Question

\[2y^{\prime\prime}+y^\prime-y=\cos x-3\sin x\]

Answer 1

\[2y_c^{\prime\prime}+y_c^\prime-y_c=0\]
\[2m^2+m-1=0\]\[(2m-1)(m+1)=0\]\[m_{1,2}=\frac12,-1\]
\[y_c=Ae^{-x}+Be^{\frac x2}\]

Answer 2

primerutevvb*(ios*beta gamma sigma

Answer 3

\[2y_{p_1}^{\prime\prime}+y_{p_1}^\prime-y_{p_1}=\cos x\]
\[y_{p_1}=C\sin x+D\cos x;\qquad y^\prime_{p_1}=C\cos x-D\sin x; \quad y^{\prime\prime}_{p_1}=-C\sin x-D\cos x\] \[-2(C\sin x+D\cos x)+(C\cos x-D\sin x)-(C\sin x+D\cos x)=\cos x\]\[-3(C\sin x+D\cos x)+(C\cos x-D\sin x)=\cos x\]\[(C-3D)\cos x-(3C+D)\sin x=\cos x\]\[C-3D=1;\qquad 3C+D=0\]\[D=-3C;\qquad C+9C=1\]\[C=\frac 1{10};\qquad D=-\frac 3{10}\]\[y_{p_1}=\frac 1{10}\left(\sin x-3\cos x\right)\]\[2y_{p_2}^{\prime\prime}+y_{p_2}^\prime-y_{p_2}=-3\sin x\]\[y_{p_2}=E\sin x+F\cos x;\qquad y^\prime_{p_2}=E\cos x-F\sin x; \quad y^{\prime\prime}_{p_2}=-E\sin x-F\cos x\]\[-2(E\sin x+F\cos x)+(E\cos x-F\sin x)-(E\sin x+F\cos x)=-3\sin x\]\[-(F+3E)\sin x+(E-3F)\cos x=-3\sin x\]\[F+3E=3;\qquad E-3F=0\]\[E=3F;\qquad 10F=3\]\[F=\frac3{10};\qquad E=\frac 1{10}\]\[y_{p_2}=\frac 1{10}\left(\sin x+3\cos x\right)\]\[y_p=\frac 1{10}\left(\sin x-3\cos x\right)+\frac 1{10}\left(\sin x+3\cos x\right)\]
\[y_p=\frac 15\sin x\]

Answer 4

i seam to have made some mistake with the particular solution ,

Answer 5

yes
the solution is
\[
\begin{array}{c}
y(x)=e^{x/2} c_1+e^{-x}
c_2+\sin (x) \\
\end{array}

\]

Answer 6

how do i get rid of the 5
?

Answer 7

why did you get rid of the -3sin(x) at the start

Answer 8

\[y_p=y_{p_1}+y_{p_2}\]

Answer 9

Can't you take the particular solution as:

\(Acosx + Bsinx\) ??

Answer 10

i did that @waterineyes

Answer 11

\[y_p = Csinx + D cosx\]\[y_p'=Ccosx-Dsinx\]\[y_p''=-Csinx-Dcosx\]
\[(C−3D)cosx−(3C+D)sinx=cosx-3sinx\]
\[C-3D = 1\]
\[3C+D = 3\]
D=0
C=1
\[y_p = sinx\]

Answer 12

so i was supposed to find the particular solution all at once, not break it up,
why didn't it work when i broke it up

Answer 13

You would have done a mistake while solving..

Answer 14

can you see it?

Answer 15

Meaning??
If you post then we can find the errors..

Answer 16

i have posted ...

Answer 17

can't find the mistake o,o... guess it can't be done like that

Answer 18

when can i use \[y_p=y_{p_1}+y_{p_2}\]
and when can't i?

Answer 19

Variation of Parameters, you can
Undetermined Coefficients, you can't

D: actually i'm not sure about this^, never splitted them

Answer 20

i was just following the method from my text book

Answer 21

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 22

You technically can split them up and you should get the same solution. You just have to realize which solutions you get for the coeffcients of each particular solution are extraneous.

Answer 23

i think i have \(y_{p_1}\) correct,
i must have made mistakes in \(y_{p_2}\)

Answer 24

Are you able to find those or not?

Answer 25

can you see my steps near the top of the page? @waterineyes

Answer 26

ok i see my error now
\[E=\frac 9{10}\neq\frac 1{10}\]

Answer 27

Yes there you have to multiply 3 with it which you have forgot to do..

Answer 28

Now you will have your answer..

Answer 29

i divided by three instead,

\[y_p=\frac 1{10}\left(\sin x-3\cos x\right)+\frac 3{10}\left(3\sin x+\cos x\right)\]\[y_p=\sin x\]\[y=y_c+y_p\]
\[y(x)=Ae^{-x}+Be^{\frac x2}+\sin x\qquad\color\red\checkmark\]

Answer 30

\[\huge \color{green}{{\checkmark}}\]

Answer 31

: )