How to convert to first order system?

Feb 2016
13
0
Hudson Ohio
Hello! I have to convert this equation into a first order system and then write the discrete equations for that system based on the Euler Method as explicitly as possible. Then I have to apply one step of Euler's method by hand with h=0.1. I think I can do the Euler step by myself, but am confused as to how to rewrite the equations as first order? Any help is appreciated. Below are the equations and ICs.

u''(t) - u3(t)*u'(t) + 3u(t) = t2​ with u(0) = 1 and u'(0) = 2
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
There is a pretty standard method for that. If you are taking differential equations, surely you have seen it?

You have the second order differential equation \(\displaystyle u''(t) - u^3(t)*u'(t) + 3u(t) = t^2\). Let v= u'. Then u''= (u')'= v' so the equation becomes \(\displaystyle v'- u^3v+ 3u= t^2\). That, together with \(\displaystyle u'= v\) gives two first order differential equations. The "initial conditions", u(0)= 1 and u'(0)= 2 become u(0)= 1 and v(0)= 2.
 
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Feb 2016
13
0
Hudson Ohio
For some reason I didn't learn how to convert systems, which is making higher level math problems a blast as you can imagine. Thanks so much! And just to double check, to convert to set up for Euler, I have y' = t2 + x3y - 3x = f(x,y) ?
 
Feb 2016
13
0
Hudson Ohio
Also, am I doing something wrong? Shouldn't the t be gone?
 
Feb 2016
13
0
Hudson Ohio
And if the t shouldn't be gone, then how do you apply Euler's method to a 3 variable problem?
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
"t" is the independent variable, not a third dependent variable. Your equation would be, as I said, \(\displaystyle u'= du/dt= v\), and \(\displaystyle v'= dv/dt= u^3v- 3u+ t^2\) so that \(\displaystyle du= vdt\) and \(\displaystyle dv= (u^3v- 3u+ t^2)dt\). To do an Euler numerical integration, one would start with u(0)= 1, v(0)= 2. Then taking, say, dt= 0.1, \(\displaystyle du= 2(0.1)= 0.2\) and \(\displaystyle dv= (1(2)- 3(1)- 0)(0.1)= -0.1\) so u(0.1)= u(0)+ du= 1+ 0.2= 1.2 and v(0.1)= v(0)+ dv= 2- 0.1= 1.99. Then \(\displaystyle du= 1.99(0.1)= 0.199\) and \(\displaystyle dv= ((1.2)^3(1.99)- 3(1.2)+ 0.1^2)(0.1)= -0.015128\) so that \(\displaystyle u(0.2)= 1.2+ 0.199= 1.399\) and \(\displaystyle v(0.2)= 1.99- 0.015129= 1.974871. etc. (You will have to decide what accuracy makes sense.\)