Q120 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Mathematics

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

  • A
    $y=1-2 e^{\left(\frac{x^2}{2}\right)}$
  • Correct answer
    B
    $y=-1+2 e^{\left(\frac{x^2}{2}\right)}$
  • C
    $y=-1-2 e^{\left(\frac{x^2}{2}\right)}$
  • D
    $y=1+2 e^{\left(\frac{x^2}{2}\right)}$

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 $y=2 e^{x^2 / 2}-1$

Hence, the correct answer is option: 'B'.

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