Question

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 that point.

Answer 1

Slope at tangent at any point = (x+xy)given
$\therefore \frac{dy}{dx}=x(1+y)or\frac{dy}{1+y}=xdx.\int \frac{dy}{1+y}=\int xdx=log(1+y)=\frac{x^2}{2}+Catx=0y=1\therefore log(1+1)=C$
or $log(1+y)=\frac{x^2}{2}+log2(1+y)=e^{(\frac{x^2}{2}+log2)}=e^{\frac{x}{2}}{e}^{log2}ory=e^{(\frac{x^2}{2}+log2)}$