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 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?

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