2nd order differential eguation

[MostlyHarmless]

Homework Statement

Here is the problem, verbatim.

Observe that y=x is a particular solution of the equation 2x^2y&#039;&#039;+xy&#039;-y=0[\tex]<br /> and find the general solution. For what values of x is the solution valid?<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> I know the answer is $y=c_1x+c_2x^{{\frac{-1}{2}}}$, but I cannot, for the life of me, figure out how on Earth I&#039;m even supposed to go about finding that solution. It should be noted that this is in the section for The Reduction of Order Method, so I assume it has to with that, however, everything we&#039;ve done thus far, since the introduction of differential operators, has only concerned Diff. Eq. w/ constant coefficients. So that&#039;s where I&#039;m hitting my main roadblock.<br /> <br /> I&#039;ve tried several things, none of which got me anywhere. For example: I tried re-writing the original expression as (2x^2D^2+xD-1)y=0 and letting $y=c_1x+c_2y_1$ I quickly realized that just takes me back to my original expression. <br /> <br /> I then tried letting y=vx, as per reduction of order method suggests. That would give $y&#039;=v&#039;x+v$ and $y&#039;&#039;=v&#039;&#039;x+2v&#039;$<br /> plugging that into the original equation: <br /> <br /> $2x^3v&#039;&#039;+4x^2v&#039;+x^2v&#039;+xv-xv=0$<br /> <br /> The xv&#039;s will subtract out. Leaving <br /> <br /> $2x^3v&#039;&#039;+5x^2v&#039;=0$ <br /> <br /> dividing by $2x^3$ gives <br /> v&amp;#039;&amp;#039;+{\frac{5}{2}}x^{-1}v&amp;#039;=0 <br /> Then I made the substitution w=v&#039;; w&#039;=v&#039;&#039; to make it first order linear in w. <br /> w&amp;#039;+{\frac{5}{2}}x^{-1}w=0 <br /> giving an integrating factor of <br /> <br /> $e^{∫{\frac{5}{2}}x^{-1}}=x^{\frac{5}{2}}$ Multiplying through I get d[wx^{\frac{5}{2}}]=0 Then that&#039;s where I get stuck. Integrating now will eventually give v=0, which is unhelpful.

 

[Dick]

Homework Statement

Here is the problem, verbatim.

Observe that y=x is a particular solution of the equation 2x^2y&#039;&#039;+xy&#039;-y=0[\tex]<br /> and find the general solution. For what values of x is the solution valid?<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> I know the answer is $y=c_1x+c_2x^{{\frac{-1}{2}}}$, but I cannot, for the life of me, figure out how on Earth I&#039;m even supposed to go about finding that solution. It should be noted that this is in the section for The Reduction of Order Method, so I assume it has to with that, however, everything we&#039;ve done thus far, since the introduction of differential operators, has only concerned Diff. Eq. w/ constant coefficients. So that&#039;s where I&#039;m hitting my main roadblock.<br /> <br /> I&#039;ve tried several things, none of which got me anywhere. For example: I tried re-writing the original expression as (2x^2D^2+xD-1)y=0 and letting $y=c_1x+c_2y_1$ I quickly realized that just takes me back to my original expression. <br /> <br /> I then tried letting y=vx, as per reduction of order method suggests. That would give $y&#039;=v&#039;x+v$ and $y&#039;&#039;=v&#039;&#039;x+2v&#039;$<br /> plugging that into the original equation: <br /> <br /> $2x^3v&#039;&#039;+4x^2v&#039;+x^2v&#039;+xv-xv=0$<br /> <br /> The xv&#039;s will subtract out. Leaving <br /> <br /> $2x^3v&#039;&#039;+5x^2v&#039;=0$ <br /> <br /> dividing by $2x^3$ gives <br /> v&amp;#039;&amp;#039;+{\frac{5}{2}}x^{-1}v&amp;#039;=0 <br /> Then I made the substitution w=v&#039;; w&#039;=v&#039;&#039; to make it first order linear in w. <br /> w&amp;#039;+{\frac{5}{2}}x^{-1}w=0 <br /> giving an integrating factor of <br /> <br /> $e^{∫{\frac{5}{2}}x^{-1}}=x^{\frac{5}{2}}$ Multiplying through I get d[wx^{\frac{5}{2}}]=0 Then that&#039;s where I get stuck. Integrating now will eventually give v=0, which is unhelpful.

