(a) Write cos 3x + i sin 3x = e^(3ix) = (cos x + i sin x)^3 by Euler's formula, expand, and equate real and imag inary parts to derive the identities, cos^(3) x = 3/4 cos x + 1/4 cos 3x, sin^(3) x = 3/4 sin x - 1/4 sin 3x (b) Use the result of part (a) to find a general solution of y'' + 4y = cos^(3) x.

Leonidas Cook

Leonidas Cook

Open question

2022-08-17

(a) Write
cos3x+isin3x=e3ix=(cosx+isinx)3
by Euler's formula, expand, and equate real and imag inary parts to derive the identities
cos3x=34cosx+14cos3x
sin3x=34sinx14sin3x
(b) Use the result of part (a) to find a general solution of
y+4y=cos3x.

Answer & Explanation

cuevamc

cuevamc

Beginner2022-08-18Added 6 answers

a) Given
The equation is cos3x+isin3x=e3ix=(cos3x+isin3x)3
Expand the equation
cos3x+isin3x=cos3x+3icos2xsinx3cosxsin2xisin3x
=cos3x+3i(1sin2x)sinx3cosx(1cos2x)isin3x
=cos3x+3isinx3isin3x3cosx+3cos3xisin3x
=4cos3x+3isinx3cosx4isin3x
=(4cos3x3cosx)+i(3sinx4sin3x)
Equate the real part:
cos3x=4cos3x3cosx
4cos3x=cos3x+3cosx
cos3x=14cos3x+34cosx
Equate the imaginary part:
sin3x=3sinx4sin3x
4sin3x=3sinxsin3x
sin3x=34sinx14sin3x
Hence proved
firestopper53sh

firestopper53sh

Beginner2022-08-19Added 1 answers

Step 4
(b)
Consider the differential equation y+4y=cos3x
From part(a),
y+4y=34cosx+14cos3x
Thus, the homogeneous equation is y+4y=0
Corresponding auxiliary equation is
r2+4=0
r=±2i
Thus, the complementary function is yc=c1cos2x+c2sin2x
Step 5
Consider the particular equation is yp=Acosx+Bcos3x
Differentiate the above function
yp=Acosx+Bcos3x
yp=Asinx3Bsin3x
yp=Acosx9Bcos3x
Step 6
Now,
Acosx9Bcos3x+4(Acosx+Bcos3x)=34cosx+14cos3x
Acosx9Bcos3x+4Acosx+4Bcos3x=34cosx+14cos3x
5Bcos3x+3Acosx=34cosx+14cos3x
Equate the coefficients:
3A=34
A=14
5B=14
B=120
Hence, yp=14cosx120cos3x
Thus, the general solution is
y(x)=yc+yp=c1cos2x+c2sin2x+14cosx120cos3x

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