# Solve the following $(x^{2}+y^{2})\;dy$ =$xy \;dx$

Toolbox:
• First order , first degree DE
• Variable separable : Variables of a DE are rearranged to separate then, ie
• $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
• Can be written as $\large\frac{g_1 (y)}{g_2(y)}$$dy=-\large\frac{f_1(x)}{f_2(x)}$$dx$
• The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
• A D.E of first order and first degree is said to be homogeneous if it can be put in the form $\large\frac{dy}{dx}=f\bigg(\large\frac{y}{x}\bigg)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)} Where f_1 and f_2 are homogeneous functions in x and y. • To solve we put y=vx and proceed. Step 1: (x^2+y^2)dy=xy \;dx \large\frac{dy}{dx}=\large\frac{xy}{x^2+y^2} This is of the form \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)} Where f_1 and f_2 are homogeneous functions in x and y. Let y=vx=>\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
Step 2:
The D.E becomes $v+ x \large\frac{dv}{dx}=\frac{x-vx}{x^2+v^2x^2}$
$\qquad=\large\frac{v}{1+v^2}$
$x \large\frac{dv}{dx}=\large\frac{v}{1+v^2}$$-v \qquad=\large\frac{ v-v(1+v^2)}{1+v^2} \qquad= \large\frac{-v^3}{1+v^2} \large\frac{1+v^2}{v^3}$$dv=\large\frac{-dx}{x}$
Step 3:
The variables are separated
$\int \large\frac{1+v^2}{v^3}$$dv=-\int \large\frac{dx}{x}$$+\log c$
$\int \bigg(\large\frac{1}{\sqrt 3}+\frac{1}{v}\bigg)$$dv=-\int \large\frac{dx}{x}$$+\log c$