How do I choose them? Of course, I match the initial condition, so at t equals 0. I'm looking for the null solutions, the solutions of the equations themselves.Īnd there I have two solutions, two coefficients to choose. There's no force term in these equations. That's superposition, adding solutions to linear equations. Any number c1 of that, plus any number c2 of that is still a solution. One solution, another solution.Īnd what do I do with linear equations? I take combinations. e to the lambda t times its eigenvector, 1 minus 3. If I put that into the equation, it will solve the equation. So the first one is e to the 6t times x, which is 1, 1. Remember, that's the picture that we're looking for. You will soon see why I expect two eigenvalues, and each eigenvalue should have an eigenvector. Right? I'm looking for inputs, the eigenvector, so that the output is a number times that eigenvector, and that number is lambda, the eigenvalue. When I did the multiplication, Ax came out to be 2 minus 6. I just do A times x to recognize the lambda, the eigenvalue. And we can find out what lambda 2 is, once I find the eigenvectors of course. And I should call this one x2 and lambda 2. I should call this first one maybe x1 and lambda 1. Again, I've worked in advance, produced this eigenvector, and I think it's 1 minus 3. If I multiply A by that eigenvector, 1, 1, do you see what happens when I multiply by 1? That gives me a 6. What are the eigenvalues, what are the eigenvectors of that matrix? And remember, I want Ax equals lambda x. Now eigenvalues and eigenvectors will solve it. Then that problem is exactly y prime, the vector, derivative of the vector, equal A times y. Those numbers, 5, 1, 3, 3, go into the matrix. Second one, linear, constant coefficient, 3 and 3. There's the first equation for y1- prime meaning derivative, d by dt, time derivative- is linear, a constant coefficient. We'll find the lambdas and the x's, and then we'll have the solution to the system of differential equations. And in this example, first of all, I'm going to spot the eigenvalues and eigenvectors without a system, just go for it in the 2 by 2 case. Even if the matrix is real, lambda could be complex. It can be real, it could be complex number, as you will see. That's no help to know that 0 is a solution. If lambda is 0, I would have Ax equals 0. Well, if lambda was 1, I would have Ax equal x. It's in the same direction as x just the length is changed. So what am I looking for? I'm looking for vectors x, the eigenvectors, so that multiplying by A- multiplying A times x gives a number times x. And I have the big equation, Ax, the matrix times my eigenvector, is equal to lambda x- the number, the eigenvalue, times the eigenvector. So I cancel e to the lambda t because it's never zero. OK Now I cancel either the lambda t, just the way I was always canceling e to the st. So do you see that substituting into the equation with this nice notation is just this has to be true. To get the derivative I include the lambda. Now, the derivative of y, the time derivative, brings down a lambda. I'm plugging in what e to the lambda tx, that's y. I plug that into the differential equation and what happens? So here's my equation. And x is just multiples of that exponential, as you'll see. All the time dependence is in the exponential, as always. This is a vector, but that does not depend on time. Now we have e to the lambda t- we changed s to lambda, no problem- but multiplied by a vector because our unknown is a vector. When we had one equation, we looked for solutions just e to the st, and we found that number s. Good.Īnd now I can tell you right away where eigenvalues and eigenvectors pay off. And A is an n by n matrix, n rows, n columns. So we have n equations, n components of y. So why is now a vector- so this is a system of equations. Let me show you the reason eigenvalues were created, invented, discovered was solving differential equations, which is our purpose. So eigenvalue is a number, eigenvector is a vector. Systems meaning more than one equation, n equations. And the reason we want those, need those is to solve systems of linear equations. So today begins eigenvalues and eigenvectors.
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