23.

Solve the following differential equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{x} ( 2\mathrm{y} - \mathrm{x} )}{\mathrm{x} ( 2\mathrm{y} + \mathrm{x})}$,   if y = 1 when x = 1

We need to solve the following differential equation

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{x} ( 2\mathrm{y} - \mathrm{x} )}{\mathrm{x} ( 2\mathrm{y} + \mathrm{x} ) } \\ \frac{{\displaystyle \mathrm{dy}}}{{\displaystyle \mathrm{dx}}} = \frac{{\displaystyle 2\mathrm{y} - \mathrm{x} }}{{\displaystyle 2\mathrm{y} + \mathrm{x} }} ........\left(1\right) \end{gathered}$$

It is a homogeneous differential equation.

Let  y = vx               ..........(2)

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \mathrm{v} + \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}} ..........\left(3\right)$

Substituting (2) and (3) in (1),  we get:

$$\begin{gathered} \mathrm{v} + \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{x} ( 2\mathrm{v} - 1 )}{\mathrm{x} ( 2\mathrm{v} + 1 )} \\ \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{ 2\mathrm{v} - 1 }{ 2\mathrm{v} + 1 } - \mathrm{v} \\ \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{ - 2{\mathrm{v}}^2 = \mathrm{v} - 1 }{ 2\mathrm{v} + 1 } \\ \left(\frac{2\mathrm{v} + 1}{- 2{\mathrm{v}}^2 + \mathrm{v} - 1 }\right) \mathrm{dv} = \left(\frac{1}{\mathrm{x}}\right) \mathrm{dx} \\ \left(\frac{2\mathrm{v} + 1}{2{\mathrm{v}}^2 - \mathrm{v} + 1 }\right) \mathrm{dv} = \left(- \frac{1}{\mathrm{x}} \right) \mathrm{dx} \end{gathered}$$

Integrating both sides,

$$\begin{gathered} \int \frac{1}{2}\left(\frac{4\mathrm{v} - 1 + 3}{2{\mathrm{v}}^2 - \mathrm{v} + 1}\right) \mathrm{dv} = \int \left(\frac{1}{\mathrm{x}}\right) \mathrm{dx} \\ \int \frac{{\displaystyle 1}}{{\displaystyle 2}}\left(\frac{{\displaystyle 4\mathrm{v} - 1 }}{{\displaystyle 2{\mathrm{v}}^2 - \mathrm{v} + 1}}\right) \mathrm{dv} + \int \frac{{\displaystyle 3}}{{\displaystyle 2}}\left(\frac{{\displaystyle 1 }}{{\displaystyle 2{\mathrm{v}}^2 - \mathrm{v} + 1}}\right) \mathrm{dv} = \int \left(- \frac{{\displaystyle 1}}{{\displaystyle \mathrm{x}}} \right) \mathrm{dx} \end{gathered}$$

$\int \frac{1}{2}\left(\frac{4\mathrm{v} - 1}{2{\mathrm{v}}^2 - \mathrm{v} + 1}\right) \mathrm{dv} + \int \frac{3}{4}\left[\frac{1}{{\mathrm{v}}^2 - {\displaystyle \frac{\mathrm{v}}{2}} + {\displaystyle \frac{1}{2}}}\right] = \int \left(-\frac{1}{\mathrm{x}}\right) \mathrm{dx}$

$$\begin{gathered} \frac{{\displaystyle 1}}{{\displaystyle 2}} \log \left| 2{\mathrm{v}}^2 - \mathrm{v} + 1\right| + \frac{{\displaystyle 3}}{{\displaystyle 4}} \int \left[\frac{{\displaystyle 1}}{{\displaystyle {\mathrm{v}}^2 - \frac{\mathrm{v}}{2} + \frac{1}{16} + \frac{7}{16}}}\right] \mathrm{dv} =- \log \left|\mathrm{x}\right| + \mathrm{C} \\ \frac{{\displaystyle 1}}{{\displaystyle 2}} \log \left| 2{\mathrm{v}}^2 - \mathrm{v} + 1\right| + \frac{{\displaystyle 3}}{{\displaystyle 4}} \int \left[\frac{{\displaystyle \mathrm{dv}}}{{\displaystyle {\left(\mathrm{v} - \frac{1}{4}\right)}^2+ {\left(\frac{\sqrt{7}}{4}\right)}^2}}\right] =- \log \left|\mathrm{x}\right| + \mathrm{C} \\ \frac{{\displaystyle 1}}{{\displaystyle 2}} \log \left| 2{\mathrm{v}}^2 - \mathrm{v} + 1\right| + \frac{{\displaystyle 3}}{{\displaystyle 4}} \mathrm{x} \frac{4}{\sqrt{7}} {\tan }^{-1}\left(\frac{\mathrm{v} - {\displaystyle \frac{1}{4}}}{{\displaystyle \frac{\sqrt{7}}{4}}}\right) = - \log \left|\mathrm{x}\right| + \mathrm{C} \\ \frac{{\displaystyle 1}}{{\displaystyle 2}} \log \left| 2{\mathrm{v}}^2 - \mathrm{v} + 1\right| + \frac{3}{\sqrt{7}} {\tan }^{-1}\left(\frac{4\mathrm{v} - 1}{{\displaystyle \sqrt{7}}}\right) = \mathrm{C}- \log \left|\mathrm{x}\right| \end{gathered}$$

