particular integral of (d^2-4)y=2cos^2x
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Find a Particular solution to the equation
- But I stuck at this point because how to solve 1/D(D+4i)x²? The solution provide this
- I still can't figure out how they do this. Any help will be appreciated
Method 2
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[tex]Find a Particular solution to the equation
![\</strong>begin{align} \sf y_p&= \sf \frac{1}{D^2+4}(4x^2\cos 2x)\\ \\ & \sf =\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right<strong>]\\ \\ \sf &= \sf 2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\</strong>frac{1}{(D-2i)^2+4}x^2\right]\\ \\ \sf &= \sf 2\left[e^{2ix}\frac{1}{D(D+4i)}<strong>x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\ \sf \end{align} \\](https://tex.z-dn.net/?f=%5C%3C%2Fstrong%3Ebegin%7Balign%7D++++++++%5Csf+++++++y_p%26amp%3B%3D++++++++%5Csf++++++%5Cfrac%7B1%7D%7BD%5E2%2B4%7D%284x%5E2%5Ccos+2x%29%5C%5C++%5C%5C+%26amp%3B++++++++%5Csf++++++%3D%5Cfrac%7B1%7D%7BD%5E2%2B4%7D%5Cleft%5B2x%5E2%28e%5E%7B2ix%7D%2Be%5E%7B-2ix%7D%29%5Cright%3Cstrong%3E%5D%5C%5C++++%5C%5C+++++++%5Csf+++++++%26amp%3B%3D++++++++%5Csf++++++2%5Cleft%5Be%5E%7B2ix%7D%5Cfrac%7B1%7D%7B%28D%2B2i%29%5E2%2B4%7Dx%5E2%2Be%5E%7B-2ix%7D%5C%3C%2Fstrong%3Efrac%7B1%7D%7B%28D-2i%29%5E2%2B4%7Dx%5E2%5Cright%5D%5C%5C+++%5C%5C+%5Csf+++++++%26amp%3B%3D++++++++%5Csf++++++2%5Cleft%5Be%5E%7B2ix%7D%5Cfrac%7B1%7D%7BD%28D%2B4i%29%7D%3Cstrong%3Ex%5E2%2Be%5E%7B-2ix%7D%5Cfrac%7B1%7D%7BD%28D-4i%29%7Dx%5E2%5Cright%5D%5C%5C++++++++%5Csf+++++++%5Cend%7Balign%7D+%5C%5C+ "\</strong>begin{align} \sf y_p&= \sf \frac{1}{D^2+4}(4x^2\cos 2x)\\ \\ & \sf =\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right<strong>]\\ \\ \sf &= \sf 2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\</strong>frac{1}{(D-2i)^2+4}x^2\right]\\ \\ \sf &= \sf 2\left[e^{2ix}\frac{1}{D(D+4i)}<strong>x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\ \sf \end{align} \\")
But I stuck at this point because how to solve 1/D(D+4i)x²? The solution provide this![</strong>\\ \begin{align} &= \sf 2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\ \\ &= \sf <strong>\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\ \\ &= \sf \frac{1}{24}[6x^2\cos2x</strong>+x(8x^2-3)\sin 2x] \end{align} \\](https://tex.z-dn.net/?f=+%3C%2Fstrong%3E%5C%5C+%5Cbegin%7Balign%7D+%26amp%3B%3D++++++++%5Csf++++++2%5Cleft%5Be%5E%7B2ix%7D%5Cfrac%7B1%7D%7BD%28D%2B4i%29%7Dx%5E2%2Be%5E%7B-2ix%7D%5Cfrac%7B1%7D%7BD%28D-4i%29%7Dx%5E2%5Cright%5D%5C%5C+%5C%5C+%26amp%3B%3D++++++++%5Csf++++++%3Cstrong%3E%5Cfrac%7B1%7D%7B4i%7D%5Cleft%5Be%5E%7B2ix%7DD%5E%7B-1%7D%5Cleft%282x%5E2%2Bix-%5Cfrac%7B1%7D%7B4%7D%5Cright%29-e%5E%7B-2ix%7DD%5E%7B-1%7D%5Cleft%282x%5E2-ix-%5Cfrac%7B1%7D%7B4%7D%5Cright%29%5Cright%5D%5C%5C+%5C%5C++%26amp%3B%3D++++++++%5Csf++++++%5Cfrac%7B1%7D%7B24%7D%5B6x%5E2%5Ccos2x%3C%2Fstrong%3E%2Bx%288x%5E2-3%29%5Csin+2x%5D+%5Cend%7Balign%7D+%5C%5C+ "</strong>\\ \begin{align} &= \sf 2\left[e^{2ix}\frac{1}{D(D+4i)}x^2+e^{-2ix}\frac{1}{D(D-4i)}x^2\right]\\ \\ &= \sf <strong>\frac{1}{4i}\left[e^{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right)\right]\\ \\ &= \sf \frac{1}{24}[6x^2\cos2x</strong>+x(8x^2-3)\sin 2x] \end{align} \\")
