Cauchy-Euler with x=e^t? Differential Equations (ODE)

[kepherax]

Homework Statement

Use Cauchy-Euler method with x=e^t to solve the following differential equation.

Relevant Equations

x^2y''(x)+xy'(x)-9y(x)=x^2

I'm fine with this up to a certain point, but I'm not certain if I'm using the substitution correctly. After finding the homogeneous solution do I plug in x= e^t in the original equation and then divide by e^2t to put it in standard form before applying variation of parameters so f=1, or do I just substitute so f= e^2t? The former is shown, but the latter gives a matching answer to plugging the original equation into symbolab if I were to replace their x with e^t. In videos I've watched and as per the book, you must always put the original equation into standard form and use the resulting f, but I feel like I'm missing something I can't put my finger on.

 

[Mark44]

Homework Statement: Use Cauchy-Euler method with x=e^t to solve the following differential equation.
Homework Equations: x^2y''(x)+xy'(x)-9y(x)=x^2

I'm fine with this up to a certain point, but I'm not certain if I'm using the substitution correctly. After finding the homogeneous solution do I plug in x= e^t in the original equation and then divide by e^2t to put it in standard form before applying variation of parameters so f=1, or do I just substitute so f= e^2t? The former is shown, but the latter gives a matching answer to plugging the original equation into symbolab if I were to replace their x with e^t. In videos I've watched and as per the book, you must always put the original equation into standard form and use the resulting f, but I feel like I'm missing something I can't put my finger on.

View attachment 251084

You seem to have an error in your particular solution. If you substitute just your particular solution into the differential equation, you get $x^2 \cdot 0 + x \cdot 0 - 9 \cdot -\frac 1 9 \ne x^2$. Since the two exponentials make up the solution to the homogeneous problem, all of the terms resulting from that part of the solution just vanish.

I got a particular solution of the nonhomogeneous problem of $y_p = -\frac 1 5 x^2$, and this checks out.

Your solution to the homogeneous problem is fine, but for the nonhomogeneous problem, I used a different technique, assuming that the solution to $x^2y'' + xy' - 9y = x^2$ was a polynomial of the form $y_p = c_0 + c_1x + c_2x^2$.

I'm pretty rusty on the variation of parameters technique, so I don't have any comments on what you did in the lower half of the page in the photo.

 

[kepherax]

You seem to have an error in your particular solution. If you substitute just your particular solution into the differential equation, you get $x^2 \cdot 0 + x \cdot 0 - 9 \cdot -\frac 1 9 \ne x^2$. Since the two exponentials make up the solution to the homogeneous problem, all of the terms resulting from that part of the solution just vanish.

I got a particular solution of the nonhomogeneous problem of $y_p = -\frac 1 5 x^2$, and this checks out.

Your solution to the homogeneous problem is fine, but for the nonhomogeneous problem, I used a different technique, assuming that the solution to $x^2y'' + xy' - 9y = x^2$ was a polynomial of the form $y_p = c_0 + c_1x + c_2x^2$.

I'm pretty rusty on the variation of parameters technique, so I don't have any comments on what you did in the lower half of the page in the photo.

Yeah, even when I use the variation of parameters method with f = e^2t, I get -1/5 e^2t as the particular solution, which works with the substitution... part of the process with Cauchy Euler is to first solve the homogeneous equation and then divide through to put in standard form for parameter variation, though, which either way (with or without sub) will result in f=1 and as pictured, so I'm confused. I may ask a tutor tomorrow morning, I saw one today but he was also rusty on ODE and couldn't clarify this for me. Thanks for trying!

 

[Mark44]

I get -1/5 e^2t as the particular solution, which works with the substitution.

And with x = e^t, your solution is the same as mine; namely, $y_p = -\frac 1 5 x^2$.