Question

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

Answer 1

If the slope at the point $(x,y)$ is $xy$, then we have

$\displaystyle\frac{dy}{dx} = xy\ \Rightarrow\ \frac{dy}{y} = x\, dx\ [y \ne 0] \Rightarrow \int \frac{dy}{y} = \int x \, dx \ \Rightarrow \\ \\ \textstyle \ln|y| = \frac{1}{2}x^2 + C.\ y(0) = 1\ \Rightarrow\ \ln 1 = 0 + C\ \Rightarrow\ C = 0. \\ \\ \text{Thus } |y| = e^{x^2 / 2}\ \Rightarrow\ y = \pm e^{x^2 / 2},\ \text{so } y = e^{x^2/2} \text{ since } y(0) = 1 \ \textgreater \ 0. \\ \\ \text{Note that y=0 is not a solution because it does not satisfy} \\ \text{the initial condition y(0) = 1 }.$

Answer 2

By solving the differential equation, we will see that the equation for the curve is:

$y = exp( (1/2)*x^2)$

How to solve the differential equation?

We have a curve of the form:

y = f(x).

And we know that the slope at the point (x, y) is given by:

y' = xy

So we need to solve this differential equation. So we have:

$\frac{dy}{dx} = xy$

This is a separable differential equation, so we can separate the variables in the next way:

$dy/y = x*dx$

Integrating in both sides, we get:

ln(y) = (1/2)*x^2 + C

Where both integration constants are included in C.

We can apply the exponential function to both sides so we get:

$exp(ln(y)) = exp( (1/2)*x^2 + C)$

$y = exp( (1/2)*x^2 + C)$

Now we also know that this curve passes through the point (0, 1), so we have:

$1 = exp( (1/2)*0^2 + C)$

$1 = exp( C)$

Then we must have C = 0.

Which means that the equation for the curve is:

$y = exp( (1/2)*x^2)$

If you want to learn more about differential equations, you can read:

https://brainly.com/question/18760518