Solving a fifth order non-homogeneous differential equation

[JMFL]

Homework Statement

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

The Attempt at a Solution

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

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

However if I plug this in, I get -b_1=x !

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?

 

[Mark44]

Homework Statement

Find the general solution of y^(5)-y(1)=x<br />

I assume you mean $y^{(5)} - y' = x$<br /> <blockquote data-attributes="" data-quote="JMFL" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> JMFL said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>The Attempt at a Solution</h2><br /> I found the complementary function by substitution of the solution form y=e^{kx}giving k=0,1,-1,i,-i, so y_{cf}=a_0+a_1e^x+a_2e^{-x}+a_3e^{ix}+a_4e^{-ix}<br /> <br /> </div> </div> </blockquote>For the last two, it's easier to use multiples of $\sin(x)$ and $\cos(x)$.<br /> <blockquote data-attributes="" data-quote="JMFL" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> JMFL said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Now for the particular integral, the general trial solution form of a forcing term of xon the right is y=b_0+b_1x<br /> </div> </div> </blockquote><br /> Your particular solution won't work, as $y_p' = b_1$. Instead, try $y_p = b_0 + b_1x + b_2x^2$<br /> <blockquote data-attributes="" data-quote="JMFL" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> JMFL said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> However if I plug this in, I get -b_1=x!<br /> <br /> 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? </div> </div> </blockquote><br /> <br /> BTW, you are not using the tex tags correctly. You have the opening tag right, but the closing tag is. You are omitting the slash character. I fixed them in your original post, but didn't in what I copied from your post.

 

[Ray Vickson]

Homework Statement

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

The Attempt at a Solution

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

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

However if I plug this in, I get -b_1=x !

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 $y(x) = C+\int_0^x z(t) \, dt.$

 

[haruspex]

However if I plug this in, I get $−b_1=x$!

So try adding a b₂x² term.

 

[JMFL]

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?

 

[haruspex]

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.

 

[pasmith]

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

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 -y_p' = x.

 

[Mark44]

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

 

[Ray Vickson]

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 $y$.