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### Messages - Hareem Naveed

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1
##### Final Exam / Re: FE-2
« on: April 18, 2013, 12:27:38 AM »

t^{2}y" - 2y = t^{4}e^{t},    t>0

Solve Euler for homogeneous, and find the two solutions to the system:

\begin{split}
x = \ln(t)\\
y"- y'- 2y = 0\\
r^{2}-r-2 = 0\\
r_1 = -1,  r_2 = 2\\
\end{split}

\begin{split}
y_1 = e^{-x} = t^{-1}\\
y_2 = e^{2x} = t^2
\end{split}

Then, using the formula for variation of parameters, you get:

y_p(t) = -\frac {t^{2}}{3} \int{te^{t}dt} + \frac{t^{-1}}{3} \int{t^{4}e^{t}dt}

y_p(t) = -\frac{t^{2}e^t}{3}(t-1) + \frac{t^{-1}e^{t}}{3}(t^{4} - 4t^{3} + 12t^{2} - 24t + 24)

Thus, the general solution is given by:

Y(t) = c_{1}t^{-1} + c_2t^{2} + y_{p}(t)

2
##### Easter and Semester End Challenge / Re: Easter challenge
« on: March 29, 2013, 11:38:46 AM »
Attached are the two phase portraits.

In terms of difference between the two; they are level curves of the following functions:
$$H_{a}(x,y) = \cos(y)+\cos(x) \\ H_{b}(x,y) = \cos(y) + 2\cos(x)$$
Level curves are also attached.

From the level curves, in a, the centres are hemmed in by 2 defined separatrices, not so in b where there is only one.

How could I formalize these statements? Intuitively, I can "see" the answer.

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