# Homogeneous equation, problem with algebra

1. Sep 24, 2008

### chaotixmonjuish

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.

2. Sep 25, 2008

### tiny-tim

Hi chaotixmonjuish!

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?

3. Sep 25, 2008

### chaotixmonjuish

The answer to this one is |y-x|=c|y+3x|^5

i'm really concerned how this actually happens

4. Sep 25, 2008

### tiny-tim

that would be because …
is wrong.

5. Sep 25, 2008

### chaotixmonjuish

Is the step completely wrong or are A and B wrong?

6. Sep 25, 2008

### tiny-tim

A and B

7. Sep 25, 2008

### matematikawan

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
$$X=\left(\begin{array}{c}x\\y\end{array}\right) , A=\left(\begin{array}{cc}2&-1\\-3&4\end{array}\right)$$

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

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

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

8. Sep 25, 2008

### HallsofIvy

Staff Emeritus
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).

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.

No.

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.

No, that's not the part you are stuck at!

9. Sep 25, 2008

### tiny-tim

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

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 … 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!
Hi matematikawan!

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!

10. Sep 25, 2008

### chaotixmonjuish

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

11. Sep 26, 2008

### tiny-tim

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.