# Differential Equation problem

1. Jun 19, 2011

### hibernator

1. The problem statement, all variables and given/known data

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 .

3. 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.

2. Jun 19, 2011

### vela

Staff Emeritus
Show us what you tried so far.

3. Jun 19, 2011

### hibernator

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. Jun 19, 2011

### HallsofIvy

Staff Emeritus
This is wrong. If y= xv then dy/dx= x dv/dx+ v, not just x dv/dx.

5. Jun 20, 2011

### hibernator

no , I have done using dy/dx= x dv/dx+ v, from the equation x2dy/dx = y2-2x2 .

6. Jun 20, 2011

### vela

Staff Emeritus
Use the properties of logarithms:
\begin{align*} \log ab &= \log a + \log b \\ b \log a &= \log a^b \end{align*}
and exponentiate to get rid of the logs.

7. Jun 20, 2011

### median27

Applying the properties of logarithm, i get
y= x(2+Ax^3)/1-Ax^3
which is supposed to be 'Ax^2?' HmMm.

8. Jun 21, 2011

### hibernator

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. Jun 21, 2011

### vela

Staff Emeritus
It's straightforward enough to check. Just plug your answer back into the original differential equation and see if it works.

10. Jun 22, 2011

### hibernator

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