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प्रश्न
\[y - x\frac{dy}{dx} = b\left( 1 + x^2 \frac{dy}{dx} \right)\]
उत्तर
We have,
\[y - x\frac{dy}{dx} = b\left( 1 + x^2 \frac{dy}{dx} \right)\]
\[ \Rightarrow y - b = \left( b x^2 + x \right)\frac{dy}{dx}\]
\[ \Rightarrow \left( \frac{1}{y - b} \right)dy = \left( \frac{1}{b x^2 + x} \right)dx\]
Integrating both sides, we get
\[\int\left( \frac{1}{y - b} \right)dy = \int\left( \frac{1}{b x^2 + x} \right)dx\]
\[ \Rightarrow \int\left( \frac{1}{y - b} \right)dy = \frac{1}{b}\int\left( \frac{1}{x^2 + \frac{1}{b}x} \right)dx\]
\[ \Rightarrow \int\left( \frac{1}{y - b} \right)dy = \frac{1}{b}\int\left( \frac{1}{x^2 + \frac{1}{b}x + \frac{1}{4 b^2} - \frac{1}{4 b^2}} \right)dx\]
\[ \Rightarrow \int\left( \frac{1}{y - b} \right)dy = \frac{1}{b}\int\frac{1}{\left( x + \frac{1}{2b} \right)^2 - \left( \frac{1}{2b} \right)^2}dx\]
\[ \Rightarrow \log \left| y - b \right| = \frac{1}{2 \times \frac{1}{2b}b}\log \left| \frac{x + \frac{1}{2b} - \frac{1}{2b}}{x + \frac{1}{2b} + \frac{1}{2b}} \right| + \log C\]
\[ \Rightarrow \log \left| y - b \right| = \log \left| \frac{bx}{bx + 1} \right| + \log C\]
\[ \Rightarrow y - b = \frac{Cbx}{bx + 1}\]
\[ \Rightarrow Cbx = \left( y - b \right)\left( bx + 1 \right)\]
\[ \Rightarrow x = k\left( y - b \right)\left( bx + 1 \right),\text{ where }k = \frac{1}{bC}\]
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