Question

(x² + 3xy + y²) dx − x² dy = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(x^2+3xy+y^2\right)dx-x^{2}dy=0 \\ \Rightarrow \frac{dy}{dx}=\frac{x^2+3xy+y^2}{x^2} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equati}\mathrm{on}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{x^2+3v{x}^2+v^2{x}^2}{x^2} \\ \Rightarrow x\frac{dv}{dx}=1+3v+v^2-v \\ \Rightarrow x\frac{dv}{dx}=1+v^2+2v \\ \Rightarrow \frac{1}{1+v^2+2v}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{1+v^2+2v}dv=\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1}{{\left(1+v\right)}^2}dv=\int \frac{1}{x}dx \\ \Rightarrow -\frac{1}{\left(1+v\right)}=\log \left|x\right|+C \\ \Rightarrow \log \left|x\right|+\frac{1}{\left(1+v\right)}=-C \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \therefore \log \left|x\right|+\frac{x}{\left(x+y\right)}={\mathrm{C}}_1 \\ \mathrm{where} \\ {C}_1=-\mathrm{C} \\ \mathrm{Hence},\log \left|x\right|+\frac{x}{\left(x+y\right)}={\mathrm{C}}_1\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$