Finding a Basis for solutions of Second Order ODE's
In the last post, said I would show you how solutions of second-order linear, differential equations behave physically. But before we can do that, I have to cover this rather dry, math intense proof, in order to make sense of one of the 3 possible solutions. We'll discuss these possibilities in the next post, so I apologise for that...
Now in the last post, we found that sin(x) and cos(x) are linearly independent functions that satisfy the second order differential equation...
![o1.png](https://steemitimages.com/DQmcnJaGMzCFbG1fCd8KS7qBcjXZ4KLxYPJvR16bQAqrQbB/o1.png)
And hence the general solution to equation (1) is a linear combination of both of these functions. So we say that sin(x) and cos(x) form the basis of solutions for equation (2) on the interval (-∞, ∞).
The general solution to (1) is...
![n5.png](https://steemitimages.com/640x0/https://steemitimages.com/DQmNvNHr3VMnPd5e9vEBHsafZXRspVhZHWsny6Lvs1MrC4K/n5.png)
Reduction of order and finding a Basis if 1 solution is known...
The standard form of second-order Ordinary Differential Equations (ODE's) is shown in equation (2) below...
![o2.png](https://steemitimages.com/DQmYT6GQZVwDo5QT7QYYXd2wEiZKkJ1JVopu4if2YSoEwx6/o2.png)
Like above with equation (1), we expect to find 2 linearly independent solutions for equation (2) in order to form a basis for the general solution.
Now let's assume we know one solution and want to find the other linearly independent solution...
To do this, we let...
![o4.png](https://steemitimages.com/DQmZnLSEHeQNnAf2fcVVYeuwh9h4Y48BnVtDmyaPfekLDUA/o4.png)
then...
![o5.png](https://steemitimages.com/640x0/https://steemitimages.com/DQmbdWVSRv88rfBLvgqx3dAqUuTTLiFtPDkTurrdtXRjetr/o5.png)
Substituting these derivatives into equation (2), we get...
![o6.png](https://steemitimages.com/DQmVGfXb2Pu3tGhSuhRxZFGp1j9Vht7uRemrHJ2vyYpy57b/o6.png)
Ok. Since is a solution to (2), the term in the last brackets is 0. Thus...
![o7.png](https://steemitimages.com/DQmagNPTiJE1ZU3tfPhyvXCnbe2u9mPuv3sVNh84aB249rz/o7.png)
Now divide equation (3) by ...
![o8.png](https://steemitimages.com/DQmexEbk3vAbb5iEgpa5GExi48fa9wdcZFgkuSgjYDE52CS/o8.png)
Therefore equation (3) has been reduced to a first order ODE for which we can separate the variables and solve by integrating both sides...
![o9.png](https://steemitimages.com/640x0/https://steemitimages.com/DQmSD8ys93kc1PSw8217NKhgDjRVqStxASz9ZxCxXQczkKS/o9.png)
We don't need a constant of integration because we are just finding the other solution to the ODE. So exponentiating both sides...
![o10.png](https://steemitimages.com/DQmRdvuyPj9CVEiMxzupQqVx68jUcrU4ehy5aCebvKxvzSM/o10.png)
Now, since and
, it follows that...
![o12.png](https://steemitimages.com/640x0/https://steemitimages.com/DQmdbJhhuDbpTHXY9nBfiSoDG5QMVjsbitEeUvVLQHivoVt/o12.png)
Let's see if it works...
Since we already know the basis of equation (1), let's see if all of this derivation works.
For equation (1), with and
...
![o16.png](https://steemitimages.com/DQmTFVdRCsw4xKu7STP76JUdSAEGXu7bVyZuYsLucRXLWG1/o16.png)
Finally...
![o17.png](https://steemitimages.com/640x0/https://steemitimages.com/DQmRhWPN36vY2oUfEMkcTGtX1nRhGbTn7MMtwwvwyEsBtAH/o17.png)
Therefore, we have found the other solution of sin(x). So, it goes to show that if we know a solution to a second-order ODE, we can find the other solution
using equation (4).
Credits:
All equations in this tutorial were created with QuickLatex
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