Particular Solution of an Inhomogeneous Second Order ODE

In summary, the conversation discusses a particular solution for the differential equation y'' + 4y = tanx and provides answer choices for it. The correct solution is found using variation of parameters and the complementary function is given by CF = c1cos2x + c2sin2x. The particular integral is then found to be -cos(x)ln|tan(x)+sec(x)|. The complete solution is given by CS = CF + PI.
  • #1
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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.
 
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  • #2
The solutions to the homogeneous part are not correct. Think cos(2x) and sin(2x).
 
  • #3
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
Metaleer said:
The solutions to the homogeneous part are not correct. Think cos(2x) and sin(2x).

SVXX said:
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:
 

1. What is the particular solution of an inhomogeneous second order ODE?

The particular solution of an inhomogeneous second order ODE is the specific function that satisfies the given differential equation, taking into account the non-homogeneous term. It is a solution that is not affected by the initial conditions and represents the steady-state solution of the system.

2. How is the particular solution of an inhomogeneous second order ODE different from the general solution?

The general solution of an inhomogeneous second order ODE includes both the homogeneous solution and the particular solution. The particular solution is a specific solution that takes into account the non-homogeneous term, while the homogeneous solution satisfies the differential equation without considering the non-homogeneous term.

3. Can the particular solution of an inhomogeneous second order ODE be found analytically?

In some cases, the particular solution of an inhomogeneous second order ODE can be found analytically by using methods such as variation of parameters or undetermined coefficients. However, in more complex cases, numerical methods may be needed to find the particular solution.

4. What is the importance of finding the particular solution of an inhomogeneous second order ODE?

The particular solution of an inhomogeneous second order ODE is important because it helps to understand the behavior and stability of a system. It also allows for the determination of the steady-state response of the system, which is useful in many applications such as control systems and circuit analysis.

5. Can there be more than one particular solution for an inhomogeneous second order ODE?

No, there can only be one particular solution for an inhomogeneous second order ODE. This is because the particular solution is a specific solution that satisfies the given non-homogeneous term, and there can only be one such solution for a given differential equation.

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