# A first order DE

1. Aug 21, 2010

### James889

Hi,

I have the first order differential equation $$y+y' = x$$
y(0) = 0
First i tried to assume a solution of the form Ax+b, that didn't quite work.

Then i tried to use the Integrating factor technique, work follows

$$e^{\int1} = e^x$$

$$\int e^x(y+y') = \int{xe^x}$$

$$ye^x = \int xe^x$$

let u = x, du=1
v = e^x dv=e^x

So i end up with:
$$ye^x = xe^x -e^x$$

which is wrong...
Any ideas?

2. Aug 21, 2010

### l'Hôpital

You forgot your " + C" at the end of your integration. : )

3. Aug 21, 2010

### James889

The answer is supposed to be $$y = ce^{-x} + x-1$$

But the answer i get when dividing both sides by e^x is y = x-1

Last edited: Aug 21, 2010
4. Aug 21, 2010

### l'Hôpital

Where is the integration constant? Where is the "+ C"? That is important because when you divide by e^x, you'll end up with the e^(-x) you were looking for.

5. Aug 21, 2010

### vela

Staff Emeritus
What didn't work when you did this? Is it the same problem where you're missing the e-x term?

6. Aug 22, 2010

### James889

Yes,

Put y = Ax + b
then y' = A

So i get: Ax + b + A = x

7. Aug 22, 2010

### willem2

What values of A and b make the left hand side also equal to x?

if y=f(x) is a single solution of the equation y'+y = x the other solutions
are y = f(x) + C g(x) where y= g(x) is a solution of y' + y = 0.

8. Aug 22, 2010

### James889

The values A=1, b=-1

But does this mean i have to guess G(x) ?

9. Aug 22, 2010

### vela

Staff Emeritus
No, you have to solve the corresponding homogeneous equation y' + y = 0 and add that solution to the other one, yp=x-1, to get the general solution.

10. Aug 22, 2010

### HallsofIvy

As l'Hopital said, you forgot the "constant of integration", C, in your original solution:
[tex]ye^x= xe^x- e^x+ C[/itex]

Now, dividing through by $e^x$ gives $y= x- 1+ Ce^{-x}$