# Solve: $2(y+3)-xy\Large\frac{dy}{dx}\normalsize =0,$given that $y(1)=-2$

Toolbox:
• A linear differential equation is of the form $\large\frac{dy}{dx}$$=f(x), can be solved by the method of seperating the variables and integrating • \int \large\frac{dx}{x+a}$$=\log |x+a|+c$
Step 1:
Given $2(y+3)-xy \large\frac{dy}{dx}$$=0 we can write this as xy \large\frac{dy}{dx}$$=2(y+3)$
Seperating the variables we get
$\large\frac{y}{y+3}$$dy=\large\frac{2}{x}$$dx$
Let us integrate on both sides
$\int \large\frac{y}{y+3}$$dy=\int\large\frac{2}{x}$$dx$
Consider $\int \large\frac{y}{y+3}$$dy Add +3 and subtract 3 to the numerator \int \large\frac{y+3-3}{y+3}$$dy$
Now seperate the terms
$\int \large\frac{y+3}{y+3}$$dy-\int \large\frac{3}{y+3}$$dy$
$=\int dy-3 \int \large\frac{1}{y+3}$$dy Integrating this we get y-3 \log |y+3| \int \large\frac{2}{x}$$dx=2 \log x$
$=\log x^2$
Hence the solution is
$y-3 \log (y+3)=\log x^2+c$
=>$y- \log (y+3)^3=\log x^2+c$
$y= \log[x^2(y+3)^3]+c$
Step 2:
Given $y(1)=-2$
This implies When $x=1,y=-2$
Hence substituting for x and y , we can evaluate the value of c
$-2=\log c[1(-2+3)^3]+c$
=>$-2=\log 1+c$ But $\log 1=0$
Therefore $c=-2$
Now substituting the value for c
$y=\log [x^2(y+3)^3]-2$
$y+2=\log_e[x^2(y+3)^3]$
Therefore $e^{\large y+2}$$=x^2(y+3)^3$