Question

Need some help with Variation of Parameters:
y'''+y'=tan x

Answer 1

So here what I'd do is integrate both sides with respect to x right away so it's only a second order instead of a third order!

Answer 2

So it would be:
\[y''+y=\ln sinx\]

Answer 3

Not quite, but almost. It will be:
\[y''+y=-\ln|\cos(x)|+C_1\]

Don't forget the constant of integration!

Answer 4

So would:
\[u _{1}=-\int\limits_{}^{} \frac{ sinx(-lncosx+c1) }{ 1 }dx\]

Answer 5

@Kainui

Answer 6

@AccessDenied

Answer 7

i don't disagree with that work, i'm mainly just thinking about whether u = y' reduction of order is a more effective method than using integration at the start and setting it up afterwards... but i feel like you just get similar not-so-pretty integrals either way

Answer 8

so in the integral, how would you integrate sin(x)C1? @AccessDenied @Kainui

Answer 9

C1 is just a constant, so it is the equivalent of:
\( \int C_1 \sin x\ dx\)

Answer 10

Oh thats right, ok..

Answer 11

Does that first integral =
\[U1= -\cos(x)*\ln \cos(x) - 2\cos(x)+c\]
@AccessDenied @Kainui

Answer 12

@dumbcow @wio

Answer 13

where did the constant of integration C_1 on \( \displaystyle \int C_1 \sin x \ dx \) go?

Answer 14

As much as I like the idea of reducing the order, I'll give it a try as given.
\[y'''+y'=0~~\Rightarrow~~r^3+r=0~~\Rightarrow~~r=0,\pm i\]
giving these three fundamental solutions to the homogeneous portion:
\[y_1=1\\y_2=\cos x\\
y_3=\sin x\]
The Wronskian would be
\[W=W(y_1,y_2,y_3)=\begin{vmatrix}1&\cos x&\sin x\\0&-\sin x&\cos x\\0&-\cos x&-\sin x\end{vmatrix}=\begin{vmatrix}-\sin x&\cos x\\-\cos x&-\sin x\end{vmatrix}=1\]
Now, the particular solution will have the form \(u_1y_1+u_2y_2+u_3y_3\), where
\[\begin{align*}u_1&=\int\frac{\tan x W_1}{W}~dx\\
&=\int\tan x\begin{vmatrix}0&\cos x&\sin x\\0&-\sin x&\cos x\\1&-\cos x&-\sin x\end{vmatrix}~dx\\
&=\int\tan x~dx\end{align*}\]
\[\begin{align*}u_2&=\int\frac{\tan x W_2}{W}~dx\\
&=\int\tan x\begin{vmatrix}1&0&\sin x\\0&0&\cos x\\0&1&-\sin x\end{vmatrix}~dx\\
&=-\int\tan x\cos x~dx\end{align*}\]
\[\begin{align*}u_3&=\int\frac{\tan x W_3}{W}~dx\\
&=\int\tan x\begin{vmatrix}1&\cos x&0\\0&-\sin x&0\\0&-\cos x&1\end{vmatrix}~dx\\
&=\int\tan x\sin x~dx\end{align*}\]