2nd order ODE to 1st Order

  • Thread starter jrv24
  • Start date
  • #1
1
0
Hi, have this strange 2nd order ODE in one of my tutorials that I am struggling to start. I am not used to dealing with derivatives of both x and y as well as a function involving t.
I was wondering if anyone may be able to point me to the starting line.

I am trying to convert them into 1st order ODEs.

Thanks
 

Attachments

  • Picture 1.png
    Picture 1.png
    5.3 KB · Views: 435

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Do you mean convert those two second order equations to four first order equations? There is no way to reduce a single second order equation to a single first order equation or two second order equations to two first order equations unless you have additional information such as already knowing a solution.

The standard method would be to define new variables, say z1, z2, z3, and z4 such that z1=x, z2= dx/dt, z3= y, and z4= dy/dt. Of course, [itex]d^2x/dt^2= d(dx/dt)dt= dz2/dt[/itex] and [itex]d^2y/dt^2= d(dy/dt)/dt= dz4/dt[/itex].

The equation [itex]d^2x/dt^2+ dy/dt- y+ x= e^t[/itex] becomes [itex]dz2/dt+ z4- z3+ z1= e^t[/itex] or [itex]dz2/dt= -z1+ z3- z4+ e^t[/itex] and [itex]d^2y/dt^2+ 2dx/dt+ y- 2x= 0[/itex] becomes [itex]dz4/dt+ 2z2+ z3- 2z1= 0[/itex] or [itex]dz4/dt= 2z1- 2z2- z3[/itex].

The other two equations are, of course, [itex]dz1/dt= z2[/itex] and [itex]dz3/dt= z4[/itex].
 
  • #3
There is no way to reduce a single second order equation to a single first order equation or two second order equations to two first order equations unless you have additional information such as already knowing a solution.

please , could you re-explain this passage.it is not uderstood for me , thank you
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
If you have, say, a second order differential equation and already know one solution you can reduce it to a first order equation for another, linearly independent, solution in much the same way that you can reduce the degree of a polynomial if you already know one of its roots.

For example, I know that one of the roots of the polynomial equation [itex]x^3- 6x^2+ 11x- 6= 0[/itex] is x= 1 and that means that the polynomial has a factor of x- 1. Dividing that polynomial by x- 1 I get \(\displaystyle x^2- 5x+ 6= 0\) as a second degree equation for the other two roots.

Similarly, if I know that [itex]y= e^x[/itex] is a solution to the differential equation y''- 3y'+ 2y= 0, I can "try" a solution of the form [itex]y= u(x)e^x[/itex]. Then [itex]y'= u(x)e^x+ u'(x)e^x[/itex] and [itex]y''=ue^x+ 2u'e^x+ u''e^x[/itex].

Putting those into the
 

Related Threads on 2nd order ODE to 1st Order

Replies
6
Views
4K
Replies
4
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
4
Views
3K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
Top