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