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**MAT244––Home Assignments / Re: 9.2 Question 19**

« **on:**April 03, 2018, 10:32:42 AM »

Thanks!

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I got that (0,0) is a saddle point and that the general solution is $x(t) =c_{1}e^{2t}[1,2]^T+ c_{2}e^{-t}[1,-1]^T$.

I just can't seem to get an expression of the form H(x,y)=c

I just can't seem to get an expression of the form H(x,y)=c

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Can someone please help me with finding H(x,y) = c for the following system?

dx/dt = y

dy/dt = 2x+y

dx/dt = y

dy/dt = 2x+y

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Will population dynamics be on the final exam? I understand the methods being used but do we need to memorize things like the logistic equation?

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In 7.8, the solutions often mention picking n1 = 0 or n2 = 0 or n3 = 0 for the generalized eigenvector. How do we decide which one to make zero? It seems that picking different ones yield different answers.

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Solving for eigenvalues:

\begin{gather*}

(4-r)(-6-r)+24 = 0\\

\implies - 24 - 4r + 6r + r^2 = 0\\

\implies r^2 + 2r = 0 \implies r(r+2) = 0

\end{gather*}

eigenvalues are $0$ and $-2$.

When $r = 0$, Nullspace of $\begin{pmatrix}

4 & -3 \\

8 & -6

\end{pmatrix}$ equals to Nullspace of $\begin{pmatrix}

4 & -3 \\

0 & 0

\end{pmatrix}$, which is equal to the span of $\begin{pmatrix}

3 \\

4

\end{pmatrix}$.

When $r = -2$, Nullspace of $\begin{pmatrix}

6 & -3 \\

8 & -4

\end{pmatrix}$, which is equal to Nullspace of $\begin{pmatrix}

6 & -3 \\

0 & 0

\end{pmatrix}$, which is a span of $\begin{pmatrix}

1 \\

2

\end{pmatrix}$.

Then

$$\mathbf{x} = c_1\begin{pmatrix}3\\4\end{pmatrix} + c_2e^{-2t}\begin{pmatrix}1\\2\end{pmatrix}$$

As $t$ goes to infinity, the solutions tend to $c_1\begin{pmatrix}3\\4\end{pmatrix}$. A sketch of the phase portrait:

\begin{gather*}

(4-r)(-6-r)+24 = 0\\

\implies - 24 - 4r + 6r + r^2 = 0\\

\implies r^2 + 2r = 0 \implies r(r+2) = 0

\end{gather*}

eigenvalues are $0$ and $-2$.

When $r = 0$, Nullspace of $\begin{pmatrix}

4 & -3 \\

8 & -6

\end{pmatrix}$ equals to Nullspace of $\begin{pmatrix}

4 & -3 \\

0 & 0

\end{pmatrix}$, which is equal to the span of $\begin{pmatrix}

3 \\

4

\end{pmatrix}$.

When $r = -2$, Nullspace of $\begin{pmatrix}

6 & -3 \\

8 & -4

\end{pmatrix}$, which is equal to Nullspace of $\begin{pmatrix}

6 & -3 \\

0 & 0

\end{pmatrix}$, which is a span of $\begin{pmatrix}

1 \\

2

\end{pmatrix}$.

Then

$$\mathbf{x} = c_1\begin{pmatrix}3\\4\end{pmatrix} + c_2e^{-2t}\begin{pmatrix}1\\2\end{pmatrix}$$

As $t$ goes to infinity, the solutions tend to $c_1\begin{pmatrix}3\\4\end{pmatrix}$. A sketch of the phase portrait:

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Will the term test be cumulative?

Will the final exam be cumulative?

Will the final exam be cumulative?

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I believe today's quiz involved something that we haven't covered in class yet. Did anyone else have this issue?

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Characteristic equation:

$r^2+8r+25=0$

r = $-4 +3i, -4-3i$ (using quadratic equation)

Homogeneous solution:

$y_c(t) = c_1e^{-4t}cos(3t) + c_2e^{-4t}sin(3t)$

Particular solutions:

First Particular:

$Y = Ae^{-4t}$

$Y' = -4Ae^{-4t}$

$Y'' = 16Ae^{-4t}$

$A(16+8(-4)+25)=9$

$9A=9$

$A=1$

Second Particular:

$Y = Asin(3t)+Bcos(3t)$

$Y' = 3Acos(3t)-3Bsin(3t)$

$Y'' = -9Asin(3t)-9Bcos(3t)$

Plug into the given equation:

sines:

$-9A+8(-3B)+25A = 104$

$16A-24B=104$

$2A-3B=13$

cosines:

$-9B+8(3A)+25B = 0$

$16B+24A=0$

$2B+3A=0$

==> $A=2, B=-3$

General solution:

$y(t) = c_1e^{-4t}cos(3t) + c_2e^{-4t}sin(3t) + e^{-4t} + 2sin(3t) -3cos(3t)$

$r^2+8r+25=0$

r = $-4 +3i, -4-3i$ (using quadratic equation)

Homogeneous solution:

$y_c(t) = c_1e^{-4t}cos(3t) + c_2e^{-4t}sin(3t)$

Particular solutions:

First Particular:

$Y = Ae^{-4t}$

$Y' = -4Ae^{-4t}$

$Y'' = 16Ae^{-4t}$

$A(16+8(-4)+25)=9$

$9A=9$

$A=1$

Second Particular:

$Y = Asin(3t)+Bcos(3t)$

$Y' = 3Acos(3t)-3Bsin(3t)$

$Y'' = -9Asin(3t)-9Bcos(3t)$

Plug into the given equation:

sines:

$-9A+8(-3B)+25A = 104$

$16A-24B=104$

$2A-3B=13$

cosines:

$-9B+8(3A)+25B = 0$

$16B+24A=0$

$2B+3A=0$

==> $A=2, B=-3$

General solution:

$y(t) = c_1e^{-4t}cos(3t) + c_2e^{-4t}sin(3t) + e^{-4t} + 2sin(3t) -3cos(3t)$

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$M_y = 8xy$

$N_x = 4xy $

$\implies$ The equation is not exact

$\frac{M_y-N_x}{N} = \frac{2}{x}$

$\frac{d\mu}{dx} = \frac{2}{x}\mu$

$\mu = x^2$

The integrating factor is $\mu = x^2$.

