Thread: 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(

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

What do you get?

For 1 you are correct. Now solve for in terms of and see what happens as you make .

For 2, the integrating factor is . Multiply both sides by and see what you can do...

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?

Your working out is a bit all over the place.

You should have





.

Go from here.

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?

Correct. Now see what happens as you make .

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.

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?

nvm. I got it haha.

Originally Posted by Taurus3

nvm. I got it haha.

Great!