Question

(2x+3)dy/dx -y+(2x+3)

Answer 1

$(2x+3)\frac{dy}{dx}-y=0\\ \Rightarrow \frac{dy}{dx}=\frac{y}{2x+3}\\ \Rightarrow \frac{dy}{y}=\frac{dx}{2x+3}$

Integrating both sides, we get:

$ln(y)=\frac{ln(2x+3)}{2}+ln(c)\\ \Rightarrow 2ln(y)=ln(2x+3)+2ln(c)\\ \Rightarrow ln(y^2)=ln(2x+3)+ln(c^2)\\ \Rightarrow ln(y^2)=ln(c^2(2x+3))\\ \therefore y^2=c^2(2x+3)\\ \Rightarrow y^2=C(2x+3)$