# Solve the following $x^{2}\large\frac{dy}{dx}$ =$y^{2}+2xy$ given that $y=1,$ when $x=1$

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)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)} Where f_1 and f_2 are homogeneous functions in x and y. • To solve we put y=vx and proceed. Step 1: x^2 \large\frac{dy}{dx}$$=y^2+2xy$ given $y=1$ when $x=1$
$\large\frac{dy}{dx}=\frac{y^2+2xy}{x^2}$ is homogeneous in x,y substitute $y=vx$=> $\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx} Step 2: The DE becomes v+ x \large\frac{dv}{dx}=\large\frac{v^2x^2+2x.vx}{x^2} \qquad= v^2 +2v \therefore x \large\frac{dv}{dx}$$=v^2+v$
$\large\frac{dv}{v(1+v)}=\large\frac{dx}{x}$
Step 3:
The variables are separated
$\int \large\frac{dv}{v(1+v)}$$=\int \large\frac{dx}{x}$$+\log c$
$\int \bigg(\large\frac{1}{v}-\frac{1}{1+v}\bigg)$$dv=\int \large\frac{dx}{x}$$+\log c$
$\log v -\log (1+v)=\log x +\log c$
$\log \large\frac{v}{x(1+v)}$$=\log c => v=x(1+v)c Substitute v=\large\frac{y}{x} \large\frac{y}{x}$$=x\bigg(1+\large\frac{y}{x}\bigg)c$
When $x=1,y=1$
$\therefore$ substitute $x=1,y=1$
$1=(1+1)c=>c=\large\frac{1}{2}$