MAT244-2014F > TT1

TT1-problem 2

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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 didâ€”-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^(âˆ«-1/x * dx) = x^(-1)
(y2/x)' = lnx + 1
y2/x = âˆ«(lnx + 1)dx = âˆ«lnxdx + x = xlnx - x + x = xlnx

Victor Ivrii:
Sang Wu,

the complete solution properly typed has been posted already. What is your point?