Find all solutions of the differential equation
![\[ (x-2)(x-3) y' + 2y = (x-1)(x-2) \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-db87596c296ded9f8165939947134155_l3.png "Rendered by QuickLaTeX.com")
on the following intervals:
-
;
-
;
-
.
Prove that all of the solutions have a finite limit as 
, and that none of the solutions have a finite limit as 
.
First, some general setup that will apply to all parts. Since on any of the three interval we have 
we can divide through by this term to obtain,
![\[ (x-2)(x-3) y' + 2y = (x-1)(x-2) \quad \implies \quad y' + \frac{2}{(x-2)(x-3)} y = \frac{x-1}{x-3}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0f56870942b59f322c2434b3a6ea60a1_l3.png "Rendered by QuickLaTeX.com")
We will then apply Theorem 8.3 (page 310 of Apostol) with
![\[ P(x) = \frac{2}{(x-2)(x-3)}, \qquad Q(x) = \frac{x-1}{x-3}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a8d3dd6df0a7ac554b7567386e3da777_l3.png "Rendered by QuickLaTeX.com")
In each part will choose a different value for 
and give the solutions in terms 
.
- On the interval we choose the initial value
. Then we have (using partial fractions to evaluate the integral)
![\[ A(x) = \int_1^x \frac{2}{(t-2)(t-3)} \, dt = 2 \left( \log |3-x| - \log |2-x| - \log 2 \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3706bfa7d9c0827ff523129e596dca4d_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\begin{align*} f(x) &= \frac{b e^{2 \log|2-x| + 2 \log 2}}{e^{2 \log |3-x|}} + \frac{e^{2 \log|2-x| + 2 \log 2}}{e^{2 \log|3-x|}} \int_1^x \frac{t-1}{t-3} \frac{e^{2 \log|3-t|}}{e^{2 \log |2-t| + 2 \log 2}} \, dt \\[9pt] &= b \frac{4 (x-2)^2}{(x-3)^2} + \frac{4(x-2)^2}{(x-3)^2} \int_1^x \frac{(t-1)(t-3)}{4(t-2)^2} \, dt \\[9pt] &= \frac{4 (x-2)^2}{(x-3)^2} \left( b + \frac{1}{4} \int_1^x \frac{t^2 - 4t + 3}{t^2 - 4t + 4} \, dt \right) \\[9pt] &= \frac{4 (x-2)^2}{(x-3)^2} \left( b + \frac{1}{4} \left( \int_1^x \, dt - \int_1^x \frac{1}{(t-2)^2} \, dt \right) \right)\\[9pt] &= \frac{4 (x-2)^2}{(x-3)^2} \left( b + \frac{1}{4} \left( x-1 + \frac{1}{t-2} \Bigr \rvert_1^x \right) \right)\\[9pt] &= \frac{4(x-2)^2}{(x-3)^2} \left( b + \frac{1}{4} x + \frac{1}{4(x-2)} \right)\\[9pt] &= \left( \frac{x-2}{x-3} \right)^2 \left( x + \frac{1}{x-2} + C \right), \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5e4d61f89f14cf53058353a8295c3f42_l3.png "Rendered by QuickLaTeX.com")
where
is the constant 
.
- On the interval the same solution is going to work, with the different initial condition getting absorbed in the constant term. We can go ahead and compute (borrowing from our computations in part (a) where appropriate). Choose
giving us
Therefore,
![\begin{align*} f(x) &= \frac{b e^{2 \log |2-x|}}{e^{2 \log|3-x|}} + \frac{e^{2 \log |2-x|}}{e^{2 \log |3-x|}} \int_{\frac{5}{2}}^x \frac{t-1}{t-3} \frac{e^{2 \log |3-t|}}{e^{2 \log |2-t|}} \, dt \\[9pt] &= b \frac{(x-2)^2}{(x-3)^2} + \frac{(x-2)^2}{(x-3)^2} \int_{\frac{5}{2}}^x \frac{(t-1)(t-3)}{(t-2)^2} \, dt \\[9pt] &= \frac{(x-2)^2}{(x-3)^2} \left( b + \int_{\frac{5}{2}}^x \frac{t^2 - 4t + 3}{t^2 - 4t + 4} \, dt \right) \\[9pt] &= \frac{(x-2)^2}{(x-3)^2} \left( b + (x - 1) + \frac{1}{t-2}\Bigr \rvert_{\frac{5}{2}}^x \right) \\[9pt] &= \frac{(x-2)^2}{(x-3)^2} \left( b + x - 1 + \frac{1}{x-2} + 2 \right) \\[9pt] &= \left(\frac{(x-2)}{(x-3)}\right)^2 \left( x + \frac{1}{x-2} + C \right), \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-40bc8ae527662b5f617cb25789abe1cd_l3.png "Rendered by QuickLaTeX.com")
where
is the constant 
.
- This will follow identically to parts (a) and (b) arriving at the same result. The different initial condition getting absorbed in the constant .
In point b) I obtained C=b-9/2
In point a) I obtained C=4b.
Same here
Same, the rest were just too similar and too tedious to even try.