graphing derivatives?

Question

why does this not have a similar graph to an exponential?
dy/dx=y(1-x)

Also, if you solve this, do you get the same answer as me?
problem: dy/dx=y(1-x) , y(0)=2 , y(x)>0
answer: y(x)=2e^(x-.5x^2)

I just want to make sure I am doing this right.

yeah, you are. I am in calc 2 but I am studying diff equations a little.

Answer 1

You have the correct solution to the diff eq. It's a separable equation, and can be rewritten as:

1/y dy = (1-x) dx

which has solution:

ln(y) = (x - (x^2)/2) + C

or y = C*exp(x - (x^2)/2).

Using the initial condition gets you C = 2.

I'm not sure what you are trying to graph, though. A plot of z = dy/dx = f(x,y) would require a 3-dimensional graph of some sort (e.g., a contour plot on the x-y plane). A plot of y(x) vs. x isn't going to look like an exponential because is isn't an y(x) isn't a simple exponential. It's actually a gaussian multiplied by an exponential:

y(x) = 2*exp(x - (x^2)/2) can be written as:

y(x) = 2*exp(x) * exp(-(x^2)/2)

The exp(x) factor is a simple exponential, while the exp(-(x^2)/2) factor is a gaussian.

Similarly, if you meant "Why doesn't the graph of dy(x)/dx look like an exponential?", it's because the function isn't anything like a simple exponential. Substituting the expression above for y(x), into the original equation gives:

dy(x)/dx = 2 * (1-x) * exp(x) * exp(- (x^2)/2)

Answer 2

Your answer is correct. You can most easily know its correct by verifying the initial conditions and the derivative.

It does have an 'exponential' derivative.

Now that you have y you can substitute y into the derivative.

y'(x) = (1-x) 2 e^[ x- x^2/2)]

The first form is the implicit form. You may learn to recognize some functions in their implicit form eventually.

Answer 3

You *are* taking a class in ordinary differential equations, aren't you? Because dy/dx = f(x,y)
implies y=F(x,y) which leads to a 'Greens Function' kind of solution.

Or am I misunderstanding the question?


Doug

Answer 4

y'(x) = (1-x) 2 e^[ x- x^2/2)]
Is correct.