Differential equation?

Question

1/(y^2-4)(dx/dy)=y find when x(0)=e and show it's really the answer.

Here is a another differential I would like to see the method of, If I can see how It's done, I think I can finish the HW, I kinda have an Idea, but an example would be awesome.

Answer 1

Put all the ys on one side and all the xs on the other:
dx = y(y^2-4) dy
Now integrate both sides:
dx, which is just 1 dx, integrates to x.
y(y^2 - 4) dy, which is just y^3 - 4y dy, integrates to y^4/4 - 2y^2.

We'll throw in a constant somewhere, so we have
x = y^4 / 4 - 2y^2 + c.
Substitute in y = 0, and we get e = c. So x = y^4 / 4 - 2y^2 + e.

Now, to check, its obvious that x(0) = e, but lets try differentiating.

dx/dy = y^3 - 4y.
So 1/(y^2 - 4) dx/dy = (y^3 - 4y)/(y^2 - 4) = y(y^2 - 4)/(y^2 - 4) = y.

Answer 2

Ok... here's a partial solution...
multiply each side by dx and you get
integral (dy/(y^2 - 4)) = integral y dx
Break down the left side by partial fractions, integrate the right side, and you get
1/4 integral [(1/(y-2)) + (1/(y+2))]dy = xy + C1
The left side integrates as 1/4 [ln (y-2) + ln (y+2)] + C2
which simplifies as 1/4 ln (y+2)(y-2) + C2 =
1/4 ln (y^2 - 4) + C2 = ln (y^2 - 4) + C2
Raise e to the power of both sides, set x(0) = e and solve for the constants.
That's about all I can do for now cos my brother is bugging me to get on the computer but that ought to get you going.

Answer 3

It's a seperable ODE, nice and simple. Break the dx/dy and solve to:
dx=y(y^2-4)dy

1. Do the indefinite integral
2. Combine the arbitrary constants into a single C
3. Plug in your boundary condition (x(0)=e) and solve for C.
4. To show it's the answer just plug your sol'n for x(y) back into the original ODE and demonstrate that it fits.