Question

Use variation of parameters to find a particular solution of y"+9y=tan3x.

Answer 1

r^2+9=0
so \[y _{p}=u _{1}\cos3x+u _{2}\sin3x\]
\[u'_{1}\cos3x+u'_{2}\sin3x=0\]
\[-3u'_{1}\sin3x+3u'_{2}\cos3x=\tan3x\]
------------------------------------
\[3u'_{2}\sin ^{2}3x+3u'_{2}\cos ^{2}3x=\sin3x\]
\[3u'_{2}=\sin3x\]
\[u'_{2}=\frac{ \sin3x }{ 3 }\]
integrate
\[u _{2}=-\frac{ \cos3x }{ 9 }\]

Answer 2

\[u'_{1}\cos3x=-u'_{2}\sin3x\]
\[u'_{1}=-\frac{ 1 }{ 3 }\tan3xsin3x\]
integrate
\[u _{1}=\frac{ \sin3x }{ 9 }-\frac{ \tan3x }{ 9 }\]
plug in,
\[y _{p}=\frac{ \sin3xcos3x }{ 9 }-\frac{ \sin3x }{ 9 }-\frac{ \sin3xcos3x }{ 9 }\]
But the answer in the book says
\[y _{p}=\frac{ -\cos3xln \left| \sec3x+\tan3x \right| }{ 9 }\]

Answer 3

can you say a little bit about this method
I'm kinda confused
or show me a website that solves in a similar way

Like are you actually using the Wronskain thingy?

Answer 4

can you tell me how you integrated -1/3 tan(3x) sin(3x) and got sin(3x)/9-tan(3x)/9 ?

Answer 5

By the method of integration by parts where u=tan3x, du=3sec^2 3x dx, dv=sin3x dx and v=-cos3x/3. Correct me if I'm wrong.

Answer 6

so you did
\[\int \tan(3x) \sin(3x)d x\]
\[ \frac{-1}{3} \cos(3x) \cdot \tan(3x)-\int \frac{-1}{3} \cos(3x) \cdot 3 \sec^2(3x)\]

\[\frac{-1}{3} \sin(3x)+\int \sec(3x) dx\]
....

Answer 7

how did you get -tan(3x)/3 from the second integral?

Answer 8

or tan(3x)/3

Answer 9

\[\int\limits \sec(u) du=\ln|\sec(u)+\tan(u)|+C\]

Answer 10

\[-\frac{ \sin3x }{ 3 }-\int\limits_{}^{}-\sec ^{2}3x dx\]
=\[-\frac{ \sin3x }{ 3 }+\int\limits_{}^{}\sec^2 3x dx\]
\[-\frac{ \sin3x }{ 3 }+\frac{ \tan3x }{ 3 }+C\]\
multiplying -1/3,
\[u _{1}=\frac{ \sin3x }{ 9 }-\frac{ \tan3x }{ 9 }\]

Answer 11

but how did you get + int sec^2(3x) dx ?

Answer 12

should be +int sec(3x) dx

Answer 13

Wait a minute, let me check my work real quick...

Answer 14

Okay, I see now. It should be sec 3x, not sec^2 3x. Let me fix it and see.

Answer 15

LOL! I don't know why I'm so rubbish...Thanks a million!

Answer 16

np

Answer 17

:)