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प्रश्न
Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.
उत्तर
According to the question,
\[\frac{dy}{dx} = x + xy\]
\[ \Rightarrow \frac{dy}{dx} = x\left( 1 + y \right)\]
\[ \Rightarrow \frac{1}{y + 1}dy = x dx\]
Integrating both sides, we get
\[\int\frac{1}{y + 1}dy = \int x dx\]
\[ \Rightarrow \log \left| y + 1 \right| = \frac{x^2}{2} + \log C\]
\[ \Rightarrow \log \left| \frac{y + 1}{C} \right| = \frac{x^2}{2}\]
\[ \Rightarrow y + 1 = C e^\frac{x^2}{2} \]
Since, the curve passes through (0, 1)
It satisfies the equation of the curve.
\[ \therefore 1 + 1 = C e^0 \]
\[ \Rightarrow C = 2\]
Puting the value of `C` in the equation of the curve, We get
\[ y + 1 = 2 e^\frac{x^2}{2} \]
\[ \Rightarrow y = - 1 + 2 e^\frac{x^2}{2}\]
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