First-Order Linear Differential Equation

[Dr.Doom]

Homework Statement

y'=1+x+y+xy, y(0)=0

Homework Equations

\frac{dy}{dx}+P(x)y=Q(x)
\rho(x)=e^\int(P(x)dx)

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)=e^x+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!

 

[Tomer]

Meaning, your P(x) is wrong.

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?

 

[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)

 

[Tomer]

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

 

[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+e^x+x2/2

 

[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.

 

[Tomer]

Or do that :-)

 

[Dr.Doom]

Thanks for the help, guys!