Question

check my work

Answer 1

?

Answer 2

pre-algebra?

Answer 3

??

Answer 4

\[(D^3+1)y=2\cos^2x\]\[D^3 \equiv \frac{d^3}{dx^3}\]
\[(D^3+1)y=\cos(2x)+1\]
Auxillary equation
\[D^3+1=0\]
\[D^3+1^3=0\]\[a^3+b^3=(a+b)(a^2-ab+b^2)\]
\[(D+1)(D^2-D+1)=0\]
when D+1=0\[D=-1\]
When
\[D^2-D+1=0\]\[D=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\sqrt{-3}}{2}=\frac{1\pm\sqrt{3}i}{2}=\frac{1}{2}\pm \frac{\sqrt{3}}{2}i\]
Thus
\[C.F.=Ae^{-x}+e^{\frac{x}{2}}(B\cos(\frac{\sqrt{3}x}{2})+C\sin(\frac{\sqrt{3}x}{2}))\]
\[P.I.=\frac{1+\cos(2x)}{D^3+1}=\frac{1}{D^3+1}+\frac{\cos(2x)}{D^3+1}\]\[P.I.=\frac{e^{0x}}{D^3+1}+\frac{\cos(2x)}{(D+1)(D^2-D+1)}\]\[\frac{1}{f(D)}e^{ax}=\frac{1}{f(a)}e^{ax} \space \space \space ; \space \space \space f(a) \neq 0\]\[f(D)=D^3+1 \space \space \space ; \space \space \space f(0)=1 \neq 0\]
\[\therefore \frac{e^{0x}}{D^3+1}=\frac{e^{0x}}{1}=1\]
\[P.I.=1+\frac{\cos(2x)}{(D+1)(D^2-D+1)}\]\[\frac{1}{f(D^2)}(\cos(ax))=\frac{1}{f(-a^2)}.(\cos(ax)) \space \space \space ; \space \space \space f(-a^2)\neq 0\]
\[f(D^2)=D^2+1\]
\[f(-(2^2))=-4+1=-3\]
\[P.I.=1+\frac{\cos(2x)}{(D+1)(-D-3)}=1-\frac{\cos(2x)}{(D+1)(D+3)}=1-\frac{\cos(2x)}{D^2+4D+3}\]
\[f(D^2)=D^2+3\]
\[f(-(2^2))=-4+3=-1 \neq0\]
\[P.I.=1-\frac{\cos(2x)}{4D-1}=1-\frac{(4D+1)\cos(2x)}{16D^2-1}\]
\[f(D)=16D^2-1\]\[f(-(2^2))=-64-1=-65\neq0\]\[P.I.=1-\frac{(4D+1)\cos(2x)}{-65}=1+\frac{1}{65}.(4D+1)\cos(2x)\]\[P.I.=1+\frac{1}{65}(-8\sin(2x)+\cos(2x))\]\[y=C.F.+P.I.\]\[y=Ae^{-x}+e^{\frac{x}{2}}(B\cos(\frac{\sqrt{3}x}{2})+C\sin(\frac{\sqrt{3}x}{2}))+..\]\[..+1+\frac{1}{65}(-8\sin(2x)+\cos(2x))\]

Answer 5

@baru thxx

Answer 6

the CF part looks right,
the PI, i learnt it slightly differently so i'm not sure, but the final answer looks right, you are supposed to get a shifted sinusoidal function with a different amplitude but same angular frequency, which you have got :)