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प्रश्न
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
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उत्तर
We have,
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
\[\Rightarrow \frac{dy}{dx}\left( x + 1 \right) = y\left( 1 - y \right)\]
\[ \Rightarrow \frac{dy}{y\left( 1 - y \right)} = \frac{dx}{\left( x + 1 \right)}\]
Integrating both sides, we get
\[\int\frac{dy}{y\left( 1 - y \right)} = \int\frac{dx}{x + 1}\]
\[ \Rightarrow \int\left( \frac{1}{y} + \frac{1}{1 - y} \right)dy = \int\frac{dx}{x + 1}\]
\[ \Rightarrow \log \left| y \right| - \log \left| 1 - y \right| = \log \left| x + 1 \right| + C . . . . . \left( 1 \right)\]
\[\text{ Since the curve passes throught the point }\left( 2, 2 \right),\text{ it satisfies the equation of the curve . }\]
\[ \Rightarrow \log \left| 2 \right| - \log \left| 1 - 2 \right| = \log \left| 2 + 1 \right| + C\]
\[ \Rightarrow C = \log \left| \frac{2}{3} \right|\]
\[\text{ Putting the value of C in }\left( 1 \right),\text{ we get }\]
\[\log \left| y \right| - \log \left| 1 - y \right| = \log \left| x + 1 \right| + \log \left| \frac{2}{3} \right|\]
\[ \Rightarrow \log \left| \frac{y}{\left( 1 - y \right)} \right| = \log \left| \frac{2\left( x + 1 \right)}{3} \right|\]
\[ \Rightarrow \left| \frac{y}{\left( 1 - y \right)} \right| = \left| \frac{2\left( x + 1 \right)}{3} \right|\]
\[ \Rightarrow \frac{y}{\left( 1 - y \right)} = \pm \frac{2\left( x + 1 \right)}{3}\]
\[ \Rightarrow \frac{y}{\left( 1 - y \right)} = \frac{2\left( x + 1 \right)}{3} or \frac{y}{\left( 1 - y \right)} = - \frac{2\left( x + 1 \right)}{3}\]
\[\text{ Here, given point }\left( 2, 2 \right)\text{ does not satisfy } \frac{y}{\left( 1 - y \right)} = \frac{2\left( x + 1 \right)}{3}\]
\[\text{ But it satisfy }\frac{y}{\left( 1 - y \right)} = - \frac{2\left( x + 1 \right)}{3}\]
\[ \therefore \frac{y}{\left( 1 - y \right)} = - \frac{2\left( x + 1 \right)}{3}\]
\[ \Rightarrow \frac{y}{\left( y - 1 \right)} = \frac{2\left( x + 1 \right)}{3}\]
\[ \Rightarrow 3y = 2\left( x + 1 \right)\left( y - 1 \right)\]
\[ \Rightarrow 3y = 2xy - 2x + 2y - 2\]
\[ \Rightarrow 2xy - 2x - y - 2 = 0\]
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