Differential Equation problem

[hibernator]

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.

 

[vela]

Show us what you tried so far.

 

[hibernator]

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 ?

 

[HallsofIvy]

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 ?

 

[hibernator]

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 .

 

[vela]

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

 

[median27]

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

 

[hibernator]

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?

 

[vela]

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

 

[hibernator]

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 ^^