Findthe equation of a curve passing through the point (0, 1). If the slope of thetangent to the curve at any point (x, y) is equal to the sum of the xcoordinate (abscissa) and the product of the x coordinate and y coordinate(ordinate) of t

Findthe equation of a curve passing through the point (0, 1). If the s

1.	Findthe equation of a curve passing through the point (0, 1). If the slope of thetangent to the curve at any point (x, y) is equal to the sum of the xcoordinate (abscissa) and the product of the x coordinate and y coordinate(ordinate) of t
Answer» 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.