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I have a system of differential equations:

$y_1'=2y_1+y_2\\ y_2'=-2y_1+5y_2$

The coefficient matrix has eigenvalues $\lambda=3$ and $\lambda=4$ corresponding to eigenvectors $\pmatrix{1\\1}$ and $\pmatrix{1\\2}$.

I have determined the general solution to the system to be

$y=c_1e^{3x}\pmatrix{1\\1}+c_1e^{4x}\pmatrix{1\\2}$.

How do I determine the particular solution that requires $\lim_{t\to\infty}e^{-4t}y_1(t)=0$ and $y_2'(0)=3$?

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1 Answer 1

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$y=c_1e^{3x}\pmatrix{1\\1}+c_2e^{4x}\pmatrix{1\\2}$.

$y'=3c_1e^{3x}\pmatrix{1\\1}+4c_2e^{4x}\pmatrix{1\\2}$.

$y'(0)=3c_1\pmatrix{1\\1}+4c_2\pmatrix{1\\2}$.

so $y'_2(0)=3c_1+8c_2=3$ and we got $c_1=1-\frac{8}3c_2$

Now, $e^{-4x}y(x)= c_1e^{-x}\pmatrix{1\\1}+c_2\pmatrix{1\\2}= (1-\frac{8}3c_2)e^{-x}\pmatrix{1\\1}+c_2\pmatrix{1\\2}$

and

$e^{-4t}y_1(t)= (1-\frac{8}3c_2)e^{-x} +c_2$

so $\lim_{t\to\infty}e^{-4t}y_1(t)=0$ gives $c_2=0$.

Substitute to get $c_1=1$ and

$y=e^{3x}\pmatrix{1\\1}$.

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  • $\begingroup$ How do you go from the third to the fourth line? $\endgroup$ Nov 29, 2015 at 14:15
  • $\begingroup$ evaluate the sum in the third line to get a vector $\pmatrix{3c_1+4c_2\\3c_1+8c_2}$, take the second\bottom component $\endgroup$ Nov 29, 2015 at 16:17

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