MAT244-2014F > TT1
TT1-problem 2
Bogdan Scaunasu:
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
Victor Ivrii:
Guys, there is no point to post solution which has been posted already by someone else (especially, as Bogdan did—-with an error)
Bogdan Scaunasu:
I apologize for posting a solution after someone else did. I was writing mine and did not notice.
Sang Wu:
a. W = x2(lnx + 1)
b. W= xy2' - y2 = x2(lnx + 1)
u = e^(∫-1/x * dx) = x^(-1)
(y2/x)' = lnx + 1
y2/x = ∫(lnx + 1)dx = ∫lnxdx + x = xlnx - x + x = xlnx
Victor Ivrii:
Sang Wu,
the complete solution properly typed has been posted already. What is your point?
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