Reduce second order diff equation into series of first order diff equations

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Homework Help Overview

The discussion revolves around reducing a second order differential equation into a series of first order equations. The specific equation under consideration is y'' - 30y' - 3y = -2, with initial conditions provided.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reduce the second order equation by introducing a substitution, y' = z, and subsequently forming a first order equation. Some participants question the correctness of this reduction and the subsequent application of the Runge-Kutta method.

Discussion Status

There is an ongoing exploration of the reduction process and the application of numerical methods. Some guidance has been offered regarding the potential for analytical solutions, and concerns have been raised about the accuracy of the numerical results obtained by the original poster.

Contextual Notes

Participants note that the steep nature of the function may complicate the evaluation using approximate methods, and there is a mention of the initial conditions and the step size used in the numerical method.

shaka091
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Hi guys I was hoping if someone could help me with this second order differential equation which i have to reduce into a series of first order equations and then solve using a fourth order runge kutta method.

The equation is

y"-30y'-3y=-2 with the initial conditions y(1)=-12 and y'(x=1)=-2

My attempt at reducing it is

I firstly did y'=z and then did dz/dx=30z+3y-2 with the conditions being x(0)=1, y(0)=-12 and z(0)=-2

could someone please tell me if this is right and if not point me in the right direction to be able to reduce it. I think i can solve it using the runge kutta. Thank you very much!
 
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That is entirely correct.

You can also solve this equation analytically without too much trouble by using an ansatz y(x) = A exp(rx) to the homogeneous equation, and finding the trivial solution y=2/3 to the full equation, so you can check that your integrator gives you correct results.
 
Thank you very much for your reply!

I followed all the steps in my lecture notes to reduce it and thought it was correct but I thought I must have got it wrong when i then put it into the runge kutta as i was getting back some pretty big values.

Evaluating y(1.2) using a step size of h=0.2

I got y(n+1)= -24.212 and z(n+1)=-377.1852 and x(n+1)= 1.2

I don't think that is right...
 
The function is really steep so it might be difficult to evaluate with approximate methods. The correct values should be higher, for example z(1.2) ~*-1337
 

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