Solving a Second-Order Differential Equation with Variable Coefficients

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SUMMARY

The forum discussion focuses on solving the second-order differential equation 2yy'' + 2xy' = 0. Participants explore various methods, including setting p = y' and referencing case 1 from an unspecified textbook. The conversation highlights the solution y = -x as a potential outcome, although it is noted that this is not the only solution. The discussion emphasizes the importance of correctly interpreting the equation and the role of y' in the solution process.

PREREQUISITES
  • Understanding of second-order differential equations
  • Familiarity with the concept of derivatives and notation (e.g., y', y'')
  • Knowledge of substitution methods in differential equations
  • Access to relevant textbooks or resources on differential equations
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  • Study the method of solving second-order differential equations with variable coefficients
  • Research the application of substitution techniques in differential equations
  • Learn about the implications of dependent and independent variables in differential equations
  • Explore case studies from textbooks on differential equations for practical examples
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Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone seeking to deepen their understanding of solving complex mathematical problems.

der.physika
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Solve the Differential Equation 2yy\prime\prime+2xy\prime=0

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

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

Did you consider the
y\prime that it's not the same as y
 
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 2yy\prime+2xy\prime=0

set p=y\prime, 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:

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

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

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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