Solve an ODE using exact methods?

  • #1

Homework Statement


dy/dx + 0.8 y = 0.6 e ^-(0.6+0.8) , y(0) = 1

Solve this ordinary differential equation subject to the given condition using exact methods and evaluate the solution y for x = 0.0 (0.05) 0.5, i.e from x = 0 to x = 0.5 in steps of 0.05).

Hi, am pretty hopeless at ODEs, I do not even understand what the question means. Would someone be able to put it simply. I am not sure what exact methods are. Is this to do with runge kutta?

Any help would be great thanks!
 

Answers and Replies

  • #2
HallsofIvy
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You do not understand what "solve the differential equation" means? It means, "find a function y(x) that satifies the differential equation and also satisfies y(0)= 1." You are then asked to evaluate that function at x= 0.05, 0.10, 0.15, 0.20, 0.25, 0.30, 0.35, 0.40, 0.45, and 0.50.

As far as "use exact methods" is concerned, they do not want you to use any approximate or numerical method.

There are many different ways to solve that equation- it is a linear first order equation with constant coefficients. One way to solve it is to find an "integrating factor"- there is a standard formula for finding integrating factors of first order linear equations. Do you know it? Or you could find the solution to the "characteristic equation". Do you know what that is?

I do have a question about "e^-(0.6+0.8)". Was there supposed to be an "x" in there? Otherwise, why not just write "e^(-1.4)"? Was it supposed to be "e^-(0.6x+ 0.8)" or "e^-(0.6+ 0.8x)"? (The second is somewhat harder than the first.)
 
  • #3
Hi sorry your probably right that it should be (1.4) there is also an x on the outside of the bracket meaning it would be e^(1.4)x. I don't know what integrating factors are. This subject is a bit of a mystery to me I've gone through the class notes about half a dozen times and I just can't seem to get my head around it I have notes on 2nd order linear odes with constant coefficients would this be similar?
 

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