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 (abscissa) and the product of the x coordinate and y coordinate (ordinate) of t

Answer 1

We know that the slope of the tangent to the curve is ` (dy)/(dx)`
` therefore (dy)/(dx) = x + xy rArr (dy)/(dx) - xy = x " " `... (i)
This is of the form ` (dy)/(dx) + Py = Q`, where ` P = -x and Q = x `.
So, the given differential equation is linear.
`IF= e ^(int Pdx) = e ^(int - x dx) = e^((-x ^(2))/( 2 ))`
Hence, the solution of the given differential equation is given by
`y xx IF = int (Q xx IF )dx +C`,
i.e., `y xx e ^((-x^(2))/(2)) = int x e^((-x^(2))/( 2))dx + C `
` = int e ^(-t) dt +C `, where ` (x^(2))/(2) = t `
` " " = - e ^(-t) + C = - e ^((-x^(2))/( 2 )) +C `
`therefore " " y= -1 +Ce ^((x ^(2))/( 2 ))" " `... (ii)
We have to find a curve satisfying (ii) and passing through (0, 1).
Putting ` x = 0 and y = 1 `in (ii), we get ` C = 2`.
Hence, ` y = -1 + 2e^((x ^(2))/( 2)) ` is the equation of the required curve.