ODE by substitution

  • #1
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Homework Statement



By means of substitution x=X+1, y=Y+2 ,shwo that the equation dy/dx=(2x-y)/(x+2y+5) can be reduced to dY/dX=(2X-Y)/(X+2Y).Hence, find the general solution of the given equation.

Homework Equations





The Attempt at a Solution



The first part is quite simple to prove.

Second part,

since its a homogenous differential equation, i would use the substitution Y=vX here.

v+X(dv/dX)=(2X-vX)/(2vX+X)

[tex]-\frac{1}{2}\int \frac{4v+2}{2v^2+2v+2} dv=\int \frac{1}{X} dx[/tex]

ln |2v^2+2v+2|=-2ln |X|-c

then v=Y/X

ln |(2Y^2)/X^2+2Y/X+2|=-2ln |X|-c'

Then substitute back again from the first part,

the solution is

ln |2(y-2)^2/(x-1)^2+2(y-2)/(x-1)+2|=-2ln |x-1|-c'

AM i correct?
 

Answers and Replies

  • #2
362
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[tex]-\frac{1}{2}\int \frac{4v+2}{2v^2+2v+2} dv=\int \frac{1}{X} dx[/tex]

check this again because i got

[tex]-\frac{1}{2}\int \frac{4v+2}{2v^2+2v-2} dv=\int \frac{1}{X} dX[/tex]
 
  • #3
438
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check this again because i got

[tex]-\frac{1}{2}\int \frac{4v+2}{2v^2+2v-2} dv=\int \frac{1}{X} dX[/tex]

ok thanks ,how bout the rest?
 
  • #4
362
0


ln |2v^2+2v+2|=-2ln |X|-c

then v=Y/X

ln |(2Y^2)/X^2+2Y/X+2|=-2ln |X|-c'

hmm, why do you change c to c' ?

and, others correct if i'm not wrong.

p/s: don't forget change the negative sign, :biggrin:
 
Last edited:
  • #5
438
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hmm, why do you change c to c' ?

and, others correct if i'm not wrong.

p/s: don't forget change the negative sign, :biggrin:

oh, maybe its not necessary. I multiplied both sides with 2 so c is now 2c and c'=2c

Is it ok to not change? Would it affect the equation?
 
  • #6
35,000
6,753


oh, maybe its not necessary. I multiplied both sides with 2 so c is now 2c and c'=2c

Is it ok to not change? Would it affect the equation?
You did the right thing even though c and c' are arbitrary constants.
 
  • #7
362
0


yea, but you already multiply both side

"ln |2v^2+2v+2|=-2ln |X|-c

then v=Y/X

ln |(2Y^2)/X^2+2Y/X+2|=-2ln |X|-c' "

and you didn't multiplied here, correct me if i'm wrong
 

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