First Differential Equation

[TG3]

Homework Statement

Find the general solution for y' +3y = t + e^(-2t) for y.

The Attempt at a Solution

At first I thought that since the equation was already separated, I could simply integrate both sides and get a solution easily:

That results in 1.5 y^2 + y = .5t^2 - .5e^(-2t)
(Unless I made a simple error, which is quite possible.)

However, I quickly realized that this is not the approach to take, since it still contains multiple powers of y.

So, I suspect that I will need to find an integrating factor to multiply by, (commonly called mu, I believe) but I'm not sure how you're supposed to find that.

 

[rock.freak667]

For an equation

$$\frac{dy}{dt}+P(t)y=Q(t)$$


an integrating factor u is given by u=e^{∫P(t) dt}

 

[Mark44]

Homework Statement

Find the general solution for y' +3y = t + e^(-2t) for y.

The Attempt at a Solution

At first I thought that since the equation was already separated, I could simply integrate both sides and get a solution easily:

It might look separated to the most casual observer, but it's not, and further, it's not separable. You have
dy/dt + 3y = t + e^-3t

As it sits, there's no way to get all of the terms involving y and dy on one side, and the other terms involving t and dt on the other side. An integrating factor, as rock.freak667 suggested, is the way to go.