Solving a fifth order non-homogeneous differential equation

In summary: Generalizing that, here you should probably look for something whose fifth derivative is zero, ie. a fourth order polynomial, so you need only worry about... a fourth order polynomial?
  • #1
JMFL
3
0

Homework Statement


Find the general solution of [itex]y^{(5)}-y(1)=x[/itex]

The Attempt at a Solution


I found the complementary function by substitution of the solution form [itex]y=e^{kx}[/itex] giving [itex]k=0,1,-1,i,-i[/itex], so [itex]y_{cf}=a_0+a_1e^x+a_2e^{-x}+a_3e^{ix}+a_4e^{-ix}[/itex]

Now for the particular integral, the general trial solution form of a forcing term of x on the right is [itex]y=b_0+b_1x[/itex]

However if I plug this in, I get [itex]-b_1=x[/itex] !

I admit that I am not too familiar with dealing with differential equations of order greater than two. It seems that the general form of particular integrals when you have higher order ODEs has to be different? Or is the problem the fact that I already have a constant term in my complementary function sot here will not be a constant term in the particular integral too?
 
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  • #2
JMFL said:

Homework Statement


Find the general solution of [itex]y^(5)-y(1)=x[itex]
I assume you mean ##y^{(5)} - y' = x##
JMFL said:

The Attempt at a Solution


I found the complementary function by substitution of the solution form [itex]y=e^{kx}[itex] giving [itex]k=0,1,-1,i,-i[itex], so [itex]y_{cf}=a_0+a_1e^x+a_2e^{-x}+a_3e^{ix}+a_4e^{-ix}[itex]
For the last two, it's easier to use multiples of ##\sin(x)## and ##\cos(x)##.
JMFL said:
Now for the particular integral, the general trial solution form of a forcing term of [itex]x[itex] on the right is [itex]y=b_0+b_1x[itex]
Your particular solution won't work, as ##y_p' = b_1##. Instead, try ##y_p = b_0 + b_1x + b_2x^2##
JMFL said:
However if I plug this in, I get [itex]-b_1=x[itex]!

I admit that I am not too familiar with dealing with differential equations of order greater than two. It seems that the general form of particular integrals when you have higher order ODEs has to be different? Or is the problem the fact that I already have a constant term in my complementary function sot here will not be a constant term in the particular integral too?

BTW, you are not using the tex tags correctly. You have the opening tag right, but the closing tag is [/itex]. You are omitting the slash character. I fixed them in your original post, but didn't in what I copied from your post.
 
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  • #3
JMFL said:

Homework Statement


Find the general solution of [itex]y^{(5)}-y(1)=x[/itex]

The Attempt at a Solution


I found the complementary function by substitution of the solution form [itex]y=e^{kx}[/itex] giving [itex]k=0,1,-1,i,-i[/itex], so [itex]y_{cf}=a_0+a_1e^x+a_2e^{-x}+a_3e^{ix}+a_4e^{-ix}[/itex]

Now for the particular integral, the general trial solution form of a forcing term of x on the right is [itex]y=b_0+b_1x[/itex]

However if I plug this in, I get [itex]-b_1=x[/itex] !

I admit that I am not too familiar with dealing with differential equations of order greater than two. It seems that the general form of particular integrals when you have higher order ODEs has to be different? Or is the problem the fact that I already have a constant term in my complementary function sot here will not be a constant term in the particular integral too?

You actually have a 4th order differential equation in the variable ##z(x) = y'(x) = dy(x)/dx##. Then ##y(x) = C+\int_0^x z(t) \, dt.##
 
  • #4
JMFL said:
However if I plug this in, I get ##−b_1=x ##!
So try adding a b2x2 term.
 
  • #5
haruspex said:
So try adding a b2x2 term.

Hi haruspex,

I was considering doing this, but I wanted to know why this form of the particular integral is not working. I did not wish to be blindly guessing particular integral forms or adding more x's without knowing why I was doing what I was doing!

Why does the particular integral form fail? Are the PI forms that I am used to only for second order ODEs?
 
