Question

At any point on a curve, the slope of the tangent is equal to the sum of abscissa and the product of ordinate and abscissa of that point. If the curve passes through (0, 1), then the equation of the curve is ______.

Options

• `"y" = 2"e"^(x^2/2) - 1`

• `"y" = 2"e"^(x^2/2)`

• `"y" = 2"e"^(-x^2)`

• `"y" = 2"e"^(-x^2) - 1`

Answer 1

At any point on a curve, the slope of the tangent is equal to the sum of abscissa and the product of ordinate and abscissa of that point. If the curve passes through (0, 1), then the equation of the curve is `underline("y" = 2"e"^(x^2/2) - 1)`.

Explanation:

`"dy"/"dx" = "x" + "xy"`

`=> 1/(1 + "y") "dy"` = x dx

Integrating on both sides, we get

`int 1/(1 + "y") "dy" = int x "dx" + "c"`   ...(i)

Since the required curve passes through (0, 1), c = log 2

`therefore log (1 + "y") = x^2/2 + log 2`    ...[From (i)]

`=> log ((1 + "y")/2) = x^2/2`

`=> "y" = 2"e"^(x^2/2) - 1`