\[2y^{\prime\prime}+y^\prime-y=\cos x-3\sin x\]
\[2y_c^{\prime\prime}+y_c^\prime-y_c=0\] \[2m^2+m-1=0\]\[(2m-1)(m+1)=0\]\[m_{1,2}=\frac12,-1\] \[y_c=Ae^{-x}+Be^{\frac x2}\]
primerutevvb*(ios*beta gamma sigma
\[2y_{p_1}^{\prime\prime}+y_{p_1}^\prime-y_{p_1}=\cos x\] \[y_{p_1}=C\sin x+D\cos x;\qquad y^\prime_{p_1}=C\cos x-D\sin x; \quad y^{\prime\prime}_{p_1}=-C\sin x-D\cos x\] \[-2(C\sin x+D\cos x)+(C\cos x-D\sin x)-(C\sin x+D\cos x)=\cos x\]\[-3(C\sin x+D\cos x)+(C\cos x-D\sin x)=\cos x\]\[(C-3D)\cos x-(3C+D)\sin x=\cos x\]\[C-3D=1;\qquad 3C+D=0\]\[D=-3C;\qquad C+9C=1\]\[C=\frac 1{10};\qquad D=-\frac 3{10}\]\[y_{p_1}=\frac 1{10}\left(\sin x-3\cos x\right)\]\[2y_{p_2}^{\prime\prime}+y_{p_2}^\prime-y_{p_2}=-3\sin x\]\[y_{p_2}=E\sin x+F\cos x;\qquad y^\prime_{p_2}=E\cos x-F\sin x; \quad y^{\prime\prime}_{p_2}=-E\sin x-F\cos x\]\[-2(E\sin x+F\cos x)+(E\cos x-F\sin x)-(E\sin x+F\cos x)=-3\sin x\]\[-(F+3E)\sin x+(E-3F)\cos x=-3\sin x\]\[F+3E=3;\qquad E-3F=0\]\[E=3F;\qquad 10F=3\]\[F=\frac3{10};\qquad E=\frac 1{10}\]\[y_{p_2}=\frac 1{10}\left(\sin x+3\cos x\right)\]\[y_p=\frac 1{10}\left(\sin x-3\cos x\right)+\frac 1{10}\left(\sin x+3\cos x\right)\] \[y_p=\frac 15\sin x\]
i seam to have made some mistake with the particular solution ,
yes the solution is \[ \begin{array}{c} y(x)=e^{x/2} c_1+e^{-x} c_2+\sin (x) \\ \end{array} \]
how do i get rid of the 5 ?
why did you get rid of the -3sin(x) at the start
\[y_p=y_{p_1}+y_{p_2}\]
Can't you take the particular solution as: \(Acosx + Bsinx\) ??
i did that @waterineyes
\[y_p = Csinx + D cosx\]\[y_p'=Ccosx-Dsinx\]\[y_p''=-Csinx-Dcosx\] \[(C−3D)cosx−(3C+D)sinx=cosx-3sinx\] \[C-3D = 1\] \[3C+D = 3\] D=0 C=1 \[y_p = sinx\]
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
You would have done a mistake while solving..
can you see it?
Meaning?? If you post then we can find the errors..
i have posted ...
can't find the mistake o,o... guess it can't be done like that
when can i use \[y_p=y_{p_1}+y_{p_2}\] and when can't i?
Variation of Parameters, you can Undetermined Coefficients, you can't D: actually i'm not sure about this^, never splitted them
i was just following the method from my text book
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.
i think i have \(y_{p_1}\) correct, i must have made mistakes in \(y_{p_2}\)
Are you able to find those or not?
can you see my steps near the top of the page? @waterineyes
ok i see my error now \[E=\frac 9{10}\neq\frac 1{10}\]
Yes there you have to multiply 3 with it which you have forgot to do..
Now you will have your answer..
i divided by three instead, \[y_p=\frac 1{10}\left(\sin x-3\cos x\right)+\frac 3{10}\left(3\sin x+\cos x\right)\]\[y_p=\sin x\]\[y=y_c+y_p\] \[y(x)=Ae^{-x}+Be^{\frac x2}+\sin x\qquad\color\red\checkmark\]
\[\huge \color{green}{{\checkmark}}\]
: )
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