find the general solution...

**slapmaxwell1** · May 5th 2011, 04:28 PM

ok here is my problem...

$\boldsymbol{X'}=\begin{pmatrix} \frac{1}{2} &0 \\ 1& \frac{-1}{2} \end{pmatrix}\boldsymbol{X}$

$\boldsymbol{X}(0)=\begin{pmatrix} 3\\5 \end{pmatrix}$

ok so when i started this problem i found that the values of lambda were 1/2 and
-1/2

so my first solution is... $\boldsymbol{X_{1}}=\begin{pmatrix} 1\\1 \end{pmatrix}e^{\frac{1}{2}t}$

this solution was for when lambda was 1/2.

my problem comes when lambda is -1/2...

i basically have 2 equations k1 +(0)k2 = 0 and k1+(0)k2 = 0? so am i to assume that k1 and k2 are 0?

any thoughts would be appreciated.

**TheEmptySet** · May 5th 2011, 04:32 PM

Originally Posted by slapmaxwell1

ok here is my problem...

$\boldsymbol{X'}=\begin{pmatrix} \frac{1}{2} &0 \\ 1& \frac{-1}{2} \end{pmatrix}\boldsymbol{X}$

$\boldsymbol{X}(0)=\begin{pmatrix} 3\\5 \end{pmatrix}$

ok so when i started this problem i found that the values of lambda were 1/2 and
-1/2

so my first solution is... $\boldsymbol{X_{1}}=\begin{pmatrix} 1\\1 \end{pmatrix}e^{\frac{1}{2}t}$

this solution was for when lambda was 1/2.

my problem comes when lambda is -1/2...

i basically have 2 equations k1 +(0)k2 = 0 and k1+(0)k2 = 0? so am i to assume that k1 and k2 are 0?

any thoughts would be appreciated.

No that tells you that

$k_1=0$

but

$k_2$

is free so you get the eigenvector

$\begin{pmatrix} 0 \\ 1\end{pmatrix}$

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