How Can I Solve This Second-Order Differential Equation?

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Discussion Overview

The discussion revolves around solving the second-order differential equation \(\frac{d^2y}{dx^2} + 4y = \sin{2x}\). Participants explore various methods for finding the solution, including trial solutions and Lagrange's method, while addressing challenges encountered during the process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in solving the equation, noting that their attempts with \(y = c \sin{2x}\) lead to cancellation, suggesting a potential complex solution.
  • Another participant suggests finding the homogeneous solution first, indicating that the characteristic equation yields roots \(r = 2i, -2i\), leading to a homogeneous solution of \(y_{homogeneous} = A \cos{2x} + B \sin{2x}\).
  • A trial solution of the form \(y = Ax \sin{2x} + Bx \cos{2x}\) is proposed, but a later reply indicates that substituting this into the equation results in zero, causing frustration.
  • One participant recommends using Lagrange's method of variation of parameters, proposing a particular solution of the form \(y_{p} = C(x) \sin{2x} + D(x) \cos{2x}\) and suggests solving for the unknown functions by substitution.
  • A participant later claims to have found a solution of \(y = -\frac{1}{4} \cos{2x}\), although they have not yet verified it.
  • There are repeated mentions of needing to multiply by \(x\) in certain trial solutions, indicating a potential misunderstanding or error in the formulation of the trial functions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to solving the differential equation, with multiple competing methods and some unresolved calculations. Disagreement exists regarding the effectiveness of the proposed trial solutions and methods.

Contextual Notes

Some participants express uncertainty about the correctness of their calculations and the appropriateness of the methods used. There are indications of missing assumptions or steps in the reasoning, particularly regarding the application of trial solutions and the method of undetermined coefficients.

Exulus
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Could anyone help me with this? I need to solve this equation:

\frac{d^2y}{dx^2} + 4y = \sin{2x}

Everything I seem to try for y ends up canceling itself out, as the second diffferential is always the negative of what you begin with, and the 2x part of the sin means you get a factor of 4 in the second differential, which then cancels with the 4y. Eg:

Trying y = c \sin{2x}

\frac{dy}{dx} = 2c \cos{2x}
\frac{d^2y}{dx^2} = -4c \sin{2x}

Substituting these into the LHS of my question gives:

-4c \sin{2x} + 4c \sin{2x} = 0

The only thing i can think of now is that its a complex solution..but we haven't been taught that yet!

Any help appreciated :(

(ps no idea why the itex stuff is coming out so small..sorry about that).
 
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You'll have to find the homogeneous solution first before applying the method of undetermined coefficients. Let (D^2+4)y=r^2+4=0. hence r=2i,-2i
Thus y(homogenous)=Acos2x+Bsin2x
Therefore, let the trial function be y=AxSin2x+BxCos2x
You should get the answer from here. =)
 
*Ammendments:Trial solution=CxSin2x+DxCos2x
 
Last edited:
Hi,

I tried your trial solution but it still comes out as 0 :(

y = Cx\sin{2x} + Dx\cos{2x}

\frac{dy}{dx} = 2Cx\cos{2x} + C\sin{2x} - 2Dx\sin{2x} + D\cos{2x}

\frac{d^2y}{dx^2} = -4Cx\sin{2x} + 2C\cos{2x} - 2C\cos{2x} - 4Dx\cos{2x} - 2D\sin{2x} - 2D\sin{2x} =
= -4x(C\sin{2x} + D\cos{2x}) - 4D\sin{2x}

Subbing back into question:

-4x(C\sin{2x} + D\cos{2x}) + 4Cx\sin{2x} + 4Dx\cos{2x} = 0

:(
 
Try Lagrange's method (of constant variation)

Assume the inhomogenous solution

y_{p}=C(x)\sin 2x+D(x)\cos 2x

And solve for the unknown functions,by plugging it in the original ODE.

Daniel.
 
doh, it looks like i did my sums wrong i think. I've now got y = -\frac{1}{4}\cos{2x} which looks like the right answer (haven't tested yet).

Cheers for the help guys! :)
 
Last edited:
If you multiplicate it by x I'll believe you...
 
Anyways,there's an alternative to 'trial solutions'.Lagrange's method is suitable for these problems.

Daniel.
 
After subbing the values of d^2y/dx^2,dy/dx and y into the differential equation, the RHS of the equation should be sin2x, then you'll have to do comparing of the coefficients to determine the value of the constants =)
 
  • #10
Palindrom said:
If you multiplicate it by x I'll believe you...
Sorry yeah i typed it wrong, whoops. I did actually have it with the x written down..honest ;)

Thanks for the other method as well, dextercioby :)
 

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