# ODE by substitution

1. Jul 1, 2010

### thereddevils

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

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.

2. Relevant equations

3. 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)

$$-\frac{1}{2}\int \frac{4v+2}{2v^2+2v+2} dv=\int \frac{1}{X} dx$$

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?

2. Jul 1, 2010

### annoymage

Re: Ode

check this again because i got

$$-\frac{1}{2}\int \frac{4v+2}{2v^2+2v-2} dv=\int \frac{1}{X} dX$$

3. Jul 1, 2010

### thereddevils

Re: Ode

ok thanks ,how bout the rest?

4. Jul 1, 2010

### annoymage

Re: Ode

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,

Last edited: Jul 1, 2010
5. Jul 1, 2010

### thereddevils

Re: Ode

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. Jul 1, 2010

### Staff: Mentor

Re: Ode

You did the right thing even though c and c' are arbitrary constants.

7. Jul 1, 2010

### annoymage

Re: Ode

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