Question

\[y^{\prime\prime}+4y=\cos2x\]

Answer 1

\[y_c^{\prime\prime}+4y_c=0\]\[m^2+4m=0\]\[m(m+4)=0\]\[m=0,-4\]\[y_c=A+Be^{-4}\]

Answer 2

\[y_p^{\prime\prime}+4y_p=\cos2x\]\[y_p=C\cos2x+D\sin2x\]\[y_p^\prime=-2C\sin2x+2D\cos2x\]\[y_p^{\prime\prime}=-4C\cos2x-4D\sin2x\]

Answer 3

\[(-4C\cos2x-4D\sin2x)+4(C\cos2x+D\sin2x)=\cos2x\]\[2C\cos2x+2D\sin2x=\cos2x\]\[2D\sin2x=0;\qquad 2C\cos2x=\cos2x\]\[D=0;\qquad C=\frac{1}{2}\]\[y_p=\frac 12\cos2x\]
\[y=y_c+y_p\]
\[y(x)=A+Be^{-4}+\frac 12\cos2x\]

Answer 4

however the answer in the back of the book was
\[ y (x)= \frac{1}4x\sin2x+A\cos2x+B\sin2x \]
where did i go wrong/

Answer 5

You plugged in the wrong thing. You put y instead of y' two posts ago.

Answer 6

ah yes i got the substitution wrong

Answer 7

wait y' ?

Answer 8

Well, I couldn't find the mistake yet..

Answer 9

was i ment to multiply the particular solution by x for some reason?

Answer 10

Wait, I think you will get complex roots...

\[y_c'' + 4y_c = 0\]

\[m^2 + 4 = 0\]


Isn't it???


\[m = \pm 2i\]

Answer 11

yeah i did make a mistak my my second lines of working, \[\cancel{m^2+4m=0}\]
thanks for spotting that @waterineyes \[m^2+4=0\]

Answer 12

Happy To Help...

Welcome dear..

Answer 13

\[y_c=A\left(E\cos 2x+F\sin 2x\right)\]

Answer 14

I think you can easily take as:

\[y_c = Acos2x + B \sin2x\]

Or not??

Answer 15

ah yes good idea

Answer 16

\[y^{\prime\prime}+4y=\cos2x\]\[y_c^{\prime\prime}+4y_c=0\]\[m^2+4=0\]\[m=\pm2i\]\[y_c=A\cos 2x+B\sin 2x\]

Answer 17

Yes you are going right...
Now solve for particular solution..

Answer 18

\[y_p^{\prime\prime}+4y_p=\cos2x\]\[y_p=C\cos2x+D\sin2x\]\[y_p^\prime=-2C\sin2x+2D\cos2x\]\[y_p^{\prime\prime}=-4C\cos2x-4D\sin2x\]\[(-4C\cos2x-4D\sin2x)+4(C\cos2x+D\sin2x)=\cos2x\]\[0=\cos2x\]
??there must be a mistake

Answer 19

How can this be possible??
I am in trouble too...

See your answers first term is \(\frac{1}{4}xsin2x\) we have to get this term but how..

Answer 20

Are you sure your question contains \(cos2x\) on Right Hand Side..??

Answer 21

yes the question in the book is
\[y^{\prime\prime}-4y=\cos2x\]

Answer 22

oh no i see +≠-,

what a mistake to make
sorry everyone

Answer 23

hmm actually i have two copies of the book and they are different

Answer 24

the latest version the one i was working with , has made a printing error, the question should have been \[y^{\prime\prime}-4y=\cos2x\]

Answer 25

\[\text{---------}\]\[y^{\prime\prime}-4y=\cos2x\]
\[y_c^{\prime\prime}-4y_c=0\]
\[m^2-4=0\]\[m=\pm2\]
\[y_c=Ae^{2x}+Be^{-2x}\]
\[y_p^{\prime\prime}-4y_p=\cos2x\]
\[y_p=C\cos2x+D\sin2x\]\[y_p^\prime=-2C\sin2x+2D\cos2x\]\[y_p^{\prime\prime}=-4C\cos2x-4D\sin2x\]
\[(-4C\cos2x-4D\sin2x)+4(C\cos2x+D\sin2x)=\cos2x\]\[-8C\cos2x-8D\sin2x=\cos2x\]\[-8D\sin2x=0;\qquad -8C\cos2x=\cos2x\]\[D=0;\qquad C=-\frac{1}{8}\]
\[y_p=-\frac 18\cos2x\]
\[y=y_c+y_p\]
\[y(x)=Ae^{2x}+Be^{-2x}-\frac 18\cos2x\]\[\text{---------}\]