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Question
At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.
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Solution
According to the question,
\[\frac{dy}{dx} = x + xy\]
\[ \Rightarrow \frac{dy}{dx} = x\left( 1 + y \right)\]
\[\Rightarrow \frac{1}{1 + y}dy = x dx\]
Integrating both sides with respect to x, we get
\[\int\frac{1}{1 + y}dy = \int x dx\]
\[ \Rightarrow \log \left| 1 + y \right| = \frac{x^2}{2} + C\]
\[\text{ Since the curve passes through }\left( 0, 1 \right),\text{ it satisfies the above equation . }\]
\[ \therefore \log \left| 1 + 1 \right| = \frac{0}{2} + C\]
\[ \Rightarrow C = \log 2\]
Putting the value of C, we get
\[\log \left| 1 + y \right| = \frac{x^2}{2} + \log 2\]
\[ \Rightarrow \log \left| \frac{1 + y}{2} \right| = \frac{x^2}{2}\]
\[ \Rightarrow \frac{1 + y}{2} = e^\frac{x^2}{2} \]
\[ \Rightarrow y + 1 = 2 e^\frac{x^2}{2} \]
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