Solving differential equations

Click For Summary

Homework Help Overview

The problem involves solving a differential equation of the form (x^2+1)dy/dx = xy, with a specific initial condition y(2) = 5. Participants are tasked with finding y(3) based on this setup.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various substitution methods and integration techniques, including the use of logarithmic properties and exponential functions. Some express confusion over the correct application of constants and simplifications.

Discussion Status

Several participants have offered corrections and clarifications regarding the integration steps and the handling of constants. There is an ongoing exploration of different interpretations of the problem, with no explicit consensus reached yet.

Contextual Notes

Participants note potential errors in previous calculations and the importance of careful manipulation of logarithmic expressions. There is acknowledgment of confusion stemming from the complexity of the substitutions and corrections being discussed.

Shannabel
Messages
74
Reaction score
0

Homework Statement


If (x^2+1)dy/dx = xy for all y>0, and y(2) = 5, then y(3) = ?


Homework Equations





The Attempt at a Solution


dy/y = xdx/(x^2+1) -- use substitution with u=x^2
lny = 2(u+1)/du
lny = 2ln(x^2+1)+c
y = e^(2ln(x^2+1)+c)
5 = e^(2ln(5)+c)
5 = e^(ln25+c)
5 = 25+c
c = 20

y(3) = e^(2ln10+20)
= e^ln100+20
= 100+20
= 120



... the answer should be 5sqrt(2), so I am very off, but I can't find why. help?
 
Physics news on Phys.org
If u = x^2, then du = 2 x dx.
So instead of
\ln y = 2(u+1)/du
you get
d(\ln y) = du/[2(u + 1)]
and, integrating,
\ln y = \ln[ 2(u + 1) ] + c.
 
Last edited:
We have exp[(1/2)*ln(x^2+ 1) + c] = k*sqrt(x^2+1), where k =exp(c). Note: x*dx/(1+x^2) = (1/2)du/(u+1).

RGV
 
Ray Vickson said:
We have exp[(1/2)*ln(x^2+ 1) + c] = k*sqrt(x^2+1), where k =exp(c). Note: x*dx/(1+x^2) = (1/2)du/(u+1).

RGV

if y = e^(1/2)ln(x^2+1)+c
5 = e^lnsqrt(5)+c
then i end up with c = 5-sqrt(5), and then
y(3) = e^1/2ln10+5-sqrt(5)
= sqrt(10)+5-sqrt(5)
... confused?
 
@CompuChip: you forgot a factor (1/2) in your last line...
 
Shannabel said:
if y = e^(1/2)ln(x^2+1)+c
5 = e^lnsqrt(5)+c
then i end up with c = 5-sqrt(5), and then
y(3) = e^1/2ln10+5-sqrt(5)
= sqrt(10)+5-sqrt(5)
... confused?

You forgot a couple of brackets...

You should use:
5 = e^{[\ln \sqrt 5+c]}
You'll see that you'll get a different value for c. :smile:
 
I like Serena said:
You forgot a couple of brackets...

You should use:
5 = e^{[\ln \sqrt 5+c]}
You'll see that you'll get a different value for c. :smile:

do you mean then i should get
5 = sqrt5 +e^c?
how do i simplify that?
 
Shannabel said:
do you mean then i should get
5 = sqrt5 +e^c?
how do i simplify that?

No, that is not right.
Let's take it one step back.
Let me put the braces slightly diffently for clarity.
5 = e^{(\ln \sqrt 5)+c}
Here you need to simplify by applying the rules for powers.
That is e^{a+b}=e^a \cdot e^b.
 
I like Serena said:
No, that is not right.
Let's take it one step back.
Let me put the braces slightly diffently for clarity.
5 = e^{(\ln \sqrt 5)+c}
Here you need to simplify by applying the rules for powers.
That is e^{a+b}=e^a \cdot e^b.

so then don't i get
5 = e^ln(sqrt5) + e^c?
and e^ln(sqrt5) = sqrt5?
 
  • #10
Shannabel said:
so then don't i get e^ln(sqrt5) = sqrt5?

Yes! :smile:


Shannabel said:
5 = e^ln(sqrt5) + e^c?

No. :frown:
 
  • #11
I like Serena said:
Yes! :smile:




No. :frown:

oh, durrr
5 = sqrt5*e^c?
 
  • #12
Shannabel said:
oh, durrr
5 = sqrt5*e^c?

Yes! :wink:
 
  • #13
I like Serena said:
Yes! :wink:

YAY!
so from there, i get
5/sqrt5 = e^c?
so c = ln(5/sqrt5)?
 
  • #14
Yes, so... :rolleyes:
 
  • #15
I like Serena said:
Yes, so... :rolleyes:

hahaha
f(3) = e^lnsqrt10+ln5/sqrt5
= sqrt10+5/sqrt5

... annnd I'm off track again lol
 
  • #16
Err... can you do the "oh, durrr" thing again? :shy:
 
  • #17
I like Serena said:
Err... can you do the "oh, durrr" thing again? :shy:

well that was dumb.
e^lnsqrt10+ln5/sqrt5 = e^lnsqrt10*e^ln5/sqrt5
= sqrt10*5/sqrt5

I GOT IT YAYYYY :):)
thanksyou :)
 
  • #18
Congratulations! :biggrin:One thing though...

You started with:
y = e^{(1/2)\ln(x^2+1)+c}
But that is not the equation you had in your first post... o:)

How did you get it?
 
  • #19
I like Serena said:
Congratulations! :biggrin:


One thing though...

You started with:
y = e^{(1/2)\ln(x^2+1)+c}
But that is not the equation you had in your first post... o:)

How did you get it?

:D

i changed the first equation to
dy/y = xdx/(x^2+1), then used the substitution u=x^2 on the right side :)
 
  • #20
Shannabel said:
:D

i changed the first equation to
dy/y = xdx/(x^2+1), then used the substitution u=x^2 on the right side :)

Yeah, I got that...
So did you follow CompuChip's correction and then my correction on CompuChip's correction? :confused:
 
  • #21
I like Serena said:
Yeah, I got that...
So did you follow CompuChip's correction and then my correction on CompuChip's correction? :confused:

i was confused by what you were saying but i did it over and realized i should have had 1/2 instead of 2 :)
 
  • #22
Shannabel said:
i was confused by what you were saying but i did it over and realized i should have had 1/2 instead of 2 :)

Oh, that's okay then! :approve:
 
  • #23
Shannabel said:
hahaha
f(3) = e^lnsqrt10+ln5/sqrt5
= sqrt10+5/sqrt5

... annnd I'm off track again lol

Why don't you read my previous response?

RGV
 
  • #24
Ray Vickson said:
Why don't you read my previous response?

RGV

i did, it confused me :(
 
  • #25
Ray Vickson said:
Why don't you read my previous response?

RGV
Shannabel said:
i did, it confused me :(
Yeah, to be honest, I already thought that step was too big. :redface:
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
3
Views
2K