$$\begin{gathered} \mathrm{Put} \mathrm{v} = \frac{\mathrm{y}}{\mathrm{x}} \\ \frac{1}{2} \log \left| 2 {\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2 - \left(\frac{\mathrm{y}}{\mathrm{x}}\right) + 1 \right| + \frac{3}{\sqrt{7}} {\tan }^{-1}\left[\frac{{\displaystyle \frac{4\mathrm{y}}{\mathrm{x}}} - 1}{\sqrt{7}}\right] = \mathrm{C} - \log \left|\mathrm{x}\right| \\ \frac{1}{2} \log \left|\frac{2{\mathrm{y}}^2 - \mathrm{xy} + {\mathrm{x}}^2}{{\mathrm{x}}^2}\right| + \frac{3}{\sqrt{7}} {\tan }^{-1} \left(\frac{4\mathrm{y} - \mathrm{x}}{\sqrt{7}\mathrm{x}}\right) = \mathrm{C} - \log \left|\mathrm{x}\right| ..........\left(4\right) \\ \mathrm{Now} \mathrm{y} = 1 \mathrm{when} \mathrm{x} = 1 \\ \frac{1}{2} \log \left|\frac{2(1{)}^2 - 1 \mathrm{x} 1 + 1^2}{1^2}\right| + \frac{3}{\sqrt{7}} {\tan }^{-1} \left(\frac{4 \mathrm{x} 1 - 1}{\sqrt{7} \mathrm{x} 1}\right) = \mathrm{C} - \log \left|1\right| \\ \frac{1}{2} \log 2 + \frac{3}{\sqrt{7}} {\tan }^{-1} \left(\frac{3}{\sqrt{7}}\right) = \mathrm{C} .................\left(5\right) \end{gathered}$$

Therefore, from (4) and (5) we get:

$$\begin{gathered} \frac{1}{2}\log \left|\frac{2{\mathrm{y}}^2 - \mathrm{xy} + {\mathrm{x}}^2}{{\mathrm{x}}^2}\right| + \frac{3}{\sqrt{7}}{\tan }^{-1} \left(\frac{4\mathrm{y} - \mathrm{x}}{\sqrt{7}{\displaystyle }{\displaystyle \mathrm{x}}}\right) \\ = \frac{{\displaystyle 1}}{{\displaystyle 2}}\log 2 + \frac{{\displaystyle 3}}{{\displaystyle \sqrt{7}}}{\tan }^{-1}\left(\frac{{\displaystyle 3}}{{\displaystyle \sqrt{7}}}\right) - \log \left|\mathrm{x}\right| \\ \frac{{\displaystyle 1}}{{\displaystyle 2}}\log \left|\frac{{\displaystyle 2{\mathrm{y}}^2 - \mathrm{xy} + {\mathrm{x}}^2}}{{\displaystyle {\mathrm{x}}^2}}\right| - \frac{{\displaystyle 1}}{{\displaystyle 2}}\log 2 + \log \left|\mathrm{x}\right| \\ = \frac{{\displaystyle 3}}{{\displaystyle \sqrt{7}}}\left[ {\tan }^{-1} \frac{{\displaystyle 3}}{{\displaystyle \sqrt{7}}} - {\tan }^{-1} \left(\frac{4\mathrm{y} - \mathrm{x}}{\sqrt{7}{\displaystyle }{\displaystyle \mathrm{x}}}\right)\right] \\ \frac{{\displaystyle 1}}{{\displaystyle 2}}\log \left|\frac{{\displaystyle 2{\mathrm{y}}^2 - \mathrm{xy} + {\mathrm{x}}^2}}{{\displaystyle {\mathrm{x}}^2}} {\mathrm{x}}^2 \right| = \frac{{\displaystyle 3}}{{\displaystyle \sqrt{7}}} {\tan }^{-1} \left(\frac{{\displaystyle \frac{3\mathrm{x} - 4\mathrm{y} + \mathrm{x}}{\sqrt{7} \mathrm{x}}}}{1 + {\displaystyle \frac{3 \mathrm{x} (4\mathrm{y} - \mathrm{x} )}{7\mathrm{x}}}}\right) \\ \frac{{\displaystyle 1}}{{\displaystyle 2}}\log \left|\frac{{\displaystyle 2{\mathrm{y}}^2 - \mathrm{xy} + {\mathrm{x}}^2}}{{\displaystyle 2}} \right| = \frac{{\displaystyle 3}}{{\displaystyle \sqrt{7}}} {\tan }^{-1} \left(\frac{{\displaystyle \frac{4 (\mathrm{x} - \mathrm{y} )}{\sqrt{7} \mathrm{x}}}}{ {\displaystyle \frac{7\mathrm{x} + 12\mathrm{y} - 3\mathrm{x} }{7\mathrm{x}}}}\right) \\ \frac{{\displaystyle 1}}{{\displaystyle 2}}\log \left|\frac{{\displaystyle 2{\mathrm{y}}^2 - \mathrm{xy} + {\mathrm{x}}^2}}{{\displaystyle 2}} \right| = \frac{{\displaystyle 3}}{{\displaystyle \sqrt{7}}} {\tan }^{-1} \left(\frac{\sqrt{7}{\displaystyle }{\displaystyle (}{\displaystyle }{\displaystyle \mathrm{x}}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle \mathrm{y}}{\displaystyle }{\displaystyle )}}{( \mathrm{x} + 3\mathrm{y} )}\right) \end{gathered}$$