Solving second order unhomogonous equations

In summary, the conversation is about solving a second order differential equation with initial conditions and finding the particular integral. Different methods such as variation of parameters and substitution are discussed, and one person suggests using a specific method with a particular solution of the form y = x(A0 + A1x)e^(ix). The conversation ends with the solution being found through further trial and error.
  • #1
sweep123
16
0

Homework Statement


solve y"+y=xsin(x) , initial conditions y(0)=0, y'(0)=1


Homework Equations





The Attempt at a Solution


I know that when right hand side equals something like sin(x), then the particular integral is xc1sin(x)+xc2cos(x). I am unsure what the particular integral should be for above. can anyone help me. I have tried (x^2)c1sin(x)+(x^2)c2cos(x), but have not gotten a solution from it. Thanks
 
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  • #2
You need a bigger family of trial functions. Try c1*x^2*sin(x)+c2*x^2*cos(x)+c3*x*sin(x)+c4*x*cos(x).
 
  • #3
instead of looking at that equation you could try to look at an equivalent equation
[tex]L[y] = y'' + y = xe^{ix}[/tex], where [tex]i = \sqrt{-1}[/tex]. then since [tex] i [/tex] is a particular root of the characteristic equation, y has a solution of the form
[tex] y = x^{2}A_{0}e^{ix} [/tex] from there it is easily shown that the complex part of the solution of [tex] y[/tex] is the solution of the original equation you proposed.

You could also try the variation of parameters method but I often avoid it because of messy integration
 
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  • #4
Dmak said:
instead of looking at that equation you could try to look at an equivalent equation
[tex]L[y] = y'' + y = xe^{ix}[/tex], where [tex]i = \sqrt{-1}[/tex]. then since [tex] i [/tex] is a particular root of the characteristic equation, y has a solution of the form
[tex] y = x^{2}A_{0}e^{ix} [/tex] from there it is easily shown that the complex part of the solution of [tex] y[/tex] is the solution of the original equation you proposed.

You could also try the variation of parameters method but I often avoid it because of messy integration

I really don't know what you are talking about, but neither the real nor the imaginary part of x^2*exp(ix) is a particular solution to y''+y=x*sin(x).
 
  • #5
Dick said:
I really don't know what you are talking about, but neither the real nor the imaginary part of x^2*exp(ix) is a particular solution to y''+y=x*sin(x).

whoops what i meant was a particular solution of the form:
[tex] y = x(A_{0} + A_{1}x)e^{xi} [/tex] you have to derive the coefficients by evaluating [tex] L[y][/tex] and matching up the coefficients [tex] A_{0}, A_{1} [/tex] with the right hand side of the equation ( [tex] xe^{ix}[/tex] ) then you use the imaginary portion of this solution, this one was more tedious than I thought and I would recommend using the variation of parameters method.
 
  • #6
thanks. that's really helpful. is there anyway of knowing what the particular integral is for any unhomogenous second order differential equation, or is it just trial and error?
 
  • #7
ive just tried substituting the particular integral
y(x)= c1(x^2)sin(x)+c2(x^2)cos(x)+xc3sin(x)+xc4cos(x) into y"+y=xsinx. I end up with
2c1sin(x)+4xc1cos(x)+2c2cos(x)-4xc2sin(x)+2c3cos(x)-2c4sin(x)=xsin(x)

This doesn't look solvable. I've checked my differentiating a couple of times and it looks okay. Any suggestions where I may be going wrong?
 
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  • #8
This would be a fine method: http://www.sosmath.com/diffeq/second/variation/variation.html

essentially find two homogeneous solutions [tex] y_{1}, y_{2} [/tex] then using those you can find a solution to the differentiall equation of the form,

[tex] \phi(x) = y_{1}u_{1} + y_{2}u_{2}[/tex]
where
[tex]u'_{1} = \frac{-sin(x)y_{2}}{W[y_{1}, y_{2}]}[/tex]
[tex]u'_{2} = \frac{sin(x)y_{1}}{W[y_{1}, y_{2}]} [/tex]
and
[tex]W[y_{1}, y_{2}] = y_{1}'y_{2} - y_{1}y_{2}' [/tex]
 
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  • #9
sweep123 said:
ive just tried substituting the particular integral
y(x)= c1(x^2)sin(x)+c2(x^2)cos(x)+xc3sin(x)+xc4cos(x) into y"+y=xsinx. I end up with
2c1sin(x)+4xc1cos(x)+2c2cos(x)-4xc2sin(x)+2c3cos(x)-2c4sin(x)=xsin(x)

This doesn't look solvable. I've checked my differentiating a couple of times and it looks okay. Any suggestions where I may be going wrong?

It is solvable. How about c1=c4=0 and c2=-1/4 and c3=1/4? You just have to try harder.
 
  • #10
oh yeah. i should have spent a bit more time on that. Thanks
 

Related to Solving second order unhomogonous equations

1. What is a second order unhomogeneous equation?

A second order unhomogeneous equation is a mathematical equation that involves a second degree polynomial and a non-zero constant term. It is a type of differential equation that contains both homogeneous and non-homogeneous terms.

2. How do you solve a second order unhomogeneous equation?

To solve a second order unhomogeneous equation, you can use the method of undetermined coefficients or variation of parameters. In the method of undetermined coefficients, you assume a particular solution based on the non-homogeneous term, and then solve for the coefficients. In variation of parameters, you use a set of fundamental solutions to find the general solution to the equation.

3. What is the difference between a homogeneous and unhomogeneous equation?

A homogeneous equation is one where all terms have the same degree, and the constant term is equal to zero. On the other hand, an unhomogeneous equation has terms with different degrees and a non-zero constant term. In other words, a homogeneous equation can be solved without any external conditions, while an unhomogeneous equation requires additional information to find a solution.

4. Can a second order unhomogeneous equation have multiple solutions?

Yes, a second order unhomogeneous equation can have multiple solutions depending on the initial conditions or external constraints given. The general solution to the equation will have a family of solutions, and the specific solution will depend on the given conditions.

5. What are some real-life applications of second order unhomogeneous equations?

Second order unhomogeneous equations have various applications in physics, engineering, and economics. For example, they can be used to model the motion of a spring-mass system, electric circuits, and population growth. They are also used in signal processing and control systems.

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