Question

Use the method discussed under “Homogeneous Equations” to solve problems 9-16. $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{y}\mathbf{(}\mathbf{lny}\mathbf{-}\mathbf{lnx}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{x}}$

Short answer

Homogeneous equation for the given equation is $\mathbf{y}\mathbf{=}{\mathbf{xe}}^{\mathbf{Cx}}$.

Step-by-step solution

General form of Homogeneous equation

If the right-hand side of the equation$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$ can be expressed as a function of the ratio$\frac{\mathrm{y}}{\mathrm{x}}$ alone, then we say the equation is homogeneous.

Evaluate the given equation

Given, $\frac{d\mathrm{y}}{d\mathrm{x}}=\frac{\mathrm{y}(\mathrm{lny}-\mathrm{lnx}+1)}{\mathrm{x}}$.

Evaluate it.

Since,$\ln \left(\frac{\mathrm{M}}{\mathrm{N}}\right)=\ln \mathrm{M}-\ln \mathrm{N}$

localid="1655201008520" $\begin{array}{rcl}\frac{d\mathrm{y}}{d\mathrm{x}}& =& \frac{\mathrm{y}(\mathrm{lny}-\mathrm{lnx}+1)}{\mathrm{x}}\\ \frac{d\mathrm{y}}{d\mathrm{x}}& =& \frac{\mathrm{y}(\ln {\displaystyle \left(\frac{\mathrm{y}}{\mathrm{x}}\right)}+1)}{\mathrm{x}}\\ & =& \frac{\mathrm{y}}{\mathrm{x}}\left(\ln \left(\frac{\mathrm{y}}{\mathrm{x}}\right)+1\right)\\ & =& \frac{\mathrm{y}}{\mathrm{x}}\ln \left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\frac{\mathrm{y}}{\mathrm{x}}\end{array}$

Substitution method

Let us take $\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}}$.

Then $\mathrm{y}=\mathrm{vx}$.

By Differentiating,

$\begin{array}{rcl}\frac{d\mathrm{y}}{d\mathrm{x}}& =& \mathrm{v}+\mathrm{x}\frac{d\mathrm{v}}{d\mathrm{x}}\\ \mathrm{vln}\mathrm{v}+\mathrm{v}& =& \mathrm{v}+\mathrm{x}\frac{d\mathrm{v}}{d\mathrm{x}}\\ \mathrm{vln}\mathrm{v}& =& \mathrm{x}\frac{d\mathrm{v}}{d\mathrm{x}}\\ \frac{1}{\mathrm{vln}\mathrm{v}}\mathrm{dv}& =& \frac{1}{\mathrm{x}}\mathrm{dx}\end{array}$

Integrate the equation

Now, integrate on both sides.

$\begin{array}{rcl}\int \frac{1}{\mathrm{vln}\mathrm{v}}\mathrm{dv}& =& \int \frac{1}{\mathrm{x}}\mathrm{dx}\\ \int \frac{1}{\mathrm{vln}\mathrm{v}}\mathrm{dv}& =& \ln \left|\mathrm{x}\right|+{\mathrm{C}}_1\end{array}$

Integrate$\begin{array}{rcl}& & \int \frac{1}{\mathrm{vln}\mathrm{v}}\mathrm{dv}\end{array}$separately.

Let us take $\mathrm{w}=\ln \mathrm{v}$. Then,$\mathrm{dv}=\mathrm{v}\mathrm{dw}$

Now,

$\begin{array}{rcl}\int \frac{1}{\mathrm{vw}}\mathrm{vdw}& =& \int \frac{1}{\mathrm{w}}\mathrm{dw}\\ & =& \ln \left|\mathrm{w}\right|\end{array}$

Substitute $\mathrm{w}=\ln \mathrm{v}$.

$\begin{array}{rcl}\int \frac{1}{v\ln v}dv& =& \ln \left|w\right|\\ & =& \ln \left|\ln v\right|\end{array}$

Then,

$\begin{array}{rcl}\ln \left|\ln \mathrm{v}\right|& =& \ln \left|\mathrm{x}\right|+{\mathrm{C}}_1\\ \ln \mathrm{v}& =& {\mathrm{e}}^{\ln \left|\mathrm{x}\right|+{\mathrm{C}}_1}\\ \ln \mathrm{v}& =& {\mathrm{xe}}^{{\mathrm{C}}_1}\\ \ln \mathrm{v}& =& {\mathrm{xC}}_2\\ \mathrm{v}& =& {\mathrm{e}}^{{\mathrm{xC}}_2}\\ \mathrm{v}& =& {\mathrm{e}}^{\mathrm{xC}}\end{array}$

Substitute $\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}}$

localid="1655200766733" $\begin{array}{rcl}\frac{\mathrm{y}}{\mathrm{x}}& =& {\mathrm{e}}^{\mathrm{Cx}}\\ \mathrm{y}& =& {\mathrm{xe}}^{\mathrm{Cx}}\end{array}$

Therefore, Homogeneous equation for the given equation is$\mathbf{y}\mathbf{=}{\mathbf{xe}}^{\mathbf{Cx}}$.