Reese Munoz

Answered

2022-01-28

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

Answer & Explanation

coolbananas03ok

Expert

2022-01-29Added 20 answers

From

$\mathrm{cos}(a+b)=\mathrm{cos}\left(a\right)\mathrm{cos}\left(b\right)-\mathrm{sin}\left(a\right)\mathrm{sin}\left(b\right)$

We have that

$u\left(x\right)+\mathrm{cos}\left(x\right){\int}_{0}^{2\pi}\mathrm{cos}\left(t\right)u\left(t\right)dt-\mathrm{sin}\left(x\right){\int}_{0}^{2\pi}\mathrm{sin}\left(t\right)u\left(t\right)dt=(\pi +1)\mathrm{cos}\left(t\right)$

But the integrals are independent of x, so they are constant! So the form of u will be

$u\left(x\right)=(\pi +1)\mathrm{cos}\left(x\right)-{c}_{1}\mathrm{cos}\left(x\right)+{c}_{2}\mathrm{sin}\left(x\right)$

And you can calculate the value of the constants easily:

${c}_{1}={\int}_{0}^{2\pi}\mathrm{cos}\left(t\right)u\left(t\right)dt$

${c}_{1}={\int}_{0}^{2\pi}\mathrm{cos}\left(t\right)[(\pi +1)\mathrm{cos}\left(x\right)-{c}_{1}\mathrm{cos}\left(x\right)+{c}_{2}\mathrm{sin}\left(x\right)]dt$

${c}_{1}=(\pi +1){\int}_{0}^{2\pi}{\mathrm{cos}}^{2}\left(x\right)dx-{c}_{1}{\int}_{0}^{2\pi}{\mathrm{cos}}^{2}\left(x\right)dx+{c}_{2}{\int}_{0}^{2\pi}\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)dx$

You can easily calculate these integrals with the double-angle formulas:

${c}_{1}=(\pi +1)\pi -{c}_{1}\pi +{c}_{2}0$

${c}_{1}(1+\pi )=(\pi +1)\pi$

${c}_{1}=\pi$

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

$u\left(x\right)=\mathrm{cos}x$

We have that

But the integrals are independent of x, so they are constant! So the form of u will be

And you can calculate the value of the constants easily:

You can easily calculate these integrals with the double-angle formulas:

And you can do the same to get

So we have that

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