# Linear First Order Equation

1. Feb 5, 2006

### Zem

I'm using an integrating factor, rho(x), to solve an equation of the form
dy/dx + P(x)y = Q(x)

I need to find the particular solution.
y' = 1 + x + y + xy; y(0) = 0
y' - y - xy = 1 + x
dy/dx + y(-1-x) = 1 + x
P(x) = (-1-x), Q(x) = (1 + x),
rho(x) = e^(-x-1/2x^2)
(Multiply both sides by rho(x))
e^(-x - 1/2x^2)(dy/dx) + e^(-x - 1/2x^2)(1-x) = e^(-x - 1/2x^2)(1+x)

Dx[y * e^(-x - 1/2x^2)] = e^(-x - 1/2x^2)(1+x)
y * e^(-x - 1/2x^2) = int_e^(-x - 1/2x^2)(1+x)dx

(Multiply both sides by the reciprocal of rho(x))
y(x) = e^(x + 1/2x^2) * int_e^(-x - 1/2x^2)(1+x)dx

(U substitution) This is where I get stuck. It seems that I need u substitution to find the integral on the right side
u = e^(-x - 1/2x^2)
du = (-x - 1) * e^(-x - 1/2x^2)
dv = 1
v = (1+x)

When I set that up in uv - int_v du it looks just as bad as the original integral.

Thanks in advance for any assistance.

Last edited: Feb 5, 2006
2. Feb 5, 2006

### d_leet

I don't have a clue what you did, that is some of the messiest math I have seen, I can't follow your steps at all. You do know how to find an integrating factor right? Just e raised to the integral of p(x), you multiply everything in the original equation by that and you should get a function that is the derivative of a product of y and some function of x. It kind of looks like you tried to raise e to the integral of the function you were supposed to be integrating.

3. Feb 5, 2006

### Zem

I have cleaned it up. I attempted to use LaTeX notation for e and the integral sign, but made typos and my edits didn't appear.
Yes, that was a mistake I made in the post that I didn't make in writing it out. This is my integrating factor...
P(x) = (-1-x), Q(x) = (1 + x),
rho(x) = e^(-x-1/2x^2)
(Multiply both sides by rho(x))
e^(-x - 1/2x^2)(dy/dx) + e^(-x - 1/2x^2)(1-x) = e^(-x - 1/2x^2)(1+x)
(Which is the product rule of...)
y * e^(-x - 1/2x^2) = int_e^(-x - 1/2x^2)(1+x)dx

I'll fix the original post.

Last edited: Feb 5, 2006
4. Feb 5, 2006

### d_leet

Ok It looks like you did it right then, and the right hand integral is just the integral of -eudu

5. Feb 5, 2006

### Zem

By -e^udu, do you mean this?
int_e^(-x - 1/2x^2) * (-x - 1) * e^(-x - 1/2x^2)

I am certain my ti-89 would just give me 2 integrals if I punch that integral into it. Is there a way I could make a less complex integral for Q(x) for my calculator?

6. Feb 5, 2006

### d_leet

You had this right int_e^(-x - 1/2x^2)(1+x)dx?

Let u = -x - 1/2x2 then du = -(1+x) so you have -eudu

7. Feb 5, 2006

### Zem

Beautiful. Thanks again!