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 \(\displaystyle x^2- 5x+ 6= 0\) 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

 

[blue_raver22]

for reaching out about your struggle with a 2nd order ODE in your tutorial. I understand the frustration that can come with facing new challenges in a field that we are not used to.

To convert a 2nd order ODE to a 1st order ODE, the first step is to introduce a new variable, let's call it u, that is equal to the derivative of y with respect to x. This means that u = dy/dx.

Next, we can rewrite the 2nd order ODE in terms of u and y, rather than x and y. This will result in a system of two 1st order ODEs.

To solve these 1st order ODEs, you can use methods such as separation of variables, substitution, or the method of integrating factors. It is important to note that the initial conditions for the 2nd order ODE will need to be translated into initial conditions for the 1st order ODEs as well.

I hope this helps you get started on converting your 2nd order ODE to a 1st order ODE. Remember to take your time and approach it step by step. If you have any further questions or need clarification, don't hesitate to reach out. Good luck!