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Particular Solution of an Inhomogeneous Second Order ODE

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Homework Statement



A particular solution of y'' + 4y = tanx

Answer choices are:

(a) 1/2*cos(2x)ln|sec(2x)+tan(2x)|
(b) -1/2*cos(2x)ln|sec(2x)+tan(2x)|
(c) 1/2*sin(2x)(ln*cos(x)+x*sec(2x))
(d) 1/2*sin(2x)(ln*cos(x)-x*sec(2x))
(e) none of the above

Homework Equations




The Attempt at a Solution



I used variation of parameters with y1=cos(x) and y2=sin(x)

Wronskian(cos(x) sin(x))=1

yp=-v1y1+v2y2

v1'=sin(x)tan(x)
=(sin(x))^2/cos(x)
=(1-(cos(x))^2)/cos(x)
=1/cos(x)-cos(x)

v1=ln|tan(x)+sec(x)|-sin(x)

v2'=cos(x)tan(x)
=sin(x)

v2=-cos(x)

yp=-cos(x)(ln|tan(x)+sec(x)|-sin(x))+sin(x)(-cos(x))
=-cos(x)ln|tan(x)+sec(x)|+cos(x)sin(x)-sin(x)cos(x)
=-cos(x)ln|tan(x)+sec(x)|

Given the answer choices, now I feel like I've missed something fundamental. It's probably obvious. So sorry.
 

Answers and Replies

  • #2
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The solutions to the homogeneous part are not correct. Think cos(2x) and sin(2x).
 
  • #3
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Why not use the normal method?
Rewriting in terms of the D operator,
(D² + 4)y = tan(x).

The auxiliary equation of this D.E. is D = ±2i, henceforth we obtain the complementary function to be -:

CF = e0[c1cos2x + c2sin2x]

For a function such as tan(x), the particular integral is given by -:

)(D%20-%202i)}tan(x)%20=%20\frac{1}{4i}[\frac{1}{D%20-%202i}%20-%20\frac{1}{D%20+%202i}](tan(x)).gif


Operating separately, we have -:

gif.latex?\frac{1}{D%20-%202i}(tan(x))%20=%20e^{2ix}%20\int%20e^{-2ix}tan(x).gif


and

gif.latex?\frac{1}{D%20+%202i}(tan(x))%20=%20e^{-2ix}%20\int%20e^{2ix}tan(x).gif


Once you compute these integrals, the complete solution of the differential equation will be given by -:

CS = CF + PI
 
  • #4
10
0
The solutions to the homogeneous part are not correct. Think cos(2x) and sin(2x).
Why not use the normal method?
Rewriting in terms of the D operator,
(D² + 4)y = tan(x).

The auxiliary equation of this D.E. is D = ±2i, henceforth we obtain the complementary function to be -:

CF = e0[c1cos2x + c2sin2x]

For a function such as tan(x), the particular integral is given by -:

)(D%20-%202i)}tan(x)%20=%20\frac{1}{4i}[\frac{1}{D%20-%202i}%20-%20\frac{1}{D%20+%202i}](tan(x)).gif


Operating separately, we have -:

gif.latex?\frac{1}{D%20-%202i}(tan(x))%20=%20e^{2ix}%20\int%20e^{-2ix}tan(x).gif


and

gif.latex?\frac{1}{D%20+%202i}(tan(x))%20=%20e^{-2ix}%20\int%20e^{2ix}tan(x).gif


Once you compute these integrals, the complete solution of the differential equation will be given by -:

CS = CF + PI


Thanks a lot. I understand now. I was not sure about the fundamental set when dealing with tan(x), but I see it now.

SVXX: Because the lesson was on variation of parameters. :wink:
 

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