# First-Order Linear Differential Equation

1. Sep 1, 2011

### Dr.Doom

1. The problem statement, all variables and given/known data
y'=1+x+y+xy, y(0)=0

2. Relevant equations
$\frac{dy}{dx}$+P(x)y=Q(x)
$\rho$(x)=e$\int(P(x)dx)$

3. The attempt at a solution
My main problem is correctly getting it into the form, $\frac{dy}{dx}$+P(x)y=Q(x). I know what to do from there.
First, I tried y'-(y+xy)=1+x, where P(x)=1+x $\rightarrow$ $\rho$(x)=e$\int(1+x)$=ex+x2/2.
When i multiply both sides by $\rho$(x), i don't come out with anything that i can easily integrate. Any suggestions would be much appreciated!

2. Sep 1, 2011

### Tomer

Notice you wrote: y' - (y+xy) = 1+x.
To understand what's P(x), you need to see what the coefficient of y is:
y' -(1+x)y = 1+x.

What's the coefficient of y?

3. Sep 1, 2011

### Dr.Doom

-(1+x) is the coefficient of y. I went back and corrected this but I am still ruining into a problem when integrating the right hand side of my function: (1+x)e-x-(x2/2)

4. Sep 1, 2011

### Tomer

I doubt this can be integrated. Maybe you can leave the final answer in an integral form?

5. Sep 1, 2011

### Dr.Doom

Well, it's an initial value problem so I'm thinking there's something more to it. The textbook gives the answer as y(x)=-1+ex+x2/2

6. Sep 1, 2011

### rude man

y' = 1 + x + y + xy
y' = 1 + x + y(1+x)
y' = (1+x)(1+y)
dy/(1+y) = (1+x)dx

Integrate both sides etc. Don't forget arb. constant which is found by y(0) = 0.

Note: always try sep. of variables first.

7. Sep 1, 2011

### Tomer

Or do that :-)

8. Sep 1, 2011

### Dr.Doom

Thanks for the help, guys!