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SupposeAeiax+Beibx=Ceicx, for some nonzero constants A, B, C, a, b, c, and for all x. Prove that a = b = cand A + B = C.

Short Answer

Expert verified

It is proved A + B = C and a = b = c.

Step by step solution

01

Expression for the amplitude and Euler equation:

Write the expression for the amplitude equation.

Aeiax+beibx=Ceicx …… (1)

Write the expression for the Euler equation.

eiθ=cosθ+isinθ

02

Prove A + B = C :

Substitute the known values in equation (1).

A(cosax+isinax)+B(cosbx+isinbx)=C(coscx+isincx)

Write the real part of the equation.

role="math" localid="1657520367793" Acosax+Bcosbx=Ccoscx

Write the real imaginary of the equation.

Asinax+Bsinbx=Csincx

Using boundary conditions at x = 0 , the left-hand side function will be equal to the right-hand side function.

Aeia(0)+Beib(0)=Ceic(0)A+B=C

03

Prove a = b = c :

Square and add the equation (2) and (3).

(Acosax+Bcosbx)2+(Asinax+Bsinbx)2=C2(cos2cx+sin2cx)

Since, cos2cx+sin2cx=1.

On further solving,

A2cos2ax+B2cos2bx+2AB(cosAx)(cosbx)+A2sin2ax+B2sinbx+2AB(sinax)(sinbx)=C2A2(cos2ax+sin2ax+B2(cos2bx+sin2bx)+2AB(cosaxcosbx+sinaxsinbx)=C2A2+B2+2ABcos(a-b)x=C2

Substitute the value of A+B=Cin the above expression.

A2+B2+2ABcos(a-b)x=(A+B)2 A2+B2+2ABcos(a-b)x=A2+B2+2AB2ABcos(a-b)x=2ABcos(a-b)x=1

Hence,

a = b

Substitute the value of a and b in equation (1).

eiax(A+B)=Ceicxeiax(A+B)=(A+B)eicxeiax=eicx

Solve further as,

a = b

= c

Therefore, it is proved A + B = C and a = b = c.

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