Can anyone help me? I have no idea .

  • Thread starter zibb3r
  • Start date
  • #1
6
0

Main Question or Discussion Point

Can anyone help me? I have no idea.....

I have no idea how to approach this; so figured it was worth a shot asking you guys!

Cheers!

Using the Principle of Superposition, fi nd the general solution of the diff erential equation:

dx/dt = (1/2)x+4 ; subject to the initial condition x(0)=1

Thanks guys; I'm really stuck, otherwise I wouldn't be asking!

:smile:
 

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
Well, what IS the Principle of Superposition?
Clarifying your mind on that will give you a clue.
:smile:
 
  • #3
6
0
I think I've just confused myself more.... ! I get what it is, but I'm not really sure how to apply it to this equations; if that makes any sense???
 
  • #4
6
0
Hang on..... Does this look ok???

dx/dt = 1/2 x + 4
dx/(1/2 x + 4) = dt
2dx/(x+8) = dt
2ln(x+8) = t + C
ln(x+8) = (t+C)/2
x+8 = Ce^(t/2)
When t = 0, x = 1
So, 1+8 = Ce^0
C = 9
x+8 = 9e^(t/2)
x = 9e^(t/2) - 8
 
  • #5
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
That is certainly the correct answer (congrats!), but not showing mastery of the Principle of Superpositon.

As the question was given, you should show a clear understanding of THAT principle.

Here's how you may do it.

The Principle of Superposition is a property of what we call linear, homogenous differential equations.
"Linear" means that for every term in the diff.eq, the unknown function "x" or its derivatives is contained AT MOST to the first power; for example the term 1/2x(t) can appear in a linear diff.eq, whereas a term x(t)^2 can not.

"Homogenous" means, basically that all non-zero terms in a linear differential equation are explicit expressions containing the unknown function "x" or its derivatives.

For example, the linear differential eq.
dx/dt-1/2x=0 (**) is homogenous; we have a "zero term", plus two terms, the one being dx/dt, the other 1/2x.
--------------------------------
For linear, homogenous diff. eqs, we have the result: Suppose X(t) and Y(t) are solutions of the diff.eq. Then, the sum Z(t)=a*X(t)+b*Y(t), for arbitrary constants a,b is ALSO a solution. (THIS is the Principle of Superposition!!)

To take the example (**) above, with X(t), Y(t) solutions, we may verify that Z(t) is ALSO a solution, by computing LHS:

dZ/dt-1/2*dZ/dt=(a*dX/dt+b*dY/dt)-1/2*(a*X+b*Y)=a*(dX/dt-1/2X)+b*(dY/dt-1/2*Y)=a*0+b*0=0,

that is, Z(t) is also a solution, given that X and Y are solutions!
-------------------------
NOW, your problem was:
dx/dt-1/2x=4 (***)

This is NOT homogenous, because the term "4" is non-zero, and does NOT contain "x" or its derivatives!

However, let X_(p)(t) be a particular solution of (***). Then, a theorem associated with the Principle of Superposition says that if X_(h)(t) is any solution of the associated homogenous diff.eq (that is, (**)), then X_(p)(t)+X_(h)(t) is ALSO a solution of (***)

Why? We insert in LHS:
dX_(p)/dt+dX_(h)/dt-1/2*(X_(p)+X_(h))=dX_(p)/dt-1/2*X_(p)+dX_(h)/dt-1/2*X_(h)=4+0=4,

that is, the sum X_(p)(t)+X_(h)(t) is ALSO a solution of (***).
------------------------------------------------
Given this background, we can now see how you should proceed to solve this problem, utilizing the Principle of Superposition:
1. Find a solution to the homogenous equation, dx/dt-1/2x=0
2. Make a lucky guess at (***)
3. Adjust integration constant from 1. so that the sum from 1.+2. satisfies initial condition.

Now, it is easy to see that the general solution of 1. is x(t)=C*e^(t/2)=X_(h)(t)
Furthermore, for 2: Since the RHS in (3) is a constant, guess at an X_(p)(t)=A, where "A" is some constant to be determined for solving (***)
Inserting X_p(t) into (***), we get: 0-1/2*A=4, that is, A=-8 is a particular solution of (***)
------
Thus, we have a general solution of (***) x(t)=C*e^(t/2)-8.
We find then, that C=9 in order to satisfy the initial condition.
 

Related Threads on Can anyone help me? I have no idea .

Replies
1
Views
1K
Replies
11
Views
2K
  • Last Post
Replies
7
Views
2K
Replies
1
Views
473
Replies
6
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Top