# Homework Help: Differential Equation (3)

1. Feb 26, 2010

### der.physika

Solve the Differential Equation $$2yy\prime\prime+2xy\prime=0$$

set $$p=y\prime$$, and then it becomes case 1 in the textbook

Last edited: Feb 26, 2010
2. Feb 26, 2010

### ideasrule

Doesn't that simplify to y=-x?

3. Feb 26, 2010

### der.physika

Did you consider the
$$y\prime$$ that it's not the same as $$y$$

4. Feb 26, 2010

### vela

Staff Emeritus
What's your point? y' is a common factor in the equation you wrote. Did you not post the problem correctly?
You do realize we don't know what textbook you're using and even less likely to have a copy, right?

5. Feb 26, 2010

### der.physika

Okay, objection 1 makes no sense... but I'll give you you're right for objection 2.

Here's what the textbook says for case 1:

$$y\prime=p, y\prime\prime=p\prime\mbox {Dependent variable y missing}$$

Last edited: Feb 26, 2010
6. Feb 26, 2010

### vela

Staff Emeritus
Really? You don't see how x=-y is a solution to 2yy'+2xy'=2y'(x+y)=0?

7. Feb 26, 2010

### der.physika

How did you get that solution besides canceling things out?

Cause, this is like a very annoying class, like what method did you use? Did you use $$p\equiv{y}\prime$$

Last edited: Feb 26, 2010
8. Feb 26, 2010

### vela

Staff Emeritus
You get that solution by canceling things out. It's not the only solution to the original equation, however.

9. Feb 26, 2010

### der.physika

could you possibly do it, by using the substitution I posted?

10. Feb 26, 2010

### vela

Staff Emeritus
I don't see what that substitution buys you. Also, it's not your textbook's case 1 as y appears in the differential equation.

11. Feb 26, 2010

### der.physika

ok I posted this wrong, i feel stupid

$$2yy\prime\prime+2xy\prime=0$$

how do I solve this?

give me a hint maybe?

12. Feb 27, 2010

### vela

Staff Emeritus
I don't know. Maybe someone else here has a suggestion.

13. Feb 27, 2010

### Whatever123

What's the point of the 2's? I would have just canceled them out from the beginning.

(Y*Y'')/Y' = -X

(Y''/Y')(Y) = -X

Set P = Y', then:

(P'/P)Integral(P) = -X

Ok, I'm sorry. This is going nowhere. Given me 24 hours to think about it.