Finding an exact solution for u(x)+∫02πcos⁡(x+t)u(t)dt=(π+1)cos⁡x

Reese Munoz

Reese Munoz

Answered

2022-01-28

Finding an exact solution for u(x)+02πcos(x+t)u(t)dt=(π+1)cosx

Answer & Explanation

coolbananas03ok

coolbananas03ok

Expert

2022-01-29Added 20 answers

From
cos(a+b)=cos(a)cos(b)sin(a)sin(b)
We have that
u(x)+cos(x)02πcos(t)u(t)dtsin(x)02πsin(t)u(t)dt=(π+1)cos(t)
But the integrals are independent of x, so they are constant! So the form of u will be
u(x)=(π+1)cos(x)c1cos(x)+c2sin(x)
And you can calculate the value of the constants easily:
c1=02πcos(t)u(t)dt
c1=02πcos(t)[(π+1)cos(x)c1cos(x)+c2sin(x)]dt
c1=(π+1)02πcos2(x)dxc102πcos2(x)dx+c202πsin(x)cos(x)dx
You can easily calculate these integrals with the double-angle formulas:
c1=(π+1)πc1π+c20
c1(1+π)=(π+1)π
c1=π
And you can do the same to get c2:
c2=02πsin(t)u(t)dt
So we have that c1=π and c2=0, which implies that
u(x)=cosx

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