Question

$\left(x^2+y^2\right)\frac{dy}{dx}=8x^2-3xy+2y^2$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(x^2+y^2\right)\frac{dy}{dx}=8x^2-3xy+2y^2 \\ \Rightarrow \frac{dy}{dx}=\frac{8x^2-3xy+2y^2}{x^2+y^2} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{8x^2-3v{x}^2+2v^2{x}^2}{x^2+v^2{x}^2} \\ \Rightarrow x\frac{dv}{dx}=\frac{8-3v+2v^2}{1+v^2}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{8-4v+2v^2-v^3}{1+v^2} \\ \Rightarrow x\frac{dv}{dx}=\frac{4\left(2-v\right)+v^2\left(2-v\right)}{1+v^2} \\ \Rightarrow x\frac{dv}{dx}=\frac{\left(4+v^2\right)\left(2-v\right)}{1+v^2} \\ \Rightarrow \frac{1+v^2}{\left(4+v^2\right)\left(2-v\right)}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1+v^2}{\left(4+v^2\right)\left(2-v\right)}dv=\int \frac{1}{x}dx.....\left(1\right) \\ \mathrm{Let}\mathrm{us}\mathrm{consider}\mathrm{the}\mathrm{left}\mathrm{hand}\mathrm{side}\mathrm{of}\left(1\right). \\ \mathrm{Using}\mathrm{partial}\mathrm{fraction}, \\ \mathrm{Let}\frac{1+v^2}{\left(4+v^2\right)\left(2-v\right)}=\frac{Av+B}{4+v^2}+\frac{C}{2-v} \\ \Rightarrow 1+v^2=Av\left(2-v\right)+B\left(2-v\right)+C\left(4+v^2\right) \\ \Rightarrow 1+v^2=2Av-A{v}^2+2B-Bv+4C+C{v}^2 \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ 2A-B=0 \\ -A+C=1 \\ \&2B+4C=1 \\ \mathrm{Solving}\mathrm{these}\mathrm{three}\mathrm{equations},\mathrm{we}\mathrm{get} \\ A=\frac{-3}{8},B=\frac{-3}{4}\mathrm{and}C=\frac{5}{8} \\ \therefore \frac{1+v^2}{\left(4+v^2\right)\left(2-v\right)}=\frac{-{\displaystyle \frac{3}{8}}v-{\displaystyle \frac{3}{4}}}{4+v^2}+\frac{{\displaystyle \frac{5}{8}}}{2-v}.....\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ \int \frac{-{\displaystyle \frac{3}{8}}v-{\displaystyle \frac{3}{4}}}{4+v^2}+\frac{{\displaystyle \frac{5}{8}}}{2-v}=\int \frac{1}{x}dx \\ \Rightarrow -\frac{3}{8}\int \frac{v}{v^2+4}dv-\frac{3}{4}\int \frac{1}{v^2+4}dv+\frac{5}{8}\int \frac{1}{2-v}dv=\int \frac{1}{x}dx \\ \Rightarrow \frac{-3}{16}\log \left|v^2+4\right|-\frac{3}{4\times 2}\tan {}^{-1}\frac{v}{2}-\frac{5}{8}\log \left|2-v\right|=\log \left|x\right|+\log C \\ \Rightarrow -\frac{3}{4\times 2}\tan {}^{-1}\frac{v}{2}=\log \left|Cx{\left(2-v\right)}^{\frac{5}{8}}{\left(v^2+4\right)}^{\frac{3}{16}}\right| \\ \Rightarrow e^{-\frac{3}{8}\tan {}^{-1}\frac{v}{2}}=C\left|x{\left(2-v\right)}^{\frac{5}{8}}{\left(v^2+4\right)}^{\frac{3}{16}}\right| \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \Rightarrow e^{-\frac{3}{8}\tan {}^{-1}\frac{y}{2x}}=C\left|x{\left(2-\frac{y}{x}\right)}^{\frac{5}{8}}{\left({\frac{y}{x^2}}^2+4\right)}^{\frac{3}{16}}\right| \\ \Rightarrow e^{-\frac{3}{8}\tan {}^{-1}\frac{y}{2x}}=\mathrm{C}\left|x\times \frac{1}{x}{\left(2x-y\right)}^{\frac{5}{8}}{\left(y^2+4x^2\right)}^{\frac{3}{16}}\right| \\ \Rightarrow e^{-\frac{3}{8}\tan {}^{-1}\frac{y}{2x}}=\mathrm{C}{\left|2x-y\right|}^{\frac{5}{8}}{\left(y^2+4x^2\right)}^{\frac{3}{16}} \\ \mathrm{Hence},{\mathrm{e}}^{-\frac{3}{8}\tan {}^{-1}\frac{\mathrm{y}}{2\mathrm{x}}}=\mathrm{C}{\left|2x-y\right|}^{\frac{5}{8}}{\left(y^2+4x^2\right)}^{\frac{3}{16}}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$