# Thread: general solution of differential equations

1. ## general solution of differential equations

I've to find the general solution and determine how the solutions behave as t-->infinity.

1. y'-2y=t^(2)e^(2t)
μy'-2μy=μt^2e^(2t)
d/dt (μ(t)y(t)) = μy'+μ'y
What we need: μ' = -2μ, which is a separable equation with μ=e^(-2t)
d/dx(e^(-2t)y)=e^(-2t)t^2e^(2t)=t^2
Integrate: e^(-2t)y=t^3/3+C
Did I do this right and how does it behave?

2. y'+y= te^(-t)+1. I couldn't figure out this one 8(

2. Originally Posted by Taurus3
I've to find the general solution and determine how the solutions behave as t-->infinity.

1. y'-2y=t^(2)e^(2t)
I don't think this is separable.

Find and integrating factor $\displaystyle I =e^{\int -2~dt}$

What do you get?

3. For 1 you are correct. Now solve for $\displaystyle y$ in terms of $\displaystyle t$ and see what happens as you make $\displaystyle t \to \infty$.

For 2, the integrating factor is $\displaystyle e^{\int{1\,dt}} = e^t$. Multiply both sides by $\displaystyle e^t$ and see what you can do...

4. ok, so for the 2nd question:
p(t)=1, g(t)=te^(-t)+1
μ(t)=e^t
So integral of μ(t)g(t) = integral of e^t(x-e^x(x+1))
Do I have it right so far?

5. Your working out is a bit all over the place.

You should have

$\displaystyle e^t\frac{dy}{dt} + e^ty = e^t(t\,e^{-t} + 1)$

$\displaystyle \frac{d}{dt}\left(e^ty\right) = t + e^t$

$\displaystyle e^ty = \int{t + e^t\,dt}$.

Go from here.

6. so the final answer for 1 is: y=(t^3/3+C)e^(2t)
2: y=(e^t+t^2/2+C)/e^t right?

7. Correct. Now see what happens as you make $\displaystyle t \to \infty$.

8. 1. The If c is non negative, then the solution grows exponentially large in magnitude. So the solutions diverge as t gets bigger. The boundary between solutions that ultimately grow negatively occurs when c is negative.

2. uggh...can't figure this one out.

9. 2. Why don't you just divide through by the exponential, term-by-term, and see what happens to each term as t goes to infinity?

10. nvm. I got it haha.

11. Originally Posted by Taurus3
nvm. I got it haha.
Great!