Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 06 (b)

MATH152 April 2013

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Question B 06 (b)
Consider the system of linear differential equations given by ${\begin{aligned}y'_{1}(t)&=3y_{1}(t)-2y_{2}(t)\\y'_{2}(t)&=4y_{1}(t)-y_{2}(t)\end{aligned}}$ Find the general solution of this system of differential equations in real form.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The general real solution of a linear ordinary differential equation $\displaystyle y'(t)=My$ is given by $\displaystyle y(t)=C_{1}\mathrm {Re} \left(v_{a}e^{\lambda _{a}t}\right)+C_{2}\mathrm {Im} \left(v_{a}e^{\lambda _{a}t}\right)$ where $\displaystyle \lambda _{a}$ is one of the eigenvalues of the matrix $\displaystyle M$ with appropriate eigenvector $\displaystyle v_{a}$.

Hint 2

For our ODE $\displaystyle y'={\begin{bmatrix}3&-2\\4&-1\end{bmatrix}}y$ do the following: Calculate the eigenvalues $\displaystyle \lambda _{a,b}$ of the matrix $\displaystyle M={\begin{bmatrix}3&-2\\4&-1\end{bmatrix}}$. Determine one of the eigenvectors $\displaystyle v_{a}$. Separate the expression $\displaystyle v_{a}e^{\lambda _{a}t}$ in real and imaginary part. Write general constans $\displaystyle C_{1},C_{2}$ as factors to real and imaginary part and remove the $\displaystyle i$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To begin, we compute the eigenvalues of the matrix from part (a) (call this matrix M). To start, compute the characteristic polynomial. ${\begin{aligned}\det(M-\lambda I)&=\det \left({\begin{bmatrix}3&-2\\4&-1\end{bmatrix}}-\lambda {\begin{bmatrix}1&0\\0&1\end{bmatrix}}\right)\\&=\det \left({\begin{bmatrix}3-\lambda &-2\\4&-1-\lambda \end{bmatrix}}\right)\\&=(3-\lambda )(-1-\lambda )-(-2)(4)\\&=-3-2\lambda +\lambda ^{2}+8\\&=\lambda ^{2}-2\lambda +5\end{aligned}}$ This polynomial has two complex roots given by ${\begin{aligned}\lambda &={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\\&={\frac {-(-2)\pm {\sqrt {(-2)^{2}-4(1)(5)}}}{2(1)}}\\&={\frac {2\pm {\sqrt {-16}}}{2}}\\&={\frac {2\pm 4i}{2}}\\&=1\pm 2i\end{aligned}}$ Using $\displaystyle \lambda _{a}=1+2i$, we have the eigenvector which can be computed by ${\begin{bmatrix}3-(1+2i)&-2\\4&-1-(1+2i)\end{bmatrix}}={\begin{bmatrix}1-i&-1\\2&-1-i\end{bmatrix}}$ Multiplying the first row with 1+i we obtain ${\begin{bmatrix}2&-1-i\\2&-1-i\end{bmatrix}}={\begin{bmatrix}2&-(1+i)\\0&0\end{bmatrix}}$ The eigenvector $\displaystyle v_{a}$ is such that $\displaystyle 2v_{a,1}-(1+i)v_{a,2}=0$. Choosing $v_{a,1}=1$ we have $v_{a,2}={\frac {2}{i+1}}={\frac {2(i-1)}{(i+1)(i-1)}}=1-i$, hence the eigenvector we find is $v_{a}={\begin{bmatrix}1\\1-i\end{bmatrix}}$ Important: Please keep in mind that this is one possible eigenvector and there are others that differ from this by a scalar multiple. Since the eigenvalues $\displaystyle \lambda _{a},\lambda _{b}$ are complex conjugate, the eigenvector $\displaystyle v_{b}$ is the complex conjugate of $\displaystyle v_{a}$, which is $v_{b}={\begin{bmatrix}1\\1+i\end{bmatrix}}$ Now we calculate by using Eulers formula, $\displaystyle e^{x+iy}=e^{x}(\cos(y)+i\sin(y))$, ${\begin{aligned}v_{a}e^{\lambda _{a}t}&={\begin{bmatrix}1\\1-i\end{bmatrix}}e^{(1+2i)t}\\&={\begin{bmatrix}1\\1-i\end{bmatrix}}e^{t}(\cos(2t)+i\sin(2t))\\&={\begin{bmatrix}e^{t}\cos(2t)+ie^{t}\sin(2t)\\e^{t}(\cos(2t)+\sin(2t))+ie^{t}(\sin(2t)-\cos(2t))\end{bmatrix}}\\&=e^{t}{\begin{bmatrix}\cos(2t)\\\cos(2t)+\sin(2t)\end{bmatrix}}+ie^{t}{\begin{bmatrix}\sin(2t)\\\sin(2t)-\cos(2t)\end{bmatrix}}\end{aligned}}$ Hence, the real solution of $\displaystyle y$ is given by $y(t)=C_{1}e^{t}{\begin{bmatrix}\cos(2t)\\\cos(2t)+\sin(2t)\end{bmatrix}}+C_{2}e^{t}{\begin{bmatrix}\sin(2t)\\\sin(2t)-\cos(2t)\end{bmatrix}}$ for constants $\displaystyle C_{1},C_{2}$. Note: For the general (complex) solution you would not drop the part $\displaystyle v_{b}e^{\lambda _{b}t}$ and the solution would be $\displaystyle y(t)=B_{1}v_{a}e^{\lambda _{a}t}+B_{2}v_{b}e^{\lambda _{b}t}$ for constants $\displaystyle B_{1},B_{2}$. But for the real solution, it is enough to only consider $\displaystyle v_{a}e^{\lambda _{a}t}$.

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