# Thread: Differential equation help.

1. ## Differential equation help.

Before I begin, I have to admit I'm hopeless at these things. Never gotten my head around them. Differentiation I'm okay with, but not these annoying buggers!

Question:
Consider the differential equation:
$\displaystyle 2xy.dy/dx = y^2 - 9$
a. Find the steady state solutions (if any) and the general solution.
b. Find the particular solution satisfying y = 5 when x = 4.

Well, this is my attempt at answering these questions:
rearrange the formula:
dy.1/(y² – 9) = dx.1/2xy
integrate to dy and dx:
∫y/y² – 9).dy = ∫1/2x.dx
½ln(y² - 9) = ln(x)/2 + c
ln(y² - 9) = ln(x) + 2c
e(ln(y² - 9)) = e(ln(x) + 2c)
y² - 9 = Ax where A = e^2c
y² = Ax + 9

when y = 5 and x = 4:
25 = 4A + 9
A = 4
ln(A) = ln(4)
c = ln(2)

Is this correct? Personally I think not based on my previous attempts at doing these sort of questions!

2. Its perrrrfect!

Dont try to find c. Since the final equation is y^2 = 4x+9. A is enough.

And moreover you need not ask anyone actually
You can just substitute it back in the differential equation and do what you like doing.... that is differentiate and see if the equation holds

3. Originally Posted by Dr Zoidburg
Before I begin, I have to admit I'm hopeless at these things. Never gotten my head around them. Differentiation I'm okay with, but not these annoying buggers!

Question:
Consider the differential equation:
$\displaystyle 2xy.dy/dx = y^2 - 9$
a. Find the steady state solutions (if any) and the general solution.
b. Find the particular solution satisfying y = 5 when x = 4.

Well, this is my attempt at answering these questions:
rearrange the formula:
dy.1/(y² – 9) = dx.1/2xy
integrate to dy and dx:
∫y/y² – 9).dy = ∫1/2x.dx
½ln(y² - 9) = ln(x)/2 + c
ln(y² - 9) = ln(x) + 2c
e(ln(y² - 9)) = e(ln(x) + 2c)
y² - 9 = Ax where A = e^2c
y² = Ax + 9

when y = 5 and x = 4:
25 = 4A + 9
A = 4
ln(A) = ln(4)
c = ln(2)

Is this correct? Personally I think not based on my previous attempts at doing these sort of questions!
Yea of little faith. Your answer to (b) is correct. Applying your OK skill with differentiation would have shown you that:

Differentiate your solution implicitly: $\displaystyle 2y \, \frac{dy}{dx} = A \Rightarrow 2xy \, \frac{dy}{dx} = Ax$.

Substitute into the DE:

$\displaystyle Ax = (Ax + 9) - 9$.

Left hand side = right hand side.

4. OMG!
I got it right?!
whoah...maybe it is all starting to sink in!
thanks for the reassurance. That's my assignment finished, tg

5. Originally Posted by Dr Zoidburg
OMG!
I got it right?!
whoah...maybe it is all starting to sink in!
thanks for the reassurance. That's my assignment finished, tg
You found the steady state solutions, right?

6. D'oh!
Any help there would be appreciated!

7. Originally Posted by Dr Zoidburg
D'oh!
Any help there would be appreciated!
So .... what is a stationary solution? How do you find them? How does that apply to the given DE?

8. Have I got this right:
dy/dx = h(y).f(x) making
h(y) = (y^2 - 9)/y
setting h(y)=0 we get
y = {-3,+3}

These are the steady state solutions. correct?

9. Originally Posted by Dr Zoidburg
Have I got this right:
dy/dx = h(y).f(x) making
h(y) = (y^2 - 9)/y
setting h(y)=0 we get
y = {-3,+3}

These are the steady state solutions. correct?
Yes.