MAT244-2013S > Term Test 1

TT1--Problem 1

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Alexander Jankowski:
I am late, but at some point, someone is going to want a coded solution that isn't rotated by 90° CCW.

Let $M(x,y) = x$ and let $N(x,y) = y + x^2y + y^3$. Then,

M_y(x,y) = 0, & N_x(x,y) = 2xy.

So the equation is not exact, and we need an integrating factor $\mu(x,y) = \mu(y)$ to make it exact.

\frac{d\mu(y)}{dy} = \frac{N_x - M_y}{M} \mu(y) = \frac{2xy}{x} \mu(y) = y\mu(y) \Longrightarrow \frac{d\mu(y)}{\mu(y)} = 2ydy \Longrightarrow \mu(y) = e^{y^2}

Now, we require a function $\psi(x,y)$ that satisfies $\psi_x = \mu M$ and $\psi_y = \mu N$.

\psi_x(x,y) = xe^{y^2} \Longrightarrow \psi(x,y) = \frac{1}{2}x^2e^{y^2} + h(y)

We differentiate the result to get

\psi_y(x,y) = x^2xe^{y^2} + h'(y) = e^{y^2}(y + x^2y + y^3) \Longleftrightarrow h'(y) = e^{y^2}(y^3 + y) \Longrightarrow h(y) = \int{(e^{y^2}(y^3 + y))}dy.

Let $u = y^2$ so that $du = 2ydy$. Then,

h(y) = \int{(y^3e^{y^2})}dy + \int{(ye^{y^2})}dy = \frac{1}{2}\int{(ue^u)}du + \frac{1}{2}\int{e^u}du = \frac{1}{2}ue^u - \frac{1}{2}\int{e^u}du + \frac{1}{2}e^u = \frac{1}{2}ue^u - \frac{1}{2}e^u + \frac{1}{2}e^u = \frac{1}{2}y^2e^{y^2}.

Therefore, the solution is implicitly given by

C = \frac{1}{2}e^{y^2}(x^2 + y^2).

Victor Ivrii:
I decided to be generous and awarded karma to all 4. Marcia definitely realized that her first scan (actually snapshot) was almost completely useless and reposted with double resolution; honestly, even her 2nd snapshot is inferior. Matthew positioned paper in the best possible way, and Yook used grayscale (better than colour; however black and white would be even better but it requires more knowledge--see my avatar for b/w) and a monster-resolution picture (but because it was grayscale file size was not much larger!)

Alexander' post is far superior (orientation is not that important, major advantage it is typed and could be easily edited and recycled so in the most strict approach (the first gets all) Matthew and Alexander would get their karma.


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