±}}\frac{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{Cx}}^{\mathbf{-}\frac{8}{3}}}}{2}$. ### Want to see more solutions like these? 