Probably a really easy problem

  • Thread starter Exulus
  • Start date
In summary, the conversation is about solving a differential equation involving a sine function. The person is struggling to find a solution and has tried using a trial function, but it still comes out as 0. Another person suggests using Lagrange's method and the conversation ends with the first person realizing their mistake and finding the correct solution.
  • #1
Exulus
50
0
Could anyone help me with this? I need to solve this equation:

[itex]\frac{d^2y}{dx^2} + 4y = \sin{2x}[/itex]

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 [itex]y = c \sin{2x}[/itex]

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

Substituting these into the LHS of my question gives:

[itex] -4c \sin{2x} + 4c \sin{2x} = 0[/itex]

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).
 
Physics news on Phys.org
  • #2
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. =)
 
  • #3
*Ammendments:Trial solution=CxSin2x+DxCos2x
 
Last edited:
  • #4
Hi,

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

[itex]y = Cx\sin{2x} + Dx\cos{2x}[/itex]

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

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

Subbing back into question:

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

:(
 
  • #5
Try Lagrange's method (of constant variation)

Assume the inhomogenous solution

[tex] y_{p}=C(x)\sin 2x+D(x)\cos 2x [/tex]

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

Daniel.
 
  • #6
doh, it looks like i did my sums wrong i think. I've now got [itex]y = -\frac{1}{4}\cos{2x}[/itex] which looks like the right answer (haven't tested yet).

Cheers for the help guys! :)
 
Last edited:
  • #7
If you multiplicate it by x I'll believe you...
 
  • #8
Anyways,there's an alternative to 'trial solutions'.Lagrange's method is suitable for these problems.

Daniel.
 
  • #9
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 :)
 

Related to Probably a really easy problem

1. What is the definition of "Probably a really easy problem"?

"Probably a really easy problem" refers to a problem or task that is expected to have a simple solution or be easily solvable.

2. How do you determine if a problem is "Probably a really easy problem"?

A problem can be considered "probably a really easy problem" if it meets certain criteria, such as having a clear and concise question or prompt, having readily available resources or information to solve it, and not requiring advanced or complex techniques to solve.

3. Can a problem that seems "Probably a really easy problem" actually be difficult?

Yes, it is possible for a problem to seem easy at first glance but turn out to be more complex or challenging upon further examination. This could be due to unforeseen obstacles or lack of knowledge or experience with the subject matter.

4. How do you approach solving a problem that is "Probably a really easy problem"?

The best approach to solving a "probably a really easy problem" is to first carefully read and understand the question or prompt, then break it down into smaller, manageable steps. It can also be helpful to gather any necessary resources or information before attempting to solve the problem.

5. Is there a specific technique or strategy for solving "Probably a really easy problem"?

There is no one specific technique or strategy for solving "probably a really easy problem", as each problem may require a different approach. However, some general tips include breaking the problem into smaller parts, using logic and critical thinking skills, and seeking help or guidance if needed.

Similar threads

  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
5
Views
1K
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
909
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
593
  • Differential Equations
Replies
3
Views
2K
  • Calculus
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
Back
Top