Help with DE and UNIT STEP FUNCTION

tung
Messages
2
Reaction score
0
I'm not sure how to solve a differential equation with unit step function, for example:

x'' + 2x' + x = 10t*u(t), where x(0)=1 and x'(0)=0

Do I just ignore the u(t) and solve it regularly by normal integration?
 
Physics news on Phys.org
Have you learned the Laplace transform yet? If you have, transform both sides of the DE, and express L(x), then find the inverse Laplace of that. The fact that they gave you x(0) and x'(0) hints strongly that you should use that.

To do it without Laplace you'd have to separate the DE for 2 separate intervals, one for which u(t) = 1 and another for u(t)=0, for different intervals of t.
 
I really dislike the Laplace Transform! Use Defennnder's second method; solve two problems:

First solve x" + 2x' + x = 0, x(0)= 1, x'(0)= 0. Call that x1(t).

Then solve x'' + 2x' + x = 10t, x(0)= 1, x'(0)= 0. Call that x2(t).

x(t)= x1(t) for t< 0 , x2 for t> 0. Of course, they are the same at t= 0.
 
Last edited by a moderator:
LOL, well I've grown used to Laplace transform. I guess it's because once you learned something new you'll always try to find ways of applying, even if it results in a less efficient way of doing things. But anyway, it looks as though this problem was catered specially for the Laplace transform.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top