 Reese Munoz

2022-01-28

Finding an exact solution for $u\left(x\right)+{\int }_{0}^{2\pi }\mathrm{cos}\left(x+t\right)u\left(t\right)dt=\left(\pi +1\right)\mathrm{cos}x$ coolbananas03ok

Expert

From
$\mathrm{cos}\left(a+b\right)=\mathrm{cos}\left(a\right)\mathrm{cos}\left(b\right)-\mathrm{sin}\left(a\right)\mathrm{sin}\left(b\right)$
We have that
$u\left(x\right)+\mathrm{cos}\left(x\right){\int }_{0}^{2\pi }\mathrm{cos}\left(t\right)u\left(t\right)dt-\mathrm{sin}\left(x\right){\int }_{0}^{2\pi }\mathrm{sin}\left(t\right)u\left(t\right)dt=\left(\pi +1\right)\mathrm{cos}\left(t\right)$
But the integrals are independent of x, so they are constant! So the form of u will be
$u\left(x\right)=\left(\pi +1\right)\mathrm{cos}\left(x\right)-{c}_{1}\mathrm{cos}\left(x\right)+{c}_{2}\mathrm{sin}\left(x\right)$
And you can calculate the value of the constants easily:
${c}_{1}={\int }_{0}^{2\pi }\mathrm{cos}\left(t\right)u\left(t\right)dt$
${c}_{1}={\int }_{0}^{2\pi }\mathrm{cos}\left(t\right)\left[\left(\pi +1\right)\mathrm{cos}\left(x\right)-{c}_{1}\mathrm{cos}\left(x\right)+{c}_{2}\mathrm{sin}\left(x\right)\right]dt$
${c}_{1}=\left(\pi +1\right){\int }_{0}^{2\pi }{\mathrm{cos}}^{2}\left(x\right)dx-{c}_{1}{\int }_{0}^{2\pi }{\mathrm{cos}}^{2}\left(x\right)dx+{c}_{2}{\int }_{0}^{2\pi }\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)dx$
You can easily calculate these integrals with the double-angle formulas:
${c}_{1}=\left(\pi +1\right)\pi -{c}_{1}\pi +{c}_{2}0$
${c}_{1}\left(1+\pi \right)=\left(\pi +1\right)\pi$
${c}_{1}=\pi$
And you can do the same to get ${c}_{2}$:
${c}_{2}={\int }_{0}^{2\pi }\mathrm{sin}\left(t\right)u\left(t\right)dt$
So we have that ${c}_{1}=\pi$ and ${c}_{2}=0$, which implies that
$u\left(x\right)=\mathrm{cos}x$

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