Homogeneous equation, problem with algebra

Click For Summary

Discussion Overview

The discussion revolves around solving the differential equation y'=(4y-3x)/(2x-y), exploring various methods and approaches, including conventional and unconventional techniques. Participants share their algebraic manipulations, express concerns about specific steps, and propose alternative solutions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents their work on the equation, expressing difficulty in progressing from a certain algebraic manipulation.
  • Another participant suggests a different interpretation of the equation, questioning the correctness of the coefficients derived from partial fractions.
  • A later reply proposes an unconventional method using a parameter t, leading to a system of differential equations with constant coefficients, and claims to arrive at a solution that satisfies a specific relationship.
  • Multiple participants express uncertainty about the correctness of coefficients obtained during the partial fraction decomposition, with conflicting values being proposed.
  • Some participants challenge earlier steps in the algebraic manipulation, indicating potential errors without reaching a consensus on the correct approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of certain algebraic steps and coefficients, with multiple competing views and ongoing uncertainty regarding the solution process.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the manipulation of logarithmic expressions and the application of partial fractions. The discussion reflects varying interpretations of the differential equation and its solutions.

Who May Find This Useful

Readers interested in differential equations, particularly those exploring different solution methods and the nuances of algebraic manipulation in mathematical reasoning.

chaotixmonjuish
Messages
284
Reaction score
0
this is the given equation

y'=(4y-3x)/(2x-y)

and here is all the work I've done so far:

(4v-3)/(2-v)=v+x*dv/dx

i moved v over and came up with this

(-3+2v+v^2)/(2-v)=x*dv/dx

did a flip

(2-v)/(-3+2v+v^2)dv=dx/x

by partial fractions I got a=-3/2 and b=1/2

so -3/2*ln|v-1|+1/2*ln|v+3|+c=ln|x|

after making the -3/2 and 1/2 powers to ln and throwing e in

(v-1)^(-3/2)+(v+3)^(1/2)+c=x

or

((y+x)/x)^(-3/2)+((y+3x)/x)^(1/2)+c=x

that's the part I'm stuck at.
 
Physics news on Phys.org
chaotixmonjuish said:
… so -3/2*ln|v-1|+1/2*ln|v+3|+c=ln|x|

after making the -3/2 and 1/2 powers to ln and throwing e in

(v-1)^(-3/2)+(v+3)^(1/2)+c=x

or

((y+x)/x)^(-3/2)+((y+3x)/x)^(1/2)+c=x

Hi chaotixmonjuish! :smile:

hmm … are you staying up too late? :zzz:

it's C((y+x)/x)-3/2((y+3x)/x)1/2=x , isn't it? :wink:
 
The answer to this one is |y-x|=c|y+3x|^5

i'm really concerned how this actually happens
 
chaotixmonjuish said:
The answer to this one is |y-x|=c|y+3x|^5

i'm really concerned how this actually happens

that would be because …
chaotixmonjuish said:
(2-v)/(-3+2v+v^2)dv=dx/x

by partial fractions I got a=-3/2 and b=1/2

is wrong. :wink:
 
Is the step completely wrong or are A and B wrong?
 
chaotixmonjuish said:
Is the step completely wrong or are A and B wrong?

A and B :smile:
 
chaotixmonjuish said:
this is the given equation

y'=(4y-3x)/(2x-y)

.

I try to solve this DE using unconventional method using parameter t.
dy/dx = dy/dt /dx/dt.
Then dy/dt = 4y-3x and dx/dt=2x-y (Is this correct? )
which is a system of DE with constant coefficients.
X' = AX
where
[tex]X=\left(\begin{array}{c}x\\y\end{array}\right) ,<br /> A=\left(\begin{array}{cc}2&-1\\-3&4\end{array}\right)[/tex]

The eigenvalues of A are 1 and 5 with eigenvectors [tex]\left(\begin{array}{c}1\\1\end{array}\right)[/tex] and [tex]\left(\begin{array}{c}1\\-3\end{array}\right)[/tex] respectively.

