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**Quiz 6 / Re: problem 2, night section**

« **on:**November 27, 2013, 10:04:38 PM »

see attachment

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By the way Mr Ivrii, is it possible for us to get 4.5 out of 5 marks on the midterm questions? When can we know the result?

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For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

no mistake. d(ln(1-y))/dt = -1/1-y

You are right! My bad

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Just wondering, for this question, do we need to draw the direction field?

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For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

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As promised:

$$ dy/dt + 2y/t = 1 \\

\mu (dy/dt) + 2y\mu/t = \mu \\

d/dt (\mu y) = d\mu/dt\cdot y + \mu\cdot dy/dt \\

\implies d\mu/dt\cdot y = 2y\mu/t \\

\implies \mu = t^2 \\

d/dt(t^2 y) = t^2 \\

\implies t^2 y = (1/3)t^3 + c \\

\implies y = (1/3)t + c/t^2

$$

$$ dy/dt + 2y/t = 1 \\

\mu (dy/dt) + 2y\mu/t = \mu \\

d/dt (\mu y) = d\mu/dt\cdot y + \mu\cdot dy/dt \\

\implies d\mu/dt\cdot y = 2y\mu/t \\

\implies \mu = t^2 \\

d/dt(t^2 y) = t^2 \\

\implies t^2 y = (1/3)t^3 + c \\

\implies y = (1/3)t + c/t^2

$$

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I will try to type this afterwards. See attachment

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I'm just wondering about the coverage on the midterm. So only chapter 1 and 2 are covered right? Because I looked at the past term test 1 and found questions on second order linear equations, so I just want to make sure if that's gonna be on the midterm or not.

And also, according to some of the past tests, there are 4 questions in total, out of 20 points?

Thanks!

And also, according to some of the past tests, there are 4 questions in total, out of 20 points?

Thanks!

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Since our midterm is coming very soon, I am just wondering if some of the past/sample midterms will be posted here? Thanks

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First I apologize that the format is bad, because it's my first time typing math equations using this forum, so I am not sure how it works.

Typing is always better than scanning. Please try to modify post to see how I did it. Observe special meaning of \$ ...\$ for inline math and \$\$...\$\$ for display math (occupying its own lines) in the source. -- V.I.

Second, wherever it says Q(x,y), it is not Q actually, it's that Greek letter phi which I don't know how to type. OK, I will change $Q$ to $\Phi$ but I am not sure why you don't like $Q$.

So here we go.

**Solution:** Rewrite ~~function~~ equation as

$$( 9x^2 + y - 1 ) + ( x - 4y )y' = 0$$

So we have $M_y(x,y) = 1 = N_x(x,y)$, so this~~function~~ equation is exact.

$$

\left\{\begin{aligned}

&\Phi_x(x,y) = 9x^2 + y - 1,\\

&\Phi_y(x,y) = x - 4y.

\end{aligned}\right.

$$

Above system was more complicated to type.

By integrating $\Phi_x(x,y)$, we have $\Phi (x,y) = 3x^3 + xy - x + h(y)$ with arbitrary $h(y)$.

By differentiating derived $\Phi (x,y)$ with respect to $y$ we have $\Phi_y(x,y) = x + h'(y)$

Note that $ \Phi_y(x,y) = x + h'(y) = x - 4y$.

Therefore we have $h'(y) = -4y$, and $h(y) = -2y^2$.

So

$$

\Phi (x,y) = 3x^3 + xy - x - 2y^2 = c.

$$

Now we plug $y(1) = 0$ into this equation, we have $3(1^3) + (1)(0) - 1 - 2(0^2) = c\implies c = 2$.

So the solution is

$$

3x^3 + xy - x - 2y^2 = 2.

$$

Good Job -- V.I.

Typing is always better than scanning. Please try to modify post to see how I did it. Observe special meaning of \$ ...\$ for inline math and \$\$...\$\$ for display math (occupying its own lines) in the source. -- V.I.

Second, wherever it says Q(x,y), it is not Q actually, it's that Greek letter phi which I don't know how to type. OK, I will change $Q$ to $\Phi$ but I am not sure why you don't like $Q$.

So here we go.

$$( 9x^2 + y - 1 ) + ( x - 4y )y' = 0$$

So we have $M_y(x,y) = 1 = N_x(x,y)$, so this

$$

\left\{\begin{aligned}

&\Phi_x(x,y) = 9x^2 + y - 1,\\

&\Phi_y(x,y) = x - 4y.

\end{aligned}\right.

$$

Above system was more complicated to type.

By integrating $\Phi_x(x,y)$, we have $\Phi (x,y) = 3x^3 + xy - x + h(y)$ with arbitrary $h(y)$.

By differentiating derived $\Phi (x,y)$ with respect to $y$ we have $\Phi_y(x,y) = x + h'(y)$

Note that $ \Phi_y(x,y) = x + h'(y) = x - 4y$.

Therefore we have $h'(y) = -4y$, and $h(y) = -2y^2$.

So

$$

\Phi (x,y) = 3x^3 + xy - x - 2y^2 = c.

$$

Now we plug $y(1) = 0$ into this equation, we have $3(1^3) + (1)(0) - 1 - 2(0^2) = c\implies c = 2$.

So the solution is

$$

3x^3 + xy - x - 2y^2 = 2.

$$

Good Job -- V.I.

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