Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.

Given slope at any point = sum of the abscissa and the product of the abscissa and the ordinate = x + xy

According to question,




We can see that it is a linear differential equation.


Comparing it with


P = – x, Q = x


I.F = e∫Pdx


= e – xdx


=


Solution of the given equation is given by


y × I.F = ∫Q × I.F dx + c


y × = ∫ x × dx + c……(1)


Let I = ∫ x × dx


Let



I = ∫ – dt = –


Now substituting the value of I in equation (1)


y + c


y = – 1 + c……(2)


As the equation passing through (0,1),


1 = – 1 + c× 1


c = 2


Putting the value of c in equation (1)


y = – 1 + 2


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