Solving a Second-Order Differential Equation with Variable Coefficients

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Homework Help Overview

The discussion revolves around solving a second-order differential equation with variable coefficients, specifically the equation 2yy'' + 2xy' = 0. Participants are exploring various approaches to tackle this problem, which appears to be challenging for the original poster.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the simplification of the equation and the implications of setting p = y'. There are questions about the correctness of the posted equation and the relevance of the textbook referenced. Some participants suggest alternative methods and express confusion about the original poster's approach.

Discussion Status

The discussion is ongoing, with various interpretations and suggestions being explored. Some participants have offered insights into the nature of the solutions, while others express uncertainty about the methods being used. There is no clear consensus on the best approach to take.

Contextual Notes

There are indications of missing information regarding the textbook being referenced, and some participants question the validity of the original problem statement. The original poster expresses frustration and seeks hints rather than complete solutions.

der.physika
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Solve the Differential Equation [tex]2yy\prime\prime+2xy\prime=0[/tex]

set [tex]p=y\prime[/tex], and then it becomes case 1 in the textbook

Can someone please help me solve this? It's a night mare
 
Last edited:
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Doesn't that simplify to y=-x?
 
ideasrule said:
Doesn't that simplify to y=-x?

Did you consider the
[tex]y\prime[/tex] that it's not the same as [tex]y[/tex]
 
der.physika said:
Did you consider the y' that it's not the same as y
What's your point? y' is a common factor in the equation you wrote. Did you not post the problem correctly?
der.physika said:
Solve the Differential Equation [tex]2yy\prime+2xy\prime=0[/tex]

set [tex]p=y\prime[/tex], and then it becomes case 1 in the textbook
You do realize we don't know what textbook you're using and even less likely to have a copy, right?
 
vela said:
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?

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:

[tex]y\prime=p, y\prime\prime=p\prime\mbox {Dependent variable y missing}[/tex]
 
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der.physika said:
Okay, objection 1 makes no sense
Really? You don't see how x=-y is a solution to 2yy'+2xy'=2y'(x+y)=0?
 
vela said:
Really? You don't see how x=-y is a solution to 2yy'+2xy'=2y'(x+y)=0?

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 [tex]p\equiv{y}\prime[/tex]
 
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You get that solution by canceling things out. It's not the only solution to the original equation, however.
 
vela said:
You get that solution by canceling things out. It's not the only solution to the original equation, however.

could you possibly do it, by using the substitution I posted?
 
  • #10
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
ok I posted this wrong, i feel stupid

[tex]2yy\prime\prime+2xy\prime=0[/tex]

how do I solve this?

give me a hint maybe?
 
  • #12
I don't know. Maybe someone else here has a suggestion.
 
  • #13
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.
 

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