The equation of a curve passing through the point $(0,1)$, given that the slope of the tangent to the curve…
The equation of a curve passing through the point $(0,1)$, given that 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 $x$ and $y$ coordinates at that point, is
$y=1-2 e^{\left(\frac{x^2}{2}\right)}$
$y=-1+2 e^{\left(\frac{x^2}{2}\right)}$
$y=-1-2 e^{\left(\frac{x^2}{2}\right)}$
$y=1+2 e^{\left(\frac{x^2}{2}\right)}$
Solution
Given, $\frac{d y}{d x}=x+x y$
$\frac{d y}{d x}=x(1+y) \Rightarrow \frac{d y}{1+y}=x d x$
On integrating, we get
$\log (y+1)=\frac{x^2}{2}+C$
$\Rightarrow \quad y+1=A e^{x^2 / 2} \Rightarrow y=A e^{x^2 / 2}-1$
Since it passes through $(0,1)$.
$\because \quad 1=A-1 \Rightarrow A=2$
$\therefore$ Equation of curve is
$y=2 e^{x^2 / 2}-1$