Question

One solution to the differential equation: \[y''-2\tan(x) \ y' -y=0 \ is \ y=\frac{ 1 }{ \cos x }\] Find another solution of the equation such that the two solutions form a fundamental set of solutions

Answer 1

Got an A in Diff Eq in the fall, can't remember anything about it now :(

Answer 2

I believe this is a problem addressing reduction of order isn't it @Wislar ?

I believe you want to start with setting:
\[\Large y=\frac{c(x)}{\cos x} \]

Answer 3

I'm sure you could do it that way, but taking the derivatives like that would take awhile. Isn't there a way to do this with Able's formula?

Answer 4

Using Abel's Formula, let \(y_2\) be the solution you're looking for. Then you set up the following equation:
\[\large W(\sec x,y_2)=Ce^{-\int (-2\tan x)~dx}\]
Which is the same as
\[\large\begin{vmatrix}\sec x&y_2\\\sec x\tan x&y_2'\end{vmatrix}=Ce^{2\int \tan x~dx}\]
Solving for \(y_2\) gives a solution in the form @Spacelimbus mentioned.

Answer 5

I tried solving it using reduction of order, but I couldn't get that to work out

Answer 6

\[y=\frac{ v }{ \cos x } \\ \\ y'=\frac{ v' \cos x +v \sin x }{ \cos ^{2}x } \\ y''=\frac{ v''\cos ^{3}x +vcos ^{3}x+2\cos ^{2}x \sin x v' + 2vcosx \sin ^{2}x }{ \cos ^{4}x }\]

Answer 7

Have you tried the method I used? It seems to work out, and I've checked the answer with WolframAlpha

Answer 8

I did try your method too and it is actually much easier than reduction of order, but I got C/cosx with yours too and that is the same thing as the current solution.

Answer 9

Evaluating the Wronskian in my last equation gives you
\[\large\sec x~y_2'-\sec x\tan x~y_2=Ce^{-2\ln|{\cos x}|}\\
\large\sec x~y_2'-\sec x\tan x~y_2=C\left(e^{\ln|{\cos x}|}\right)^{-2}\\
\large\sec x~y_2'-\sec x\tan x~y_2=C\left(\cos x\right)^{-2}\\
\large\sec x~y_2'-\sec x\tan x~y_2=C\sec^2x\\
\large \cos x~y_2'-\sin x~y_2=C\]
Note that the LHS is the result of a differentiation of a product:
\[\large \color{red}{\cos x~y_2'-\sin x~y_2}=C\\
\large \color{red}{\frac{d}{dx}\left[\cos x~y_2\right]}=C\]
Integrating both sides, you get
\[\large\cos x~y_2=Cx\\
\large y=\frac{Cx}{\cos x}\]

Answer 10

I see what happened. I forgot to integrate the C. Thank you so much!

Answer 11

You're welcome!