find an equation of the curve that passes through the point (0, 1) and whose slope at (x, y) is 9xy.

Explanation:

Step1: Set up the differential equation

The slope of the curve at ((x,y)) is (\frac{dy}{dx}). Given that the slope is (9xy), we have the differential equation (\frac{dy}{dx}=9xy).

Step2: Separate the variables

Separate the variables: (\frac{dy}{y}=9x dx).

Step3: Integrate both sides

Integrate: (\int\frac{dy}{y}=\int9x dx). Using the integral formulas (\int\frac{1}{u}du=\ln|u| + C) and (\int x^n dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)), we get (\ln|y|=\frac{9x^{2}}{2}+C).

Step4: Solve for (y)

Exponentiate both sides: (y = e^{\frac{9x^{2}}{2}+C}=e^{C}e^{\frac{9x^{2}}{2}}). Let (A = e^{C}), so (y = Ae^{\frac{9x^{2}}{2}}).

Step5: Use the initial condition

The curve passes through the point ((0,1)). Substitute (x = 0) and (y = 1) into (y = Ae^{\frac{9x^{2}}{2}}): (1=Ae^{0}), so (A = 1).

Answer:

(y=e^{\frac{9x^{2}}{2}})