Question

[\left(D^4+5D^2+4\right)y=\cos x\]

Answer 1

\[\left(D^4+5D^2+4\right)y_c=0\]
\[m^4+5m^2+4=0\]\[(m^2+1)(m^2+4)=0\]\[m^2=-1,-4\]\[m=i,2i\]
\[y_c=A\cos x+B\sin x+C\cos2x+D\sin2x\]

Answer 2

how am i to find the particular solution?
\[\left(D^4+5D^2+4\right)y_p=\cos x\]

Answer 3

\[y_p=(Ex+F)\cos x +(Gx+H)\sin x\]?

Answer 4

yp=x(A sin x+ B cos x)

Answer 5

ah, i havent tried that yet

Answer 6

i get
\[\large y_{p}=\frac{x}{6} \ \sin x\]

Answer 7

sweet , that is my target,

Answer 8

\[y_{p}=x(A \sin x+ B \cos x) \\y_{p}''=2(A \sin x- B \cos x)+x(-A \sin x- B \cos x)\\ y_{p}''''=2(-A \sin x+ B \cos x)+x(A \sin x+ B \cos x) \]

then u have

\[6A \cos x-6B \sin x=\cos x \\ then \\ A=\frac{1}{6} \ \ \ and \ \ \ B=0\]

Answer 9

\[y_p=x(C\cos x+D\sin x)\]
\[y_p^\prime=x(-C\sin x +D\cos x)+C\cos x+D\sin x\]
\[y_p^{\prime\prime}=x( -C\cos x-D\sin x)-C\sin x+ D\cos x+(-C\sin x+ D\cos x)\]\[\qquad=x( -C\cos x-D\sin x)-2C\sin x+ 2D\cos x\]

Answer 10

sorry

yp''''=4(...********

Answer 11

ah ok, looks like im on the right track,

Answer 12

my choice of constants wasn't a great one
please note \(D≠D\)

Answer 13

D≠D ???

Answer 14

well in the original equation
\[\left(D^4+5D^2+4\right)y_c=0\]the book is using \[D=\frac{\text d}{\text dx}\]and i have been using \(A, B, C, D, ...\) for the constants,

Answer 15

ahhh....ok

Answer 16

its hard to be confused in context,
but it was a poor choice on my part for the constants, but im gonna stick with it and an instead write operator D like this
\[\text D=\frac{\text d}{\text dx}\]
\[\text D≠D\]

Answer 17

i think one of us might have made two sign errors
\[y_p=x(C\cos x+D\sin x)\]
\[y_p^\prime=x(-C\sin x +D\cos x)+C\cos x+D\sin x\]
\[y_p^{\prime\prime}=x( -C\cos x-D\sin x)-C\sin x+ D\cos x+(-C\sin x+ D\cos x)\]\[\qquad=x( -C\cos x-D\sin x)-2C\sin x+ 2D\cos x\]
\[y_p^{\prime\prime\prime}=x(C\sin x-D\cos x)-C\cos x-D\sin x-2C\cos x-2D\sin x\]\[\qquad=x(C\sin x-D\cos x)-3C\cos x-3D\sin x\]
\[y_p^{\text {iv}}=x(C\cos x+D\sin x)+C\sin x-D\cos x+3C\sin x-3D\cos x\]\[\qquad=x(C\cos x+D\sin x)+4C\sin x-4D\cos x\]

Answer 18

\[\left(\text D^4+5\text D^2+4\right)y=\cos x\]
\[\begin{array}{ccc} \left[x(C\cos x+D\sin x)+4C\sin x-4D\cos x\right]\\+5\left[x( -C\cos x-D\sin x)-2C\sin x+ 2D\cos x\right]\\+4\left[x(C\cos x+D\sin x)\right]\end{array}=\cos x\]
\[\begin{array}{ccc} x(C\cos x+D\sin x)+4(C\sin x-D\cos x)\\-5x( C\cos x+D\sin x)-10(C\sin x- D\cos x)\\+4x(C\cos x+D\sin x)\qquad\qquad\qquad\qquad\end{array}=\cos x\]
\[-6C\sin x +6D\cos x=\cos x\]
\[C=0:\qquad D=\frac 16\]
\[y_p=\frac x6\sin x\]

\[y(x)=A\cos x+B\sin x+C\cos2x+D\sin2x+\frac x6\sin x\qquad\checkmark\]

Answer 19

thanks for your help mukushla

Answer 20

your welcome my friend