<br /> <br /> That&#039;s a Euler equation. You can tell by the way the powers of x match up with the order of the differentiation. The appropriate substitution to try is $y=x^n$. Try to solve for n.

 

[LCKurtz]

Homework Statement

Here is the problem, verbatim.

Observe that y=x is a particular solution of the equation 2x^2y&#039;&#039;+xy&#039;-y=0[\tex]<br /> and find the general solution. For what values of x is the solution valid?<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> I know the answer is $y=c_1x+c_2x^{{\frac{-1}{2}}}$, but I cannot, for the life of me, figure out how on Earth I&#039;m even supposed to go about finding that solution. It should be noted that this is in the section for The Reduction of Order Method, so I assume it has to with that, however, everything we&#039;ve done thus far, since the introduction of differential operators, has only concerned Diff. Eq. w/ constant coefficients. So that&#039;s where I&#039;m hitting my main roadblock.<br /> <br /> I&#039;ve tried several things, none of which got me anywhere. For example: I tried re-writing the original expression as (2x^2D^2+xD-1)y=0 and letting $y=c_1x+c_2y_1$ I quickly realized that just takes me back to my original expression. <br /> <br /> I then tried letting y=vx, as per reduction of order method suggests. That would give $y&#039;=v&#039;x+v$ and $y&#039;&#039;=v&#039;&#039;x+2v&#039;$<br /> plugging that into the original equation: <br /> <br /> $2x^3v&#039;&#039;+4x^2v&#039;+x^2v&#039;+xv-xv=0$<br /> <br /> The xv&#039;s will subtract out. Leaving <br /> <br /> $2x^3v&#039;&#039;+5x^2v&#039;=0$ <br /> <br /> dividing by $2x^3$ gives <br /> v&amp;#039;&amp;#039;+{\frac{5}{2}}x^{-1}v&amp;#039;=0 <br /> Then I made the substitution w=v&#039;; w&#039;=v&#039;&#039; to make it first order linear in w. <br /> w&amp;#039;+{\frac{5}{2}}x^{-1}w=0 <br /> giving an integrating factor of <br /> <br /> $e^{∫{\frac{5}{2}}x^{-1}}=x^{\frac{5}{2}}$ Multiplying through I get d[wx^{\frac{5}{2}}]=0 Then that&#039;s where I get stuck. Integrating now will eventually give v=0, which is unhelpful.

<br /> <br /> Don&#039;t give up, you are almost there. $(wx^{\frac 5 2})&#039; = 0$ gives $wx^{\frac 5 2}=C$ or $w = Cx^{-\frac 5 2}$. So what is $v$? Then what do you get for your second solution $xv$? And you don&#039;t need to worry about the multiplying constant since any constant times a solution is a solution.

 

[LCKurtz]

That's a Euler equation. You can tell by the way the powers of x match up with the order of the differentiation. The appropriate substitution to try is $y=x^n$. Try to solve for n.

Don't you think that misses the point of the question? They are learning reduction of order.

 

[Dick]

Don't you think that misses the point of the question? They are learning reduction of order.

Yes, I did gloss over that point. I just picked the easiest way to a solution. Apologies.

 

[MostlyHarmless]

So just do the same method I'm doing, but rather than y=vx use $y=x^n$? Or is there a specific way of solving a Euler equation? I've not studied them before, but I can find out how to do them fairly quickly if you wouldn't mind giving me one more nudge.

 

[MostlyHarmless]

Oh ok, I can't believe I forgot my constant!

 

[MostlyHarmless]

Alright, I got: $∫v'=∫c_1x^{\frac{-5}{2}}dx$ Which gives $v=c_1x^{\frac{-3}{2}}+c_2$ multiplying through by x to get back to y gives: $y=c_1x^{\frac{-1}{2}}+c_2x$ which was the correct answer.

Thank you guys. I'd been struggling with that for awhile, I was just forgetting the constant!

 

[Dick]

Alright, I got: $∫v'=∫c_1x^{\frac{-5}{2}}dx$ Which gives $v=c_1x^{\frac{-3}{2}}+c_2$ multiplying through by x to get back to y gives: $y=c_1x^{\frac{-1}{2}}+c_2x$ which was the correct answer.

Thank you guys. I'd been struggling with that for awhile, I was just forgetting the constant!

And recognizing it as a Euler equation does give you a huge shortcut which doesn't use reduction of order. Put y=x^n cancel the x^n's and just solve for n. Try it. It's a pretty good trick to know.