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**Quiz-6 / Re: Q6--T0601**

« **on:**March 17, 2018, 08:37:00 PM »

I added some lines to it now.

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By adjusting the sliders.

I have edited my post to include the plot using pplane.

I have edited my post to include the plot using pplane.

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I got it from here: http://mathlets.org/mathlets/linear-phase-portraits-matrix-entry/

I couldn't find one that worked better. Do you have any suggestions of what to use?

I couldn't find one that worked better. Do you have any suggestions of what to use?

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a)

First we find the eigenvalues

$$det\begin{bmatrix}4-r & -3\\8 & -6-r\end{bmatrix} = (r-4)(r+6)-24 = (r^2+2r) = r(r+2)$$

$$r_1 = 0, r_2 = -2$$

The associated eigenvector for $r_1$ is:

$$r_1=0: Null\begin{bmatrix}4 & -3\\8 & -6\end{bmatrix} = Null\begin{bmatrix}4 & -3\\0 & 0\end{bmatrix} \implies 4\xi_1 = 3\xi_2 \implies \xi^{(1)} = \begin{bmatrix}3\\4\end{bmatrix} $$

The associated eigenvector for $r_2$ is:

$$r_2=-2: Null\begin{bmatrix}6 & -3\\8 & -4\end{bmatrix} = Null\begin{bmatrix}2 & -1\\0 & 0\end{bmatrix} \implies 2\xi_1 = \xi_2 \implies \xi^{(2)} = \begin{bmatrix}1\\2\end{bmatrix} $$

This gives us our general solution:

$$X(t) = c_1\begin{bmatrix}3\\4\end{bmatrix} + c_2e^{-2t}\begin{bmatrix}1\\2\end{bmatrix}$$

b)

The plot follows the same idea as 1. e) in this handout: (www.math.toronto.edu/courses/mat244h1/20181/LN/Ch7-LN9.pdf)

The plot approaches the vector $\begin{bmatrix}3\\4\end{bmatrix}$ as t approaches infinity.

First we find the eigenvalues

$$det\begin{bmatrix}4-r & -3\\8 & -6-r\end{bmatrix} = (r-4)(r+6)-24 = (r^2+2r) = r(r+2)$$

$$r_1 = 0, r_2 = -2$$

The associated eigenvector for $r_1$ is:

$$r_1=0: Null\begin{bmatrix}4 & -3\\8 & -6\end{bmatrix} = Null\begin{bmatrix}4 & -3\\0 & 0\end{bmatrix} \implies 4\xi_1 = 3\xi_2 \implies \xi^{(1)} = \begin{bmatrix}3\\4\end{bmatrix} $$

The associated eigenvector for $r_2$ is:

$$r_2=-2: Null\begin{bmatrix}6 & -3\\8 & -4\end{bmatrix} = Null\begin{bmatrix}2 & -1\\0 & 0\end{bmatrix} \implies 2\xi_1 = \xi_2 \implies \xi^{(2)} = \begin{bmatrix}1\\2\end{bmatrix} $$

This gives us our general solution:

$$X(t) = c_1\begin{bmatrix}3\\4\end{bmatrix} + c_2e^{-2t}\begin{bmatrix}1\\2\end{bmatrix}$$

b)

The plot follows the same idea as 1. e) in this handout: (www.math.toronto.edu/courses/mat244h1/20181/LN/Ch7-LN9.pdf)

The plot approaches the vector $\begin{bmatrix}3\\4\end{bmatrix}$ as t approaches infinity.

5

Divide everything by $cos(t)$ to get $y''$ by itself.

$$y'' + {sin(t)\over cos(t)}y' - {t\over cos(t)}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(t)dt}$ where c is a constant and $p(t)$ is $sin(t)\over cos(t)$ in this case. Now we solve the integral:

$$ce^{-\int{sin(t)\over cos(t)}dt}$$

Using the substitution $u=cos(t)$ and $du=-sin(t)dt$ we get

$$ce^{\int{1\over u}du}=ce^{ln(u)+C}=ce^{ln(cos(t)+C}=ce^Ccos(t)$$

But $ce^Ce^2$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {c(cos(t))}$$

$$y'' + {sin(t)\over cos(t)}y' - {t\over cos(t)}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(t)dt}$ where c is a constant and $p(t)$ is $sin(t)\over cos(t)$ in this case. Now we solve the integral:

$$ce^{-\int{sin(t)\over cos(t)}dt}$$

Using the substitution $u=cos(t)$ and $du=-sin(t)dt$ we get

$$ce^{\int{1\over u}du}=ce^{ln(u)+C}=ce^{ln(cos(t)+C}=ce^Ccos(t)$$

But $ce^Ce^2$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {c(cos(t))}$$

6

Divide everything by $t^2$ to get $y''$ by itself.

