# Homework Help: Homogenous Differential Equation Solution

1. Sep 25, 2014

### _N3WTON_

1. The problem statement, all variables and given/known data
Solve the differential equation:
$\frac{dy}{dx} = x + y$

2. Relevant equations
Integrating factor equation

3. The attempt at a solution
Ok, I am aware that in order to solve this equation, I need to make a substitution:
$v = x + y$
However, at this point I am unsure about what to do. Unfortunately, my DiffEq textbook does not cover this technique thoroughly. I am hoping that perhaps somebody can explain how this technique works, thanks.

2. Sep 25, 2014

### Ray Vickson

The substitution $v = x+y$ leads to another inhomogeneous DE, but simpler than the original. Have you tried it?

I cannot believe that your textbook does not cover the methods for solving
$$\frac{dy}{dx} + f(x) y = g(x)$$
You say it does not cover this technique thoroughly, but surely it covers it at least a little bit. You need to show your work, and explain where/why you are stuck.

3. Sep 25, 2014

### LCKurtz

@N3WTON: Alternatively, doesn't your text discuss constant coefficient homogeneous and non-homogeneous equations and the method of undetermined coefficients for NH equations?

4. Sep 25, 2014

### ehild

With the substitution, y=v-x, y'=v'-1 and the differential equation becomes v'-1=v, that is, v'=v+1. Is it not a separable equation?

ehild

5. Sep 26, 2014

### _N3WTON_

It does discuss constant coefficient homogenous and non-homogenous equations, however; no mention of the method of undetermined coefficients for NH equations. Perhaps we have just not reached that point in the text yet

6. Sep 26, 2014

### _N3WTON_

I have tried it but I don't understand what I need to do. The equation you listed could be solved using the integrating factor method, no? I'm confused about how that relates to the equation I listed...

7. Sep 26, 2014

### _N3WTON_

ok...so picking up from there I did:
$\int\frac{v'}{v+1} = \int 0$
$ln(v+1) = C$
$v + 1 = e^c$
$v = e^{c} - 1$
$x + y = e^{c} - 1$
Here I am stuck because the solution in the book is:
$y = 2e^x - x - 1$
Also the book notes that this problem can be "solved quite easily" using the integrating factor method, but I do not see it...

8. Sep 26, 2014

### pasmith

This should be $$\int \frac{v'}{v +1}\,dx = \int 1\,dx.$$

You appear to have omitted an initial condition (perhaps y(0) = 1) from your problem; otherwise '2' should instead be an arbitrary constant.

9. Sep 26, 2014

### ehild

What is v+1 divided by (v+1)???????
ehild

10. Sep 26, 2014

### _N3WTON_

nvm

11. Sep 26, 2014

### _N3WTON_

12. Sep 26, 2014

### Orodruin

Staff Emeritus
You seem to have assumed that $v'/(v+1) = 0$. This is not the case so I suggest you recheck that algebra.

Another option is to find a particular solution (there is a fairly obvious one) and then add the solution to the homogeneous equation to that (which works since the ODE is linear).

13. Sep 26, 2014

### _N3WTON_

you're right, I do apologize for that....I'll make the edit now

14. Sep 26, 2014

### _N3WTON_

ok, I see the mistake now...thank you

15. Sep 26, 2014

### pasmith

You are integrating both sides with respect to $x$. It is crucial to state what variable you are integrating with respect to, because shortly you will use substitution to change $\int \frac{v'}{v +1}\,dx$ into an integral with respect to $v$.

16. Sep 26, 2014

### _N3WTON_

thank you, so here is my second attempt:
$\int\frac{v'}{v+1} dx = \int dx$
$ln(v+1) = x + C$
$v+1 = e^x + C$
$x + y = e^x + C$

17. Sep 26, 2014

### ehild

the last line is wrong. It should be $v+1 = e^{x + C}$ instead.

18. Sep 26, 2014

### _N3WTON_

thank you....

19. Sep 26, 2014

### Ray Vickson

In your example, is it not the case that $f(x) = -1$ and $g(x) = x$?

20. Sep 26, 2014

### ehild

And $v+1 = e^{x + C} = Ae^x$ . The constants C or A depend on the initial condition.

ehild

21. Sep 26, 2014

### _N3WTON_

ok, the initial condition was f(0) = 1...you arrived at that answer just by bringing the constant C in front of the exponential?

Last edited: Sep 26, 2014
22. Sep 26, 2014

### _N3WTON_

ok so if i were to use the integrating factor method I would set:
$p(x) = -1$ ?
I'm still a bit confused about how I could do this problem using integrating factor method...I understand how to use separation of variables..

23. Sep 26, 2014

### Ray Vickson

There is a standard formula; just apply it to your problem.

24. Sep 26, 2014

### ehild

You know that $e^{{x+C}}=e^C e^x$, but eC is also a constant, I denoted it by A. So v+1=y+x+1=Aex.. Applying the condition y(0)=1, 1+1=A.

25. Sep 27, 2014

### Orodruin

Staff Emeritus
Just for completeness, since we have a linear ODE with constant coefficients and a polynomial inhomogeneity, we could simply have assumed a particular solution of the form $y_p = ax + b$ and arrived at $a = b = -1$ and thus $y_p = -x-1$, $y'_p = -1$. We then write $y = y_p + y_h$ where $y_h$ satisfies the homogeneous ODE $y'_h = y_h$, which I believe anyone familiar with differential equations can solve ... This approach is what I was trying to hint at in post #12.