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# Solve the following $(x^{2}+y^{2})dx+3xy dy$=$0$

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A)
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)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)} Where f_1 and f_2 are homogeneous functions in x and y. • To solve we put y=vx and proceed. Step 1: (x^2+y^2)dx+3xydy=0 \large\frac{dy}{dx} =\frac{-(x^2+y^2)}{3xy} is homogeneous in x,y y=vx=> \large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx}$
Step 2:
The D.E becomes
$v+ x \large\frac{dv}{dx}=\large\frac{-(x^2+v^2x^2)}{3x.vx}$
$v+ x \large\frac{dv}{dx}=\large\frac{-(1+v^2)}{3.v}$
$x \large\frac{dv}{dx}=-\bigg(\large\frac{1+v^2}{3v}\bigg)$$-v \qquad= \large\frac{-1-v^2-3v^2}{3v} \qquad= \large\frac{-(1+4v^2)}{3v} \large\frac{3v}{1+4v^2}$$dv=\large\frac{-dx}{x}$
Step 3:
The variables are separated
$\int \large\frac{3v}{1+4v^2}$$dv=-\int \large\frac{dx}{x}$$+\log c_1$
$\large\frac{3}{8} \int \large\frac{8v}{1+4v^2}$$dv = - \int\large\frac{dx}{x}$$+\log c_1$
$\large\frac{3}{8}$$\log (1+4v^2)+\log x=\log c_1 (1+4v^2)^{3/8}x=c_1 Substitute v=\large\frac{y}{x} \bigg(1+4 \large\frac{y^2}{x^2}\bigg)^{3/8}$$x=c_1$
$(x^2+4y^2)^{3/8}. x^{1/4}=c_1$
$(x^2+4y^2)^3.x^2=c$ is the G.S