Question

(ODE) Setting up a variation of parameters problem shortly. Having an issue somewhere down the line possibly with the Wronskian. Prompt posted below, along with workings so far.

Answer 1

\[y''-y=\sec(x).\]

Answer 2

\[\text{Characteristic Equation:} \ m^2-m=0; \ \ \ m(m-1)=0.\]

Answer 3

\[m=0, \ m=-1. \]

Answer 4

\[y_{h}=c_{1}e^0+c_{2}e^{-x}=c_{1}+c_{2}e^{-x}.\]

Answer 5

\[
W(e^{-x},1)=
\begin{vmatrix}
c^{-x}&1\\
-e^{-x}&0
\end{vmatrix}
\]

Answer 6

\[W=-e^{-x} \neq 0,\]Thus, the solutions are linearly independent, at least. Now just to find them:

Answer 7

\[
W_1=
\begin{vmatrix}
0&1\\
sec(x)&0
\end{vmatrix}
=-sec(x).\]

Answer 8

\[
W(y_1,y_2)=
\begin{vmatrix}
e^{-x}&0\\
-e^{-x}&sec(x)
\end{vmatrix}
=e^{-x} sec (x)\]

Answer 9

Whoops, wasn't supposed to have that notation on the left but w/e.

Answer 10

@ganeshie8

Answer 11

\[\frac{W_{1}}{W}=u'(x)=\frac{-\sec(x)}{e^{-x}}=-e^{-x}\sec(x)\]

Answer 12

\[\frac{W_{2}}{W}=\frac{e^{-x}\sec(x)}{e^{-x}}=\sec(x)\]

Answer 13

heyy

Answer 14

Integrating both of those: \[\int\limits_{}^{}\sec(x)=\ln|(\sec(x)+\tan(x)|+C\]

Answer 15

Yoooo (lol)

Answer 16

What, did I make a mistake somewhere?

Answer 17

\[\int\limits_{}^{}e^{-x}\sec(x)= \text{(Integration by parts)} \dots\]

Answer 18

Says that the antiderivative can't be found...by an integral calculator.

Answer 19

@hartnn

Answer 20

OH, nevermind, beginning characteristic equation was wrong.

Answer 21

\[y''+y=\sec(x); \ \ \ m^2+1=0. \]

Answer 22

\[m = \pm \sqrt{-1}=\pm i\]