91Ó°ÊÓ

Given,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)\text{ }â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)$

Let the given equation is homogeneous then,

localid="1663934599031" $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{F}\left(\frac{y}{x}\right)\text{ }â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(2\right)$

And we have,

$\mathbf{f}\left(\mathrm{tx},\mathrm{ty}\right)\mathbf{=}\mathbf{f}\left(x,y\right)â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(3\right)$

Substitute $\mathbf{t}\mathbf{=}\frac{1}{x}$in the equation (3),

$\mathbf{f}\left(\left(\frac{1}{x}\right)\mathbf{x}\mathbf{,}\text{ }\left(\frac{1}{x}\right)\mathbf{y}\right)\mathbf{=}\mathbf{f}\left(x,y\right)$

Simplify the above equation,

$\begin{array}{l}\mathbf{f}\left(\mathbf{1}\mathbf{,}\text{ }\left(\frac{1}{x}\right)\mathbf{y}\right)\mathbf{=}\mathbf{f}\left(x,y\right)\\ \mathbf{f}\left(x,y\right)\mathbf{=}\mathbf{f}\left(\frac{y}{x}\right)â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(4\right)\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\end{array}$

From the equation (4) and (2),

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)$

Hence proved,

The equation$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)$ is homogeneous.

The equation $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)$ is homogeneous then,

L.H.S

$\begin{array}{l}\mathbf{f}\left(\mathrm{tx},\mathrm{ty}\right)\mathbf{=}\mathbf{f}\left(\frac{\mathrm{ty}}{\mathrm{tx}}\right)\\ \mathbf{f}\left(\mathrm{tx},\mathrm{ty}\right)\mathbf{=}\mathbf{f}\left(\frac{y}{x}\right)\end{array}$

From the equation (4),

$\mathbf{f}\left(\mathbf{tx}\mathbf{,}\text{ }\mathbf{ty}\right)\mathbf{=}\mathbf{f}\left(\mathbf{x}\mathbf{,}\text{ }\mathbf{y}\right)$

Hence proved,

$\begin{array}{l}\mathbf{M}\left(\mathbf{tx}\mathbf{,}\text{ â¶Ä‰}\mathbf{ty}\right)\mathbf{=}{\mathbf{t}}^n\mathbf{M}\left(\mathbf{x}\mathbf{,}\text{ â¶Ä‰}\mathbf{y}\right)â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \text{ }\left(1\right)\\ \mathbf{N}\left(\mathbf{tx}\mathbf{,}\text{ â¶Ä‰}\mathbf{ty}\right)\mathbf{=}{\mathbf{t}}^n\mathbf{N}\left(\mathbf{x}\mathbf{,}\text{ â¶Ä‰}\mathbf{y}\right)â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \text{ }\left(2\right)\end{array}$

Substitute $\mathbf{t}\mathbf{=}\frac{1}{x}$in the equation (1) and (2),

$\begin{array}{l}\mathbf{M}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\mathbf{=}\frac{1}{{\mathbf{x}}^n}\mathbf{M}\left(\mathbf{x}\mathbf{,}\text{ â¶Ä‰}\mathbf{y}\right)\\ \mathbf{M}\left(\mathbf{x}\mathbf{,}\text{ â¶Ä‰}\mathbf{y}\right)\mathbf{=}{\mathbf{x}}^n\mathbf{M}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \text{ }\left(3\right)\\ \mathbf{N}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\mathbf{=}\frac{1}{{\mathbf{x}}^n}\mathbf{N}\left(\mathbf{x}\mathbf{,}\text{ â¶Ä‰}\mathbf{y}\right)\\ \mathbf{N}\left(\mathbf{x}\mathbf{,}\text{ â¶Ä‰}\mathbf{y}\right)\mathbf{=}{\mathbf{x}}^n\mathbf{N}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\text{ }â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(4\right)\end{array}$

Now one has,

$\mathbf{M}\left(x,y\right)\mathbf{dx}\mathbf{+}\mathbf{N}\left(x,y\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Substitute the value of equation (3) and (4) in the above equation,

$\begin{array}{c}\mathbf{M}\left(x,y\right)\mathbf{dx}\mathbf{+}\mathbf{N}\left(x,y\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ {\mathbf{x}}^n\mathbf{M}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\mathbf{dx}\mathbf{+}{\mathbf{x}}^n\mathbf{N}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\end{array}$

Solve the above equation,

$\begin{array}{c}\mathbf{M}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\mathbf{dx}\mathbf{+}\mathbf{N}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \mathbf{M}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\mathbf{dx}\mathbf{=}\mathbf{-}\mathbf{N}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)\mathbf{dy}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\frac{\mathbf{M}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)}{\mathbf{N}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)}\end{array}$

one knows that,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{F}\left(\frac{y}{x}\right)\text{ â¶Ä‰â€‰}$

Then

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\frac{\mathbf{M}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)}{\mathbf{N}\left(\mathbf{1}\mathbf{,}\text{ â¶Ä‰}\frac{y}{x}\right)}\mathbf{=}\mathbf{F}\left(\frac{y}{x}\right)$

Hence proved,

$\mathbf{M}\left(x,y\right)\mathbf{dx}\mathbf{+}\mathbf{N}\left(x,y\right)\mathbf{dy}\mathbf{=}\mathbf{0}$is homogeneous.