How can I convert a 2nd order ODE to a 1st order ODE?

[jrv24]

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

 

[HallsofIvy]

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, d^2x/dt^2= d(dx/dt)dt= dz2/dt and d^2y/dt^2= d(dy/dt)/dt= dz4/dt.

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

The other two equations are, of course, dz1/dt= z2 and dz3/dt= z4.

 

[eng leopard]

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

 

[HallsofIvy]

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 x^3- 6x^2+ 11x- 6= 0 is x= 1 and that means that the polynomial has a factor of x- 1. Dividing that polynomial by x- 1 I get [math]x^2- 5x+ 6= 0[/math] as a second degree equation for the other two roots.

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

Putting those into the