Differential Equation problem

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Homework Statement



Substituition y=vx , differential equation x2dy/dx = y2-2x2 can be show in the form x dv/dx = (v-2)(v+1)

Hence , solve the differential equation x dv/dx = (v-2)(v+1) ,expressing answer in the form of y as a function of x in the case where y > 2x > 0 .


The Attempt at a Solution



I can only show the equation ,but can't solve the equation as the answer is y =x(2+Ax2) / (1-Ax2)

TQ.
 

Answers and Replies

  • #2
vela
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Show us what you tried so far.
 
  • #3
Show us what you tried so far.

I start after from my showing.
x dv/dx = (v-2)(v+1)
dv/dx = (v-2)(v+1) / x
dx/dv = x / (v-x)(v+1)
1/x dx = 1 / (v-2)(v+1) dv

Then I start to integrate , I am still new , so I don't know how to use to type the symbol.Sorry for the inconvenience.

1/x dx = 1/3 ( (1 / (v-2) )- ( 1/ (v+1) ) -----(1/3 from partial fraction )
ln x + c = 1/3 ln (v-2) - ln (v+1)
ln x + c = 1/3 ln (v-2)/(v+1)

Then I forgot know how to continue as i left my exercise book at home. I can only done so far.Then how could I continue ?
 
  • #4
HallsofIvy
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I start after from my showing.
x dv/dx = (v-2)(v+1)
This is wrong. If y= xv then dy/dx= x dv/dx+ v, not just x dv/dx.

-dv/dx = (v-2)(v+1) / x
dx/dv = x / (v-x)(v+1)
1/x dx = 1 / (v-2)(v+1) dv

Then I start to integrate , I am still new , so I don't know how to use to type the symbol.Sorry for the inconvenience.

1/x dx = 1/3 ( (1 / (v-2) )- ( 1/ (v+1) ) -----(1/3 from partial fraction )
ln x + c = 1/3 ln (v-2) - ln (v+1)
ln x + c = 1/3 ln (v-2)/(v+1)

Then I forgot know how to continue as i left my exercise book at home. I can only done so far.Then how could I continue ?
 
  • #5
This is wrong. If y= xv then dy/dx= x dv/dx+ v, not just x dv/dx.

no , I have done using dy/dx= x dv/dx+ v, from the equation x2dy/dx = y2-2x2 .
 
  • #6
vela
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Use the properties of logarithms:
[tex]\begin{align*}
\log ab &= \log a + \log b \\
b \log a &= \log a^b
\end{align*}[/tex]
and exponentiate to get rid of the logs.
 
  • #7
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Applying the properties of logarithm, i get
y= x(2+Ax^3)/1-Ax^3
which is supposed to be 'Ax^2?' HmMm.
 
  • #8
Applying the properties of logarithm, i get
y= x(2+Ax^3)/1-Ax^3
which is supposed to be 'Ax^2?' HmMm.

Same answer as mine.I got y= x(2+Ax^3)/1-Ax^3 .But the textbook's anwser is saying that 'Ax^2.The textbook probably wrong?
 
  • #9
vela
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It's straightforward enough to check. Just plug your answer back into the original differential equation and see if it works.
 
  • #10
It's straightforward enough to check. Just plug your answer back into the original differential equation and see if it works.

I will try , thank you so much for your help ^^
 
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