Question

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

Short answer

\(y = {e^{{x^2}/2}}{\rm{. }}\)

Step-by-step solution

Given data

It is given that the slope of the curve is\(xy\).

The curve passes through\((0,1)\).

Form equation

Since slope is\(xy\)we get:

\(\begin{aligned}\frac{{dy}}{{dx}} &= xy\\\frac{1}{y}dy &= xdx\end{aligned}\)

Integrate

Now we can integrate both sides of the above equation, and we get:

\(\int {\frac{1}{y}} dy = \int x dx + C\)

\(\ln |y| = \frac{{{x^{1 + 1}}}}{{1 + 1}} + C\;\;\;\) Using formulas\(\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1;\int {\frac{1}{x}} dx = \ln |x|\)

\(\ln |y| = \frac{1}{2}{x^2} + C\)

Apply initial condition

Using initial condition\(y(0) = 1\) i.e; \(x = 0,y = 1\)

\(\begin{aligned}\ln 1 &= \frac{1}{2}{(0)^2} + C\\0 &= 0 + C\\C &= 0\end{aligned}\)

Find solution

Substitute\(C = 0\)& solve for\(y\).

\(\begin{aligned}\ln |y| &= \frac{1}{2}{x^2}|y|\\ &= {e^{\frac{1}{2}{x^2}}}\\y &= \pm {e^{\frac{1}{2}{x^2}}}\end{aligned}\)

Since, \(y(0) = 1 > 0\)we have\(y = {e^{{x^2}/2}}\)

Since it will not satisfy that the function passes the point \((0,1)\,\,y = 0\)is not a solution.

Therefore, the solution of given the differential equations is\(y = {e^{{x^2}/2}}{\rm{. }}\)