MATH 308 Lecture 30

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Lecture Notes

Review Day for exam on Wednesday

Exercise 11

Consider the system

$X'(t)={\begin{bmatrix}1&1\\4&1\end{bmatrix}}\,X(t)$

1. Show that the vectors $X_{1}(t)={\begin{bmatrix}1\\2\end{bmatrix}}\,\mathrm {e} ^{3t}$ and $X_{2}(t)={\begin{bmatrix}1\\-2\end{bmatrix}}\,\mathrm {e} ^{-t}$ are solutions
2. Are they linearly independent? Describe all solutions to the system
3. Find the solution to the initial value problem $X(0)={\begin{bmatrix}3\\2\end{bmatrix}}$

Solution for 1:

${\begin{aligned}X_{1}'(t)&={\begin{bmatrix}3\mathrm {e} ^{3t}\\6\mathrm {e} ^{3t}\end{bmatrix}}\,{\begin{bmatrix}1&1\\4&1\end{bmatrix}}\,{\begin{bmatrix}\mathrm {e} ^{3t}\\2\mathrm {e} ^{3t}\end{bmatrix}}&={\begin{bmatrix}3\mathrm {e} ^{3t}\\6\mathrm {e} ^{3t}\end{bmatrix}}X_{2}'(t)&={\begin{bmatrix}-\mathrm {e} ^{-t}\\2\mathrm {e} ^{-t}\end{bmatrix}}{\begin{bmatrix}1&1\\4&1\end{bmatrix}}\,{\begin{bmatrix}\mathrm {e} ^{-t}\\-2\mathrm {e} ^{-t}\end{bmatrix}}&={\begin{bmatrix}\mathrm {e} ^{-t}\\2\mathrm {e} ^{-t}\end{bmatrix}}\end{aligned}}$

$X_{1}$ and $X_{2}$ are solutions

Solution for 2:

Note that all solutions are of form $c_{1}\,X_{1}+c_{2}\,X_{2}=0$

$W(X_{1},X_{2})={\begin{vmatrix}\mathrm {e} ^{3t}&\mathrm {e} ^{-t}\\2\mathrm {e} ^{3t}&-2\mathrm {e} ^{-t}\end{vmatrix}}=-4\mathrm {e} ^{2t}\neq 0$

Therefore, they are linearly independent.

Solution for 3:

Solution for initial value problem satisfies $c_{1}\,X_{1}+c_{2}\,X_{2}=0$

Therefore

${\begin{aligned}X(t)&=c_{1}{\begin{bmatrix}\mathrm {e} ^{3t}\\2\mathrm {e} ^{3t}\end{bmatrix}}+c_{2}{\begin{bmatrix}\mathrm {e} ^{-t}\\-2\mathrm {e} ^{-t}\end{bmatrix}}\\X(0)={\begin{bmatrix}3\\2\end{bmatrix}}&=c_{1}\,{\begin{bmatrix}1\\2\end{bmatrix}}+c_{2}\,{\begin{bmatrix}1\\-2\end{bmatrix}}\\c_{1}&=2\\c_{2}&=1\end{aligned}}$

Exercise 10.2

Find the eigenvalues and eigenvectors of the given matrix $A={\begin{bmatrix}3&-2\\4&-1\end{bmatrix}}$

${\begin{aligned}0={\begin{vmatrix}3-\lambda &-2\\4&-1-\lambda \end{vmatrix}}&=(3-\lambda )(-1-\lambda )-(4)(-2)\\0&=\lambda ^{2}-2\lambda +5\\\lambda &=1\pm 2i\\\end{aligned}}$

Eigenvector for $1+2i$

• Solve $A\,X=(1+2i)\,X$, or
• Solve $(A-(1+2i))\,X=0$

${\begin{aligned}{\begin{bmatrix}2-2i&-2\\4&-2-2i\end{bmatrix}}\,{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}&={\begin{bmatrix}0\\0\end{bmatrix}}\\{\begin{bmatrix}1&-{\frac {1}{2}}-{\frac {1}{2}}i\\0&0\end{bmatrix}}\,{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}&={\begin{bmatrix}0\\0\end{bmatrix}}\\{\vec {x}}&=\left\langle \alpha ,(1-i)\alpha \right\rangle \end{aligned}}$