  • #6
JMFL said:
Hi haruspex,

I was considering doing this, but I wanted to know why this form of the particular integral is not working. I did not wish to be blindly guessing particular integral forms or adding more x's without knowing why I was doing what I was doing!

Why does the particular integral form fail? Are the PI forms that I am used to only for second order ODEs?
I don't know where your rule of just using a linear term comes from. I've always regarded finding PIs as a combination of experience and imagination. If you want a general rule it would be more like plugging in an entire power series with unknown coefficients and seeing if anything useful results. Either a simple equation or a recurrence relation.
 
  • #7
JMFL said:
Hi haruspex,

I was considering doing this, but I wanted to know why this form of the particular integral is not working. I did not wish to be blindly guessing particular integral forms or adding more x's without knowing why I was doing what I was doing!

Why does the particular integral form fail? Are the PI forms that I am used to only for second order ODEs?

The suggestion of [itex]y_p(x) = ax + b[/itex] in the case of [itex]y'' + py' + qy = x[/itex] has the property that [itex]y''_p = 0[/itex], so you need only worry about satisfying [itex]py'_p + qy_p = x[/itex].

Generalizing that, here you should probably look for something whose fifth derivative is zero, ie. a fourth order polynomial, so you need only worry about satisfying [itex]-y_p' = x[/itex].
 
  • #8
haruspex said:
I don't know where your rule of just using a linear term comes from. I've always regarded finding PIs as a combination of experience and imagination. If you want a general rule it would be more like plugging in an entire power series with unknown coefficients and seeing if anything useful results. Either a simple equation or a recurrence relation.
The technique of annihilators provides a more direct way of dealing with nonhomogeneous linear differential equations. I wrote a two part Insights series on using the annihilator method. You can find them by searching amongst the Insights articles, under Tutorials.

There is also this short article: http://jekyll.math.byuh.edu/courses/m334/handouts/annihilator.pdf
 
  • #9
JMFL said:
Hi haruspex,

I was considering doing this, but I wanted to know why this form of the particular integral is not working. I did not wish to be blindly guessing particular integral forms or adding more x's without knowing why I was doing what I was doing!

Why does the particular integral form fail? Are the PI forms that I am used to only for second order ODEs?

If you had paid attention to post #3 you would have seen the solution immediately: the equation for ##z = y'## is ##z^{(4)} - z = x##. Isn't ##z = -x## an obvious solution? Now integrate to get ##y##.
 

Related to Solving a fifth order non-homogeneous differential equation

1. What is a fifth order non-homogeneous differential equation?

A fifth order non-homogeneous differential equation is a mathematical equation that involves a function and its derivatives up to the fifth order, where the function is not equal to zero and the equation is not in the standard form of a homogeneous equation.

2. How do you solve a fifth order non-homogeneous differential equation?

To solve a fifth order non-homogeneous differential equation, you need to follow a specific process called the method of undetermined coefficients. This involves finding a particular solution and adding it to the complementary function, which is the solution to the associated homogeneous equation.

3. What are the steps for solving a fifth order non-homogeneous differential equation?

The steps for solving a fifth order non-homogeneous differential equation are:
1. Rewrite the equation in standard form
2. Find the complementary function by solving the associated homogeneous equation
3. Guess a particular solution based on the non-homogeneous part of the equation
4. Substitute the particular solution into the original equation and solve for the coefficients
5. Add the particular solution to the complementary function to get the general solution

4. What are the challenges of solving a fifth order non-homogeneous differential equation?

One of the main challenges of solving a fifth order non-homogeneous differential equation is finding the particular solution. This requires some guesswork and may not always be obvious. Additionally, the process can be time-consuming and tedious, especially for more complex equations.

5. Can technology be used to solve a fifth order non-homogeneous differential equation?

Yes, technology such as computer software or online solvers can be used to solve a fifth order non-homogeneous differential equation. These tools use algorithms and numerical methods to find the solution, which can save time and reduce the chances of human error. However, it is still important to understand the steps and concepts involved in solving these equations.

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