Question

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.

Answer 1

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

\[ y + 1 = 2 e^\frac{x^2}{2} \]

\[ \Rightarrow y = - 1 + 2 e^\frac{x^2}{2}\]