Question

(2x² y + y³) dx + (xy² − 3x³) dy = 0

Answer 1

We have, 
\[ \left( 2 x^2 y + y^3 \right) dx + \left( x y^2 - 3 x^3 \right) dy = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2 x^2 y + y^3}{3 x^3 - x y^2}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{2v x^3 + v^3 x^3}{3 x^3 - v^2 x^3}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{2v + v^3}{3 - v^2}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{2v + v^3}{3 - v^2} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{2v + v^3 - 3v + v^3}{3 - v^2}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{2 v^3 - v}{3 - v^2}\]
\[ \Rightarrow \frac{3 - v^2}{2 v^3 - v}dv = \frac{1}{x}dx\]
Integrating both sides, we get 
\[\int\frac{3 - v^2}{2 v^3 - v}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow 3\int\frac{1}{2 v^3 - v}dv - \int\frac{v^2}{2 v^3 - v}dv = \int\frac{1}{x}dx . . . . . (1)\]
\[\text{ Considering }\frac{1}{2 v^3 - v} = \frac{1}{v\left( 2 v^2 - 1 \right)}, \]
\[\text{ let }\frac{1}{v\left( 2 v^2 - 1 \right)} = \frac{A}{v} + \frac{Bv + C}{2 v^2 - 1} . . . . . (2)\]
\[1 = A\left( 2 v^2 - 1 \right) + \left( Bv + C \right) v\]
\[ \Rightarrow 1 = 2A v^2 - A + B v^2 + Cv\]
Comparing the coeficients of both sides, we get
\[ \therefore 2A + B = 0 , C = 0\text{ and }A = - 1\]
\[ \Rightarrow - 2 + B = 0\]
\[ \Rightarrow B = 2\]
\[\text{Substituting }A = - 1, B = 2\text{ and }C = 0\text{ in }(2),\text{ we get }\]
\[\frac{1}{v\left( 2 v^2 - 1 \right)} = - \frac{1}{v} + \frac{2v}{2 v^2 - 1} . . . . . (3)\]
From (2) and (3), we get
\[3\int\left( - \frac{1}{v} + \frac{2v}{2 v^2 - 1} \right)dv - \int\frac{v}{2 v^2 - 1}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - 3\int\frac{1}{v}dv + 5\int\frac{v}{2 v^2 - 1}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - 3 \log \left| v \right| + \frac{5}{4}\log \left| 2 v^2 - 1 \right| = \log \left| x \right| + \log C\]
\[ \Rightarrow \frac{12 \log \left| \frac{1}{v} \right| + 5 \log \left| 2 v^2 - 1 \right|}{4} = \log \left| Cx \right|\]
\[ \Rightarrow \log \left| \frac{1}{v^{12}} \times \left( 2 v^2 - 1 \right)^5 \right| = \log \left| C^4 x^4 \right|\]
\[ \Rightarrow \left| \frac{1}{v^{12}} \times \left( 2 v^2 - 1 \right)^5 \right| = \left| C^4 x^4 \right|\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \left| \frac{x^{12}}{y^{12}} \times \left( \frac{2 y^2}{x^2} - 1 \right)^5 \right| = \left| C^4 x^4 \right|\]
\[ \Rightarrow \left| \left( \frac{2 y^2 - x^2}{x^2} \right)^5 \right| = \left| C^4 x^4 \times \frac{y^{12}}{x^{12}} \right|\]
\[\text{ Hence, }C^4 x^2 y^{12} = \left| \left( 2 y^2 - x^2 \right) \right|^5\text{ is the required solution .}\]