# Differential equation: y'-y=(1/2)x-1

1. Mar 15, 2008

### false_alarm

I know this looks really easy, but trying to solve this is amazingly difficult. I couldnt do it, i kept getting wrong answers. Any body got any idea how to solve this?

2. Mar 15, 2008

### John Creighto

When methods have you leart because there is more then one way to solve the above equation.

3. Mar 15, 2008

### false_alarm

I just used the basic linear form, got an integrating factor, and then just took the antiderivative of what was left

4. Mar 15, 2008

### John Creighto

If no one else helps you out I'll try to solve it with that method tomorrow. First though you have to show me the steps you tried.

5. Mar 15, 2008

### false_alarm

y'-y=(1/2)x-1
(1/2)x-1=g(x)
y'-y=g(x)....linear form
d(x)y'-s(x)y=(d(x)y)'.....product rule
only works if d(x)'=s(x)
the only function that is its own integral, that i could think of, is e$$^{x}$$
but since the y is negative, it would have to be e$$^{-x}$$

so then u could say,

e$$^{-x}$$y'-e$$^{-x}$$y=e$$^{-x}$$g(x)

(e$$^{-x}$$y)'=e$$^{-x}$$g(x)....use the product rule
e$$^{-x}$$y=antiderivative(e$$^{-x}$$g(x))........took antiderivative
y=antiderivative(e$$^{-x}$$((1/2)x-1))/e$$^{-x}$$.......divide both sides by e$$^{-x}$$

I think im screwing up in the simplifying part. But, the integration becomes tricky

6. Mar 15, 2008

### false_alarm

Hey I just figured it out. The first time i did it i made a really small error in the beginning that made it unsolvable. Thanks anyways. In case your wondering though,

y=(-1/2)x+(1/2)

7. Mar 15, 2008

### Kukurio

You know that the solution isn't unique, right? You can add e^x times any constant to your answer and still have a valid solution.

8. Apr 12, 2008

### Bork

Yeah, the solution will only be unique if you specify an initial condition such as y(0). Otherwise your integrating factor method should still leave you with a term Ce^x for some undetermined constant C.