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Solve the initial value problem:

$$\frac{du}{dt}= \pmatrix{1&2\\-1&1}u, u(0) = \pmatrix{1\\0}$$

What I got is,

Eigenvalues : $\lambda = 1 \pm \sqrt{2}i$

Eigenvectors : $v_1 = \pmatrix{1\\-\frac{1}{\sqrt {2}i}},v_2 = \pmatrix{1\\\frac{1}{\sqrt {2}i}}$

Then, $$[e^tcos\sqrt{2}t + ie^tsin\sqrt{2}t]\pmatrix{1\\-\frac{1}{\sqrt {2}i}}$$ the problem I am having is the initial conditions. Can anyone show me how to do it?

The answer to this problem is

$$u(t) = \pmatrix{e^tcos\sqrt{2}t\\-\frac{1}{\sqrt {2}}e^tsin\sqrt{2}t}$$ I do not know how they got that.

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1 Answer

up vote 0 down vote accepted

You write "eigenvalue" in the singular, but you (correctly) give two eigenvalues; then you give only a single eigenvector. The two different eigenvalues have different eigenvectors. You can only satisfy the initial conditions by using both eigenvectors.

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I assumed it was already implied since it is a conjugate. I will revise it. –  Q.matin Nov 28 '12 at 7:09
    
@Q.matin: You only changed the eigenvectors; the solution in the next line still only uses the one eigenvector. (You also didn't correct the singulars.) –  joriki Nov 28 '12 at 9:13
    
Joriki, I dont understand what you mean by singular. Also, I thought that if you have complex conjugates eigenvalues then you dont have to write both the eigenvectors and just use the equation $[e^vtcosut + ie^vtsinut]$, where $\lambda = v \pm ui$ –  Q.matin Nov 29 '12 at 5:49
    
@Q.matin: "singular" is a grammatical term for a word form that refers to a single object, as opposed a "plural", which refers to several objects. That is, "vector" is a singular and "vectors" is a plural. You have "Eigenvalue:" and "Eigenvector:" as singulars even though both are followed by two values/vectors. Regarding the sine and cosine, the general idea that you can use those to avoid dealing with both conjugates is right, but not in the way you're doing it, where you multiply that with one of the two eigenvectors; I'll try to add something about that to my answer later. –  joriki Nov 29 '12 at 5:56
    
Okay, I am going to reattempt the problem with your suggestion and I will fix my grammar mistake. –  Q.matin Nov 29 '12 at 6:04
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