Particular Solution of 2nd Order ODE: x^2y"-4xy'+6y=ln(x)

In summary, the textbook has examples of the right side function and the corresponding form of the particular solution, but it is faster to use undetermined coefficients if the equation has a non-constant coefficient in front of the y'' term. Euler's method can be used to solve cauchy-euler equations, but it is better to use variation of parameters if the equation has a radical on the right hand side.
  • #1
november1992
120
0
I was wondering what a guess would be for the particular solution of the right hand side of an equation if it looked like this:

[itex]x^{2}[/itex]y" - 4xy' + 6y = ln(x)

My textbook has some specific examples of the right side function along with the corresponding form of the particular solution.

http://i.imgur.com/BT8AF.pngAlso, when is it best to use variation of parameters and the undetermined coefficients method to solve the equation?
 
Physics news on Phys.org
  • #2
You should really only use undetermined coefficients if the function you're solving for is either sin, cos, e, or a polynomial.

Variation of parameters could also be used, but generally undetermined coefficients is faster in those cases.

The thing is, for that equation, notice that you have a non-constant coefficient in front of your y'' term and perhaps using series would be a lot better for this equation...
 
  • #3
rewrite this as
x^4 (y/x^2)''=log(x)
 
  • #4
If you let x = et then the differential can be transformed into

z''-5z'+6z = t

It's called Euler's method I believe. Also notice the coefficient in front of the first derivative term has been reduced by 1.
 
  • #5
Zondrina said:
You should really only use undetermined coefficients if the function you're solving for is either sin, cos, e, or a polynomial.

Variation of parameters could also be used, but generally undetermined coefficients is faster in those cases.

The thing is, for that equation, notice that you have a non-constant coefficient in front of your y'' term and perhaps using series would be a lot better for this equation...

That's what I assumed, but my teacher believes it's still faster to use variation of parameters. I don't like using that method because I tend to make more mistakes when I integrate e and a trig function.

Mentallic said:
If you let x = et then the differential can be transformed into

z''-5z'+6z = t

It's called Euler's method I believe. Also notice the coefficient in front of the first derivative term has been reduced by 1.

Would this method still be plausible if the right hand side was like this:

ln(x)*cos(3x)-[itex]3x^{2}[/itex]

I was also wondering what to do in if I had a radical on the right hand side like this:

[itex]\sqrt{1-x^{2}}[/itex][itex]*e^{x}[/itex]*sin(x)
 
  • #6
november1992 said:
Would this method still be plausible if the right hand side was like this:

ln(x)*cos(3x)-[itex]3x^{2}[/itex]

I was also wondering what to do in if I had a radical on the right hand side like this:

[itex]\sqrt{1-x^{2}}[/itex][itex]*e^{x}[/itex]*sin(x)

Yes the method still works, but finding your particular solution would be hard or it might not even have an exact algebraic solution.

But then again, you'd run into the same problem simply solving a second order linear ODE with those functions on the RHS.
 
  • #7
I have another question. Can the undetermined coefficients method be used to solve cauchy-euler equations, or does variation of parameters have to be used?
 
  • #8
^yes you can use undetermined coefficients
just work with x^a polynomials and trig functions of log(x) like
log(x)^2 sin(log(x) and x^17
cauchy-euler equations and linear constant coefficients are the same thing in different forms
try to solve
x2y" - 4xy' + 6y =(log(x))2sin(log(x))

with and without changing variables and see which you prefer, I think most people prefer to change it to linear constant coefficient form.
 
  • #9
Oh, I actually meant could it be solved without changing the equation, but I guess that's a no.
 
  • #10
It can be solved without changing, you just might not want to. I suggested that you do one both ways and compare. The two ways are exactly equivalent.
 

FAQ: Particular Solution of 2nd Order ODE: x^2y"-4xy'+6y=ln(x)

What is the general form of the particular solution for this 2nd order ODE?

The general form of the particular solution for this 2nd order ODE is y = C1x^2 + C2x + ln(x), where C1 and C2 are constants determined by initial conditions.

How do you find the particular solution for this 2nd order ODE?

To find the particular solution, you first need to find the complementary solution by solving the homogeneous equation (setting the right side of the equation to 0). Then, you use the method of undetermined coefficients to find a particular solution that satisfies the non-homogeneous equation (keeping the right side of the equation as it is).

Can you give an example of finding the particular solution for this 2nd order ODE?

Sure, for example, if we have the ODE x^2y"-4xy'+6y=ln(x) with initial conditions y(1)=2 and y'(1)=3, we first solve the homogeneous equation x^2y"-4xy'+6y=0, which gives us the complementary solution y(x) = C1x^2 + C2x. Then, we use the method of undetermined coefficients to find a particular solution, which in this case would be y(x) = (1/3)ln(x) + (2/3)x^2 - (1/3)x. Finally, we plug in the initial conditions to find the values of C1 and C2, giving us the particular solution y(x) = (1/3)ln(x) + (2/3)x^2 - (1/3)x + (7/3).

Is there a shortcut method for finding the particular solution for this 2nd order ODE?

Yes, there is a shortcut method called the method of variation of parameters. This method involves finding a particular solution in the form of y(x) = u1(x)y1(x) + u2(x)y2(x), where y1(x) and y2(x) are the two linearly independent solutions of the homogeneous equation and u1(x) and u2(x) are functions to be determined. This method is usually faster and more efficient than the method of undetermined coefficients.

What are some real-life applications of solving 2nd order ODEs using the particular solution method?

The particular solution method for solving 2nd order ODEs has many applications in physics, engineering, and other sciences. Some examples include analyzing the motion of a simple harmonic oscillator, predicting the behavior of an electrical circuit, or modeling the growth of a population over time. In general, this method can be used to solve any problem that involves a system with a known governing equation and initial conditions.

Back
Top