Thank you so much. I was deriving an equation for a PIAB in three dimensions (cartesian coordinates) for my pchem midterm tonight and this concept was honestly the hardest part.
Probably a little late now but anyways. It’s an ordinary differential equation in which case you are just looking for a X(x) that relates to it’s derivatives. In this situation it is not too hard to see that sin and cos are heavily related to their second derivatives. They are just the negatives, so if you plug in either sin or cos for X(x) you’ll see it works out however using just one isn’t the whole answer. Hence why he uses both sin and cos with an arbitrary constant “a”,”b” this allows for all solutions to be covered in the singular answer. The reason this works is because the sum of two solutions to an ODE is in itself a solution to the same ODE. That proof has to do with some linear algebra but I hope this helps.
@@jordanlaforce2370 Probablya to late now but anyways. You get the sin cos solution by taking the test function e^{lambda * x}. When solving for your constant you get a complex solution and use eulers identity which gives a cos and sin solution.
Im sorry, may i ask something? Why you choose -lambda^2 as a constany, which is the constant is negatif. Why you not choose a constant positif or constanta 0, please tell me why? Thx before
Because when you find the general solution of the two ODE's you have to find the roots of the equations by square rooting and if it's just lambda or k instead of lambda^2 or k^2 you end up with a more complicated square root problem. It just easier to work with k^2 or lambda^2, than k or lambda.
The constant can be named anything you want. In this case, I knew that eventually I wanted solutions like sin(kx) with k real. In practice doing it yourself, you would likely first name the constant "C" or something and then realize later that sqrt(-C) is what appears naturally in your solutions. So you would rename it then.