Istill can't figure out how they do this. Any help will beappreciated![______________________________[/</strong>tex]<strong>Method 2</strong>[tex] \\ \sf \begin{align} & \sf =\frac{1}{4i}\left[e^<strong>{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right</strong>)\right]\\ \\ &= \sf \frac{1}{4i}\left[e^{2ix}\left(2\frac{x^3}{3}+i\frac{x^2}{2}-\frac{x}{4}\<strong>right)-e^{-2ix}\left(2\frac{x^3}{3}-i\frac</strong>{x^2}{2}-\frac{x}{4}\right)\right]\\ \\ &= <strong> \sf \frac{1}{4i}\left[\frac{2x^3}{3}(e^{2ix}-e^{-2ix})+\frac{x^2i}{2}(e^{2ix}-</strong>e^{-2ix})-\frac{x}{4}(e^{2ix}-e^{-2ix})\<strong>right]\\ \\ &= \sf \frac{1}{4i}\left[\frac{2x^3}{3}2\cos 2x+\frac{x^2i}{2}2\cos</strong> 2x-\frac{x}{4}2\cos 2x\right] \end{align} \\](https://tex.z-dn.net/?f=______________________________%5B%2F%3C%2Fstrong%3Etex%5D%3Cstrong%3EMethod+2%3C%2Fstrong%3E%5Btex%5D+%5C%5C+++++++++%5Csf++++++%5Cbegin%7Balign%7D+%26amp%3B++++++++%5Csf++++++%3D%5Cfrac%7B1%7D%7B4i%7D%5Cleft%5Be%5E%3Cstrong%3E%7B2ix%7DD%5E%7B-1%7D%5Cleft%282x%5E2%2Bix-%5Cfrac%7B1%7D%7B4%7D%5Cright%29-e%5E%7B-2ix%7DD%5E%7B-1%7D%5Cleft%282x%5E2-ix-%5Cfrac%7B1%7D%7B4%7D%5Cright%3C%2Fstrong%3E%29%5Cright%5D%5C%5C++%5C%5C+%26amp%3B%3D++++++++%5Csf++++++%5Cfrac%7B1%7D%7B4i%7D%5Cleft%5Be%5E%7B2ix%7D%5Cleft%282%5Cfrac%7Bx%5E3%7D%7B3%7D%2Bi%5Cfrac%7Bx%5E2%7D%7B2%7D-%5Cfrac%7Bx%7D%7B4%7D%5C%3Cstrong%3Eright%29-e%5E%7B-2ix%7D%5Cleft%282%5Cfrac%7Bx%5E3%7D%7B3%7D-i%5Cfrac%3C%2Fstrong%3E%7Bx%5E2%7D%7B2%7D-%5Cfrac%7Bx%7D%7B4%7D%5Cright%29%5Cright%5D%5C%5C+%5C%5C++%26amp%3B%3D+++++++%3Cstrong%3E+%5Csf++++++%5Cfrac%7B1%7D%7B4i%7D%5Cleft%5B%5Cfrac%7B2x%5E3%7D%7B3%7D%28e%5E%7B2ix%7D-e%5E%7B-2ix%7D%29%2B%5Cfrac%7Bx%5E2i%7D%7B2%7D%28e%5E%7B2ix%7D-%3C%2Fstrong%3Ee%5E%7B-2ix%7D%29-%5Cfrac%7Bx%7D%7B4%7D%28e%5E%7B2ix%7D-e%5E%7B-2ix%7D%29%5C%3Cstrong%3Eright%5D%5C%5C+%5C%5C++%26amp%3B%3D++++++++%5Csf++++++%5Cfrac%7B1%7D%7B4i%7D%5Cleft%5B%5Cfrac%7B2x%5E3%7D%7B3%7D2%5Ccos+2x%2B%5Cfrac%7Bx%5E2i%7D%7B2%7D2%5Ccos%3C%2Fstrong%3E+2x-%5Cfrac%7Bx%7D%7B4%7D2%5Ccos+2x%5Cright%5D+%5Cend%7Balign%7D+%5C%5C+ "______________________________[/</strong>tex]<strong>Method 2</strong>[tex] \\ \sf \begin{align} & \sf =\frac{1}{4i}\left[e^<strong>{2ix}D^{-1}\left(2x^2+ix-\frac{1}{4}\right)-e^{-2ix}D^{-1}\left(2x^2-ix-\frac{1}{4}\right</strong>)\right]\\ \\ &= \sf \frac{1}{4i}\left[e^{2ix}\left(2\frac{x^3}{3}+i\frac{x^2}{2}-\frac{x}{4}\<strong>right)-e^{-2ix}\left(2\frac{x^3}{3}-i\frac</strong>{x^2}{2}-\frac{x}{4}\right)\right]\\ \\ &= <strong> \sf \frac{1}{4i}\left[\frac{2x^3}{3}(e^{2ix}-e^{-2ix})+\frac{x^2i}{2}(e^{2ix}-</strong>e^{-2ix})-\frac{x}{4}(e^{2ix}-e^{-2ix})\<strong>right]\\ \\ &= \sf \frac{1}{4i}\left[\frac{2x^3}{3}2\cos 2x+\frac{x^2i}{2}2\cos</strong> 2x-\frac{x}{4}2\cos 2x\right] \end{align} \\")
Hope you like my answer.[/tex]
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