Question

find the equation of a curve that passes thru (1,3)and has slope of y/x^2 at any point of x/y
f(x) = sqrt (x) + 3; g(x) = (x+3)/2

Answer 1

what are f and g for ?

Answer 2

anyway looks like you set slope equal to y/x^2
\[\frac{dy}{dx} = \frac{y}{x^{2}}\]
separate variables and integrate both sides
\[\int\limits_{ }^{} \frac{dy}{y} = \int\limits_{}^{} \frac{dx}{x^{2}}\]
\[\ln y = -\frac{1}{x} + C\]
take e raised to both sides to cancel log
\[y = k e^{-1/x}\]
where k is constant e^C
plug in point (1,3) to find k
--> k = 3e
\[\large y = 3e^{1-\frac{1}{x}}\]