Question Number 194913 by cortano12 last updated on 19/Jul/23
Answered by witcher3 last updated on 19/Jul/23
![(f(x)+f′′(x))cos(x)=g(x) ∫^(π/2) _0 f′′cos(x)dx=f′cos(x)]_0 ^(π/2) +∫f′sin(x)dx =−f′(0)+fsin(x)]_0 ^(π/2) −∫_0 ^(π/2) fcosx)dx ⇔∫_0 ^(π/2) (f+f′′)cos(x)dx=f((π/2))−f′(0)=2 f′(0)=−1](Q194922.png)
$$\left(\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}''\left(\mathrm{x}\right)\right)\mathrm{cos}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\left.\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{2}}} \mathrm{f}''\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{f}'\mathrm{cos}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int\mathrm{f}'\mathrm{sin}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\left.=\left.−\mathrm{f}'\left(\mathrm{0}\right)+\mathrm{fsin}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{fcosx}\right)\mathrm{dx} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{f}+\mathrm{f}''\right)\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{f}\left(\frac{\pi}{\mathrm{2}}\right)−\mathrm{f}'\left(\mathrm{0}\right)=\mathrm{2} \\ $$$$\mathrm{f}'\left(\mathrm{0}\right)=−\mathrm{1} \\ $$$$ \\ $$
Answered by dimentri last updated on 20/Jul/23
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