Hence the solution is
x=c1et + c2e5t
y=c1et - 3c2e5t

which satisfies |y-x|=c|y+3x|^5.
 
chaotixmonjuish said:
this is the given equation

y'=(4y-3x)/(2x-y)

and here is all the work I've done so far:

(4v-3)/(2-v)=v+x*dv/dx
It would have been better to tell us HOW you got that! I assume that, because this is a homogenous equation you let v= y/x so y= xv and y'= xv'+ v. (4y-3x)/(2x-y), dividing both numerator and denominator by x, is equal to [4(y/x)- 3]/[2- (y/x)]= (4v- 3)/(2- v).

i moved v over and came up with this

(-3+2v+v^2)/(2-v)=x*dv/dx

did a flip

(2-v)/(-3+2v+v^2)dv=dx/x

And v^2+ 2v- 3= (v- 1)(v+3). Now, let (2-v)/(-2+ 2v+ v^2)= A/(v-1)+ B/(v+3). Multiplying on both sides by (v-1)(v+3), 2- v= A(v+3)+ B(v-1). If v= 1, 2-1= 1= 4A. A= 1/4. If v= -3, 2-(-3)= 5= -4B and B= -5/4.

by partial fractions I got a=-3/2 and b=1/2
No.

so -3/2*ln|v-1|+1/2*ln|v+3|+c=ln|x|

after making the -3/2 and 1/2 powers to ln and throwing e in

(v-1)^(-3/2)+(v+3)^(1/2)+c=x
Even with the correct coefficients, e^(ln(a)+ ln(b)) is NOT a+ b. ln(a)+ ln(b)= ln(ab) so e^(ln(a)+ ln(b))= e^(ln(ab))= ab.

or

((y+x)/x)^(-3/2)+((y+3x)/x)^(1/2)+c=x

that's the part I'm stuck at.
No, that's not the part you are stuck at!
 
HallsofIvy said:
It would have been better to tell us HOW you got that! I assume that, because this is a homogenous equation you let v= y/x so y= xv and y'= xv'+ v. (4y-3x)/(2x-y), dividing both numerator and denominator by x, is equal to [4(y/x)- 3]/[2- (y/x)]= (4v- 3)/(2- v).

oh come on … it was obvious that v was y/x :smile: …

frankly I'd rather solutions were set out as sparingly as possible, so that I can see what's going on …

every line has to be checked anyway :rolleyes: … and by definition there's a mistake somewhere, or the OP wouldn't be asking …

sometimes, too much detail is both off-putting and confusing! :confused:
matematikawan said:
I try to solve this DE using unconventional method using parameter t.
dy/dx = dy/dt /dx/dt.
Then dy/dt = 4y-3x and dx/dt=2x-y (Is this correct? )
which is a system of DE with constant coefficients.
…
Hence the solution is
x=c1et + c2e5t
y=c1et - 3c2e5t

which satisfies |y-x|=c|y+3x|^5.

Hi matematikawan! :smile:

ooh, I've never seen that before …

looks good to me … you've defined a perfectly legitimate parameter t, and got a standard linear equation, leading to a parametric solution which is actually easier to graph than the book solution! yay! :biggrin:
 
  • #10
i got as my two coefficients -5/4 and 1/4, are those correcti'm still not sure how to move things around to as to get the book answer
 
  • #11
chaotixmonjuish said:
i got as my two coefficients -5/4 and 1/4, are those correct

i'm still not sure how to move things around to as to get the book answer

Yes, -5/4 and 1/4.

And that gives you C((y+x)/x)-5/4((y+3x)/x)1/4 = x , doesn't it?

Then just raise both sides to the 4th power. :smile:
 

Similar threads

Undergrad Solving Differential Equation Using Reduction of Order
  • · Replies 2 ·
Replies
2
Views
4K
Undergrad Stuck on solving a non homogenous (linear) PDE
  • · Replies 36 ·
2
Replies
36
Views
6K
Undergrad How did the author transform the original equation into the Legendre equation?
  • · Replies 4 ·
Replies
4
Views
2K
MHB How to Solve This Homogeneous Differential Equation?
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad How to do Magnus expansion for time-dependent companion matrix?
  • · Replies 7 ·
Replies
7
Views
3K
Undergrad I with differential equation(3x^2-2y^2)(dy/dx)=2xy
  • · Replies 9 ·
Replies
9
Views
9K
Undergrad Understanding Legendre differential equation
  • · Replies 2 ·
Replies
2
Views
6K
MHB B.2.1.4 trig w/ integrating factor
  • · Replies 1 ·
Replies
1
Views
1K
MHB -b.2.2.33 - Homogeneous first order ODEs, direction fields and integral curves
  • · Replies 18 ·
Replies
18
Views
2K
Undergrad Energy of moving Sine-Gordon breather
  • · Replies 2 ·
Replies
2
Views
3K