These are all homogeneous solutions that is they are

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These are all “homogeneous solutions,” that is, they are solutions of the homogeneous equation. Very often the solutions x 1 and x 2 we start with are exponential. In the example on Friday, x 1 = e t and x 2 = e 4 t . These are the only exponential solutions. The general solution is “homogeneous” but not exponential. 3. A pair of vectors in the plane is “linearly independent” if neither one is a multiple of the other. For any such pair of vectors, { v 1 , v 2 } , any vector in the plane is a “linear combination” c 1 v 1 + c 2 v 2 . This is because v 1 and v 2 point in different directions, and using the right amount of each you can get anywhere. The collection of all solutions to ¨ x + bx ˙ + kx = 0 looks exactly the same way. Given any linearly independent pair of solutions, the general solution is a linear combination of them. 2 above shows that any linear combination is a solution. The converse, that any solution is such a linear combination, is harder, and I won’t try to justify it in more detail than to say that it comes down to the claim that an initial condition in the form x ( t 0 ) , x ˙( t 0 ) uniquely determines a solution.
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