$$y'' - {t+2\over t}y' + {t+2\over t^2}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(t)dt}$ where c is a constant and $p(t)$ is $- {t+2\over t}$ in this case. Now we solve the integral:

$$ce^{\int{t+2\over t}dt}$$

Using the substitution $u = t+2$ and $du =dt$ we get

$$ ce^{\int{u\over u-2}du}$$

Which we can split up using partial fraction decomposition giving us

$$ce^{\int({2\over u-2}+1)du} = ce^{2\int({1\over u-2})du +\int{du}} = ce^{2ln(u-2)+u+C} = ce^{C}e^{2}e^{ln(t^2)}e^{t} = ce^{C}e^{2}t^2e^{t}$$

But $ce^Ce^2$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {ct^2e^t}$$

$$y'' - {t+2\over t}y' + {t+2\over t^2}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(t)dt}$ where c is a constant and $p(t)$ is $- {t+2\over t}$ in this case. Now we solve the integral:

$$ce^{\int{t+2\over t}dt}$$

Using the substitution $u = t+2$ and $du =dt$ we get

$$ ce^{\int{u\over u-2}du}$$

Which we can split up using partial fraction decomposition giving us

$$ce^{\int({2\over u-2}+1)du} = ce^{2\int({1\over u-2})du +\int{du}} = ce^{2ln(u-2)+u+C} = ce^{C}e^{2}e^{ln(t^2)}e^{t} = ce^{C}e^{2}t^2e^{t}$$

But $ce^Ce^2$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {ct^2e^t}$$

7

Divide everything by $(1-x^2)$ to get $y''$ by itself.

$$y'' - {2x\over {1 - x^2}}y' + {α(α + 1)\over {1 - x^2}}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(x)dx}$ where c is a constant and $p(x)$ is $- {2x\over {1 - x^2}}$ in this case. Now we solve the integral:

$$ce^{\int{2x\over {1 - x^2}}dx}$$

Using the substitution $u = 1-x^2$ and $du =-2xdx$ we get

$$ ce^{-\int{1\over u}du} = ce^{-ln(u)+C} = ce^{-ln(1-x^2)} = c(1-x^2)^{-1}e^C = {ce^C\over {1-x^2}}$$

But $ce^C$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {c\over {1-x^2}}$$

$$y'' - {2x\over {1 - x^2}}y' + {α(α + 1)\over {1 - x^2}}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(x)dx}$ where c is a constant and $p(x)$ is $- {2x\over {1 - x^2}}$ in this case. Now we solve the integral:

$$ce^{\int{2x\over {1 - x^2}}dx}$$

Using the substitution $u = 1-x^2$ and $du =-2xdx$ we get

$$ ce^{-\int{1\over u}du} = ce^{-ln(u)+C} = ce^{-ln(1-x^2)} = c(1-x^2)^{-1}e^C = {ce^C\over {1-x^2}}$$

But $ce^C$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {c\over {1-x^2}}$$

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Find the Wronskian of two solutions of the given differential equation without solving the equation.

$$x^2y'' + xy' + (x^2-v^2)y = 0$$

Divide everything by $x^2$ to get $y''$ by itself.

$$y'' + {1\over x}y' + {(x^2-v^2)\over x^2}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(x)dx}$ where c is a constant and $p(x)$ is $1\over x$ in this case. Now we solve the integral:

$$ce^{-\int{1\over x}dx} = ce^{-ln(x)+C} = ce^{ln(x^{-1})+C} = cx^{-1}e^C$$

But $ce^C$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {c\over x}$$

$$x^2y'' + xy' + (x^2-v^2)y = 0$$

Divide everything by $x^2$ to get $y''$ by itself.

$$y'' + {1\over x}y' + {(x^2-v^2)\over x^2}y = 0$$

Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(x)dx}$ where c is a constant and $p(x)$ is $1\over x$ in this case. Now we solve the integral:

$$ce^{-\int{1\over x}dx} = ce^{-ln(x)+C} = ce^{ln(x^{-1})+C} = cx^{-1}e^C$$

But $ce^C$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:

$$W = {c\over x}$$

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Ahh thank you, I didn't realize that that text was expandable.

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I know we have a quiz this week and was wondering what content will be covered up to. As in, is all material up until the quiz date considered fair game for the quiz?

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