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

MATH152 April 2013

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[hide] 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.

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[show] 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 $\displaystyle v_{a}$ .

[show] 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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[show] Solution
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 $\displaystyle v_{a}e^{\lambda _{a}t}$ .