# First order differentials

1. Nov 25, 2008

### vikkisut88

1. The problem statement, all variables and given/known data
Solve: (2t+x) dx/dt + t = 0

2. Relevant equations
y' +p(X)y = q(x)
and y(x) = ($$\int$$u(x)q(x) + c)/u(x)
where u(x) = e$$\int$$p(x)dx
Note this u(x) is 2 to the power of the integral of p(x)

3. The attempt at a solution
(2t+x) dx/dt + t = 0 becomes:
dx/dt + t/(2t+x) = 0 by dividing through by (2t+x)

However, now I don't know how to separate t and x in the second term.

2. Nov 25, 2008

### jeffreydk

It looks like it is first-order homogeneous. Make the substitution z=x/t by dividing by t throughout your t/(2t+x), and then x=tz and x'=z+tz' by the product rule. Now it's in a form you want it for the relevant equations you listed. Just make the substitution the other way when you've solved for z.

3. Nov 25, 2008

### HallsofIvy

Staff Emeritus
These apply only to a linear first order d.e. And this d.e. is NOT linear because of the x multiplying dx/dt.

4. Nov 25, 2008

### vikkisut88

Okay so does that mean that what jeffreydk would not work? If so what can i use instead? I have only been taught about linear first order d.e so really don't know what to do!

5. Nov 25, 2008

### jeffreydk

Yea it is non-linear I didn't notice that. What I said wont work.

6. Nov 25, 2008

### HallsofIvy

Staff Emeritus
No, being "homogeneous" has nothing to do with being linear. I would consider it very strange that you have "only been taught about linear first order d.e.". Usually separable d.e. comes before that.

In any case, as jeffreydk said (and I won't allow him to take it back!) making the substitution z= x/t so that tz= x and z+ t dz/dt= dx/dt. Dividing the equation by t gives (2+ x/t)dx/dt+ 1= 0.

(2+ z)(z+ t dz/dt)+ 1= 0 or t(2+z) dz/dt= 1-(2+ z2)= z2- 1 which is a separable equation.