91影视

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.

Given,$(3{\mathrm{x}}^2-{\mathrm{y}}^2)\mathrm{dx}+(\mathrm{xy}-{\mathrm{x}}^3{\mathrm{y}}^{-1})\mathrm{dy}=0$.

Evaluate it.

$\begin{array}{rcl}(3{\mathrm{x}}^2-{\mathrm{y}}^2)\mathrm{dx}+(\mathrm{xy}-{\mathrm{x}}^3{\mathrm{y}}^{-1})\mathrm{dy}& =& 0\\ (\mathrm{xy}-{\mathrm{x}}^3{\mathrm{y}}^{-1})\mathrm{dy}& =& -(3{\mathrm{x}}^2-{\mathrm{y}}^2)\mathrm{dx}\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& -\frac{3{\mathrm{x}}^2-{\mathrm{y}}^2}{\mathrm{xy}-{\mathrm{x}}^3{\mathrm{y}}^{-1}}\\ & =& \frac{{\mathrm{y}}^2-3{\mathrm{x}}^2}{\mathrm{xy}-{\mathrm{x}}^3{\mathrm{y}}^{-1}}\end{array}$

Now take LCM.

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{{\mathrm{y}}^3-3{\mathrm{yx}}^2}{{\mathrm{xy}}^2-{\mathrm{x}}^3}\\ & =& \frac{{\displaystyle \frac{{\mathrm{y}}^3}{{\mathrm{x}}^3}}-{\displaystyle \frac{3\mathrm{y}}{\mathrm{x}}}}{{\displaystyle \frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}}-1}\\ & =& \frac{{\left({\displaystyle \frac{\mathrm{y}}{\mathrm{x}}}\right)}^3-3\left({\displaystyle \frac{\mathrm{y}}{\mathrm{x}}}\right)}{{\left({\displaystyle \frac{\mathrm{y}}{\mathrm{x}}}\right)}^2-1}\end{array}$

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

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

By Differentiating,

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{{\mathrm{v}}^3-3\mathrm{v}}{{\mathrm{v}}^2-1}& =& \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{{\mathrm{v}}^3-3\mathrm{v}}{{\mathrm{v}}^2-1}-\mathrm{v}& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\\ \frac{{\mathrm{v}}^3-3\mathrm{v}-{\mathrm{v}}^3+\mathrm{v}}{{\mathrm{v}}^2-1}& =& \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}\end{array}$

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

Now, integrate on both sides.

$\begin{array}{rcl}鈭�\left(\mathrm{v}-\frac{1}{\mathrm{v}}\right)\mathrm{dv}& =& -鈭�\frac{2}{\mathrm{x}}\mathrm{dx}\\ \frac{{\mathrm{v}}^2}{2}-\ln \left|\mathrm{v}\right|& =& -2\ln \left|\mathrm{x}\right|+\mathrm{C}\\ {\mathrm{v}}^2-2\ln \left|\mathrm{v}\right|& =& -4\ln \left|\mathrm{x}\right|+2\mathrm{C}\\ {\mathrm{v}}^2-\ln \left|{\mathrm{v}}^2\right|+\ln \left|{\mathrm{x}}^4\right|& =& \mathrm{C}\\ & & \end{array}$

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

$\begin{array}{rcl}{\left(\raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\right)}^2-\ln \left|{\left(\raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\right)}^2\right|+\ln \left|{\mathrm{x}}^4\right|& =& \mathrm{C}\\ \frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}-\ln \left|\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}\right|+\ln \left|{\mathrm{x}}^4\right|& =& \mathrm{C}\\ \frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}-\ln \left|\frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}\frac{1}{{\mathrm{x}}^4}\right|& =& \mathrm{C}\\ \frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}+\ln \left|\frac{{\mathrm{x}}^6}{{\mathrm{y}}^2}\right|& =& \mathrm{C}\end{array}$

Therefore, Homogeneous equation for the given equation is $\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{ln}\mathbf{\left|}\frac{{\mathbf{x}}^{\mathbf{6}}}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right|}\mathbf{}\mathbf{=}\mathbf{}\mathbf{C}$.