Question

Solve the following rational equations
$2(y+3)-xy\frac{dy}{dx}=0$ given that $y\left(1\right)=-2$

Answer 1

Given that, $2(y+3)-xy\frac{dy}{dx}=0$
$\Rightarrow 2(y+3)=xy\frac{dy}{dx}$
$\Rightarrow 2\frac{dx}{x}=\left(\frac{y}{(y+3)}\right)dy$
$\Rightarrow 2\frac{dx}{x}=\left(\frac{y+3-3}{(y+3)}\right)dy$
$\Rightarrow 2\frac{dx}{x}=(1-\frac{3}{(y+3)})dy$
On integrating both sides, we get
$2\frac{dx}{x}=\int (1-\frac{3}{(y+3)})dy$
$2\log x=y-3\log (y+3)+C...\left(i\right)$
When $x=1andy=-2$, then
$2\log 1=-2-3\log (-2+3)+C$
$\Rightarrow 2.0=-2-3.0+C$
$\Rightarrow C=2$
On substituting the value of $C$ in Eq. $\left(i\right)$, we get
$2\log x=y-3\log (y+3)+2$
$\Rightarrow 2\log x+3\log (y+3)=y+2$
$\Rightarrow \log x^2+\log (y+3{)}^3=(y+2)$
$\Rightarrow \log x^2(y+3{)}^3=y+2$
$\Rightarrow x^2(y+3{)}^3=e^{y+2}$