New equation: $4x^3y^2 + 3x^2 lnx + x^2 + 2x^4yy'=0$

$M_y = 8x^3y = N_x$ (The equation is exact)

$\phi_x = M$

$\phi = x^4y^2 + x^3\ln x + h(y)$

$\phi_y = 2x^4y + h'(y) = N$

$h'(y)=0$

$h(y)=C$

Thus, $\phi= x^4y^2 + x^3\ln x = C$

For particular solution passing (1,1):

$1^41^2 + 1^2\ln 1 = C$

$C=1$

Thus, $\phi= x^4y^2 + x^2\ln x = 1$

$N_x = 4xy $

$\implies$ The equation is not exact

$\frac{M_y-N_x}{N} = \frac{2}{x}$

$\frac{d\mu}{dx} = \frac{2}{x}\mu$

$\mu = x^2$

The integrating factor is $\mu = x^2$.

New equation: $4x^3y^2 + 3x^2 lnx + x^2 + 2x^4yy'=0$

$M_y = 8x^3y = N_x$ (The equation is exact)

$\phi_x = M$

$\phi = x^4y^2 + x^3\ln x + h(y)$

$\phi_y = 2x^4y + h'(y) = N$

$h'(y)=0$

$h(y)=C$

Thus, $\phi= x^4y^2 + x^3\ln x = C$

For particular solution passing (1,1):

$1^41^2 + 1^2\ln 1 = C$

$C=1$

Thus, $\phi= x^4y^2 + x^2\ln x = 1$

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As promised, the typed up solution:

Characteristic Equation:

$r^2-5r+6=0$

$(r-2)(r-3) = 0$

Roots: 3 and 2

Homogeneous Equation: $y_{c}(t)=c_{1}e^{3t}+c_{2}e^{2t}$

Particular solutions:

First particular solution:

$4e^{t}$

$Y(t) = Ae^{t} = Y'(t) = Y''(t)$

$Ae^{t} - 5Ae^{t} + 6Ae^{t} = 4e^{t}$

$A - 5A + 6A = 4$

$A = 2$

Second particular solution:

$e^{2t}$

$Y(t) = Ate^{2t}$

$Y'(t) = Ae^{2t} + 2Ate^{2t}$

$Y''(t) = 2Ae^{2t} + 2Ae^{2t} + 4Ate^{2t} = 4Ae^{2t} + 4Ate^{2t}$

$4Ae^{2t} + 4Ate^{2t} - 5(Ae^{2t} + 2Ate^{2t}) + 6Ate^{2t} = e^{2t}$

$-Ae^{2t} = e^{2t}$

$A = -1$

General Solution:

$y(t)=c_{1}e^{3t} + c_{2}e^{2t} + 2e^{t} -te^{2t}$

Solving for a solution satisfying $y(0)=0, y'(0) = 0$

$y(0) = c_{1}+c_2 + 2 =0$

$y'(0) = 3c_1 + 2c_2 + 1 = 0$

$c_1 = 2, c_2 = -3$

Thus, $y(t)=2e^{3t} -3e^{2t} + 2e^{t} -te^{2t}$

Characteristic Equation:

$r^2-5r+6=0$

$(r-2)(r-3) = 0$

Roots: 3 and 2

Homogeneous Equation: $y_{c}(t)=c_{1}e^{3t}+c_{2}e^{2t}$

Particular solutions:

First particular solution:

$4e^{t}$

$Y(t) = Ae^{t} = Y'(t) = Y''(t)$

$Ae^{t} - 5Ae^{t} + 6Ae^{t} = 4e^{t}$

$A - 5A + 6A = 4$

$A = 2$

Second particular solution:

$e^{2t}$

$Y(t) = Ate^{2t}$

$Y'(t) = Ae^{2t} + 2Ate^{2t}$

$Y''(t) = 2Ae^{2t} + 2Ae^{2t} + 4Ate^{2t} = 4Ae^{2t} + 4Ate^{2t}$

$4Ae^{2t} + 4Ate^{2t} - 5(Ae^{2t} + 2Ate^{2t}) + 6Ate^{2t} = e^{2t}$

$-Ae^{2t} = e^{2t}$

$A = -1$

General Solution:

$y(t)=c_{1}e^{3t} + c_{2}e^{2t} + 2e^{t} -te^{2t}$

Solving for a solution satisfying $y(0)=0, y'(0) = 0$

$y(0) = c_{1}+c_2 + 2 =0$

$y'(0) = 3c_1 + 2c_2 + 1 = 0$

$c_1 = 2, c_2 = -3$

Thus, $y(t)=2e^{3t} -3e^{2t} + 2e^{t} -te^{2t}$

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*typed solutions coming soon* (I have class until 9)

*typed solutions coming soon*

*typed solutions coming soon*

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*Typed solutions to come* (I have class until 9pm today)

*Typed solutions to come*

*Typed solutions to come*

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*I will type up the solutions soon*

*Typed solutions to come soon*

*Typed solutions to come soon*

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Is it just me or does the 2014 past midterm cover a lot of things we haven't done yet? I was only able to do question 1.

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