Question

Does the IVP\[(y-7)\frac{dy}{dx}=3x,\]\[y(0)=7\]have a unique solution?

I rearranged it to\[\frac{dy}{dx}=\frac{3x}{y-7}\]and differentiated the RHS with respect to 'y' to obtain\[-\frac{3x}{(y-7)^2}.\]Obviously, both functions are not continuous at the point (0,7) [division by zero], and therefore, the IVP has no unique solution.

Why did the professor mark me wrong on that?

Answer 1

i am not sure what an ivp is. but this seems to me like a differentia equation that you would need to separate and integrate to solve.

Answer 2

why did you not use separation of variables here?

Answer 3

I don't really have to solve the ODE to determine whether it has a unique solution or not, but it turns out that the aforementioned IVP has two solutions:\[y(x)=7-\sqrt{3}\sqrt{x^2},\]\[y(x)=\sqrt{3}\sqrt{x^2}+7.\]I just don't get why I got it marked wrong.

Answer 4

I believe it may have been my wording; I made it sound like the IVP has no solution at all. Anyway... thanks for the help!

Answer 5

I think you have to separate variables first to see if there is any discontinuity in the problem, What do you have to do to view the continuity or domain?

Answer 6

To check for the existence and uniqueness of a solution to an IVP of a first order ODE, there's a theorem which states that, given\[\frac{dy}{dx}=f(x,y),\]\[y(x_0)=y_0,\]if\[f(x,y)\]and\[\frac{\partial f}{\partial y}\]are both found to be continuous on a region R containing the point\[(x_0,y_0),\]then it's guaranteed to have a unique solution.

Answer 7

do yes solve for dy/dx and look for breaks in the domain.

Answer 8

and then take the partial derivative with respect to y of the solution and check for breaks in the domain also...