Struggling with a Second Order Differential Equation? Let Us Help!

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Discussion Overview

The discussion revolves around solving the second-order differential equation y'' + y' + y = sin(x). Participants explore various methods for finding solutions, including variation of parameters, undetermined coefficients, and trial solutions. The conversation includes attempts to clarify the approach and share insights on the complexity of the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Some participants express difficulty in solving the equation and seek quicker methods than variation of parameters.
  • One participant suggests that the equation resembles a forced oscillator and proposes using forms like Ae^(iy) as a solution.
  • Another participant mentions the importance of initial conditions for finding a solution.
  • Several participants discuss the homogeneous part of the solution and the use of trial solutions like y_p = A cos(x) + B sin(x) to find a particular solution.
  • One participant indicates that they have found the general solution to the associated homogeneous equation but finds the integration challenging.
  • Another participant questions the correctness of their approach and calculations regarding the trial solution and its derivatives.
  • There is a discussion about finding L(e^x) and L(Xe^x) in relation to another differential equation, with participants sharing their assumptions and calculations.
  • One participant points out that the solution provided by another does not satisfy the associated homogeneous equation.
  • Another participant provides a breakdown of their calculations for the trial solution and checks if it satisfies the original equation.

Areas of Agreement / Disagreement

Participants generally do not reach a consensus on the best method to solve the equation, and multiple competing views and approaches remain throughout the discussion.

Contextual Notes

Some participants express uncertainty about their calculations and the correctness of their approaches, indicating that there may be missing assumptions or unresolved steps in their reasoning.

zina_6
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I need help solving this equation y''+y'+y= sinx

I know it looks simple but It seems to be getting sticky! I have been trying to solve it using variation of perimeters,maybe there is a quicker way? If anyone can help please...:bugeye:
 
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zina_6 said:
I need help solving this equation y''+y'+y= sinx

I know it looks simple but It seems to be getting sticky! I have been trying to solve it using variation of perimeters,maybe there is a quicker way? If anyone can help please...:bugeye:


Have you learned how to use Euler's merthod for second order systems with constant coefficients yet?
 
ty no, not yet Please explain
 
Last edited:
The form reminds me of a forced oscillator where sin (y) is your forcing function. These aren't too bad. Look at forms along the lines of Ae^yi power as a solution, where i is the imaginary number.

Note e^i*y= cos(y)+i*sin(y), and since the forcing function you are looking for is sin(y) you will want to look at only the imaginary portion of this equation.


-----

There might be faster way to solve this, but I am going to need to know if you have any intial conditions.
 
Well, the homogenous part of the solution should be easy to find, to find a particular solution, just insert a trial solution y_{p}=A\cos(x)+B\sin(x) and see what happens.
 
You say you tried variation of parameters so presumably you have already found the general solution to the associated homogenous equation, y"+ Y'+ y= 0. Yes, the solutions to that are a bit complicated:
e^{-\frac{x}{2}}\left(C cos(\frac{\sqrt{3}}{2}x)+ D sin(\frac{\sqrt{3}}{2}x)\right)
so that while "variation of parameters" is possible, it involves some complicated algebra.

Since the function on the right of the equation, sin(x), is one of the solutions one "expects" for linear equations with constant coefficients, do as arildno suggests- use "undetermined coefficients".
 
Yes ,That is exactly what i got HallsofIvy, but the integration is very difficult.
 
I also tried Yp: A cosx + b cos x , after taking the derivatives they seem to cancel out and leave :A sinx + b cosx = sinx A=1, and B=0 it does not seem to work ? I must be missing something or working it wrong?
 
How about this problem did i do this correct ?
--------------------------------------------------------------------------------


L(y)= y''-3y'+2y find L (e^x) and L (Xe^x)


Hence solve the equation y''-3y'+2y=3e^x



This is how I started to work out this problem.


y1=e^x This is what i am assuming to be y1 and y2 ?
y2= Xe^x

then my W = e^2x

Yp = 3x^2e^x/2


Then the general solution Y= c1 e^x + c2 Xe^x + (9X^2e^x)/2

Did I go about this the wrong way ?

Should I have plugged the value of L (e^x) and L (Xe^x)
 
Last edited:
  • #10
OK then let's see what you got ! let me see you break it down for me .
 
  • #11
Let y= A sin x+ B cos x
Then y'= A cos x- B sin x
and y"= -A sin x- B cos x

y"+ y'+ y= (A- B- A)sin x+ (B+ A- B)cos x= sin x
or -B sin x+ A cos x = sin x

You must have B= -1, A= 1.

zina 6 said:
How about this problem did i do this correct ?
--------------------------------------------------------------------------------


L(y)= y''-3y'+2y find L (e^x) and L (Xe^x)


Hence solve the equation y''-3y'+2y=3e^x



This is how I started to work out this problem.


y1=e^x This is what i am assuming to be y1 and y2 ?
y2= Xe^x

then my W = e^2x

Yp = 9x^2e^x/2


Then the general solution Y= c1 e^x + c2 Xe^x + (9X^2e^x)/2

Did I go about this the wrong way ?

Should I have plugged the value of L (e^x) and L (Xe^x)
The problem did ask you to solve the differential equation, but only after finding L (e^x) and L (Xe^x). Do that first and perhaps you will see how it helps you solve the equation! The solution you give is not correct. In particular, xex is not a solution to the associated homogeneous equation.
 
  • #12
the formula is A cosx + b sinx
yp: A cos x +b sinx
y'p: -A sinx +bcosx
y"p: -A cosx -b sinx


(A-A+b)cosx +(-A+b-b)sinx= sinx
B cosx + -Asinx = sinx
A=-1
b=0

Yp: -1 cosx

y'p : sinx

y"p : cosx

now insert them back in gives you y"+y'+y= sinx
cosx +sinx -cosx =sinx
sinx = sinx

Y: e^-x/2[c1cos radical 3/2x + c2 sinx radical 3/2x] -cosx
??
 
  • #13
that should do it.
 

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