# Homework Help: Differential Equation Roadblock

1. Oct 19, 2008

### jinksys

1. The problem statement, all variables and given/known data
Find the solution of the differential equation yy'(1+x2)-x(1+y2)=0 if y(0)=Sqrt(3)

3. The attempt at a solution

I let y=Sqrt(3) and x=0 and solve for y'.
y'=0
Integrate both sides.
y=x +c
Plug my points in to find C
Sqrt(3)=0 +c
C=Sqrt(3)
My solution is y=x+Sqrt(3).

Check:
Plug x+Sqrt(3) in for y, and 1 for y' into the DE.
End up with Sqrt(3)=0, so somewhere I went wrong.
Any help appreciated, Thanks.

2. Oct 19, 2008

### Staff: Mentor

You are confusing y(0) with y(x). y(0) represents the y-value when x = 0. y(x) represents the y-value for an arbitrary x-value. If you were to plot the graph of y as a function of x, all you know is that the graph goes through (0, sqrt(3)).

In your attempt at a solution, you said:
How did you get y' = 0? This would have been correct if y(x) = sqrt(3). IOW if the graph of y were a horizontal line sqrt(3) units above the x-axis.

If it were true that y' = 0 (it actually isn't true, though), how did you get y = x + c? If you integrate 0 with respect to x, you don't get x + C.

The DE you are given is a separable DE. That means it is possible to get all of the things that involve y (including dy) on one side, and all of the things that involve x (and dx) on the other side. That's the approach you should take with this problem.

3. Oct 19, 2008