I need to find particular solution for the diferental equation: 2y+dy/dx -2 = 2 x>=0 y(0)=1 I have tried many things, Scorpsel on #bit-tech was helping me out but im till having trouble. I have got a bunch of these to do I have done several that are regarded as more difficult but this one is being a bitch... Any help would be great!! Thanks in advance 'doc

y = 2 - e^-2x Don't have time to go through all the steps, but basically you use an integrating factor of e^2x, so your equation becomes: d/dt(ye^2x) = 4e^2x From there, simple integration and division.

That is the correct answer, thankyou I'm a bit confused on what you mean by intergrating factor. 'doc

Now that I'm done with my classes for the day (lucky for you, I'm taking Diff Eq this quarter), I'll go in detail what I did: The general way to solve diff eqs with the form y' + v(t)y = s(t) is to use what is known as the integrating factor, usually called µ(t), which is simply e raised to the power of the integral of v(t). So now the equation becomes: d/dt(y*µ(t)) = s(t)µ(t) To picture how it works, take the derivative of ye^µ(t). Since y is a function of t, you have to use the chain rule, which would give you: y*µ'(t) + y'µ(t) = s(t)µ(t) However, we established already that µ'(t) = v(t): y*v(t)*µ(t) + y'µ(t) = s(t)µ(t) Dividing both sides by µ(t): y' + yv(t) = s(t) Our original equation. Anyway, back to solving. Now you have this: d/dt(yµ(t)) = s(t)µ(t) Integrating both sides, you get: yµ(t) = integral s(t)µ(t) The integral should be fairly easy to solve. After that, all you need to do is divide by µ(t) to get y(t). If there's an initial value problem, you'll need to solve for the constant that you got from integrating s(t)µ(t).