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

In summary, the conversation discusses solving the equation y''+y'+y= sinx using variation of parameters and undetermined coefficients. The participants also mention using Euler's method and finding the general solution to the associated homogeneous equation. Ultimately, the correct solution is found using the formula A cosx + b sinx for y'p and y"p. The final solution is e^-x/2[c1cos radical 3/2x + c2 sinx radical 3/2x] -cosx.
  • #1
zina_6
8
0
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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  • #2
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?
 
  • #3
ty no, not yet Please explain
 
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  • #4
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.
 
  • #5
Well, the homogenous part of the solution should be easy to find, to find a particular solution, just insert a trial solution [itex]y_{p}=A\cos(x)+B\sin(x)[/itex] and see what happens.
 
  • #6
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:
[tex]e^{-\frac{x}{2}}\left(C cos(\frac{\sqrt{3}}{2}x)+ D sin(\frac{\sqrt{3}}{2}x)\right)[/tex]
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".
 
  • #7
Yes ,That is exactly what i got HallsofIvy, but the integration is very difficult.
 
  • #8
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?
 
  • #9
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)
 
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  • #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.
 

1. What is a second order differential equation?

A second order differential equation is a mathematical equation that involves a function and its first and second derivatives. It is used to model physical systems and describe their behavior over time.

2. How is a second order differential equation different from a first order differential equation?

A first order differential equation involves only the first derivative of a function, while a second order differential equation involves both the first and second derivatives. This means that a second order differential equation is more complex and can provide more information about the behavior of a system.

3. What are some common applications of second order differential equations?

Second order differential equations are used in a variety of fields, including physics, engineering, and economics. They can be used to model the motion of objects, describe the growth and decay of populations, and analyze electrical circuits, among other applications.

4. How do you solve a second order differential equation?

The general solution to a second order differential equation involves finding the roots of the characteristic equation and using them to form a linear combination of solutions. This process can be complex and may require advanced mathematical techniques such as Laplace transforms or power series methods.

5. Are there any real-world limitations to using second order differential equations?

While second order differential equations can be powerful tools for modeling physical systems, they do have limitations. These equations may not accurately reflect all aspects of a system, and they can be difficult to solve in some cases. Additionally, they may not account for external factors or unpredictable events that can affect the behavior of a system.

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