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### Messages - Bogdan Scaunasu

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1
##### TT1 / Re: TT1-problem 2
« on: October 09, 2014, 04:23:33 PM »
I apologize for posting a solution after someone else did. I was writing mine and did not notice.

2
##### TT1 / Re: TT1-problem 2
« on: October 09, 2014, 12:44:36 PM »
2a) Changing the equation to:
$$y'' - \frac{2 \ln x + 3}{x (\ln x + 1)} y' + \frac{2 \ln x + 3}{x ^ 2 (\ln x + 1)} y = 0$$
Let $-p = \frac{2 \ln x + 3}{x (\ln x + 1)}$. Then:
\begin{aligned}
-\int p dx & = \int \frac{2 \ln x + 3}{x (\ln x + 1)} dx \\
& = \int \frac{2 \ln x + 2}{x (\ln x + 1)} dx + \int \frac{1}{x (\ln x + 1)} dx \\
& = 2 \int \frac{1}{x} dx + (\ln(\ln(x) + 1) + \ln C) \\
-\int p dx & = 2 \ln x + \ln(\ln(x) + 1) + \ln C
\end{aligned}
Using Abel's Theorem:
$$W = c e ^ {- \int p dx}$$
\begin{aligned}
W & = c e ^ {- \int p dx} \\
& = c e ^ {2 \ln x + \ln(\ln(x) + 1) + \ln C} \\
& = c (C x ^ {2} (\ln(x) + 1))
\end{aligned}
Therefore:
$$\boxed{W = C x ^ {2} (\ln(x) + 1)}$$

** fixed minus sign

3
##### TT1 / Re: TT1-problem 1
« on: October 09, 2014, 12:24:11 PM »
Let $M = y$ and $N = 2xy-e^{-2y}$. Then:
$$\frac{\partial M}{\partial y} = 1 \neq 2y = \frac{\partial N}{\partial x}$$
Therefore we must introduce an integrating factor $\mu$ such that:
$$\frac{\partial \mu M}{\partial y} = \frac{\partial \mu}{\partial y} M + \mu \frac{\partial M}{\partial y} \\ \frac{\partial \mu N}{\partial x} = \frac{\partial \mu}{\partial x} N + \mu \frac{\partial N}{\partial x}$$
Assume $\mu_x = 0$ so as to make calculations easier. Then:
$$\frac{\partial \mu M}{\partial y} = \frac{\partial \mu N}{\partial x} \\ \frac{\partial \mu}{\partial y} M + \mu \frac{\partial M}{\partial y} = \mu \frac{\partial N}{\partial x} \\ \mu \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{\partial \mu}{\partial y} \\ \frac{\partial \mu}{\partial y} = \frac{2y - 1}{y} \mu = (2 - y^{-1}) \mu \\ \int \frac{\partial \mu}{\mu} = \int (2 - y^{-1}) dy \\ \ln \mu = 2y - \ln y + ln C \\ \boxed{\mu = C y^{-1} e ^ {2y}}$$
Find some function $\Psi$ such that:
\begin{aligned}
\Psi_x & = \mu M = e ^ {-2y} \\
\Psi_y & = \mu N = 2 x e ^ {2 y} - y ^ {-1} \\
\end{aligned}
Then integrating $\Psi_x$:
\begin{aligned}
\int \Psi_x dx & = \int e ^ {-2y} dx \\
& = e ^ {-2y} \int dx \\
\Psi & = xe ^ {-2y} + V
\end{aligned}
Then integrating $\Psi_y$:
\begin{aligned}
\int \Psi_y dy & = \int (2 x e ^ {2 y} - y ^ {-1}) dy \\
& = 2 x \int e ^ {2 y} dy - \int y ^ {-1} dy \\
\Psi & = x e ^ {2 y} - \ln y + U
\end{aligned}
Therefore we have:
$$\Psi = x e ^ {2 y} - \ln y + C = 0$$
To solve the IVP $y(1) = -2$, we substitute $x = 1$ and $y = -2$:
$$(1) e ^ {2 (-2)} - \ln |-2| + C = 0 \\ e ^ {-4} - \ln 2 + C = 0 \\ \boxed{C = \ln 2 - e ^ {-4}}$$
$$\boxed{x e ^ {2 y} - \ln |y| = e ^ {-4} - \ln 2}$$
** fixed as per Professor Ivrii's remark

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