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Question Number 194913 by cortano12 last updated on 19/Jul/23

Answered by witcher3 last updated on 19/Jul/23

$$(\mathrm{f}\left(\mathrm{x}\right)+{\mathrm{f}}^{″}\left(\mathrm{x}\right))\cos \left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)$$$${{\underset{0}{\int }}^{\frac{\pi }{2}}{\mathrm{f}}^{″}\cos \left(\mathrm{x}\right)\mathrm{dx}={\mathrm{f}}^{\prime }\cos \left(\mathrm{x}\right)]}_0^{\frac{\pi }{2}}+\int {\mathrm{f}}^{\prime }\sin \left(\mathrm{x}\right)\mathrm{dx}$$$$={-{\mathrm{f}}^{\prime }\left(0\right)+\mathrm{fsin}\left(\mathrm{x}\right)]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\mathrm{fcosx})\mathrm{dx}$$$$\iff {\int }_0^{\frac{\pi }{2}}(\mathrm{f}+{\mathrm{f}}^{″})\cos \left(\mathrm{x}\right)\mathrm{dx}=\mathrm{f}\left(\frac{\pi }{2}\right)-{\mathrm{f}}^{\prime }\left(0\right)=2$$$${\mathrm{f}}^{\prime }\left(0\right)=-1$$$$$$

Answered by dimentri last updated on 20/Jul/23

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