Solving Linear Equations: y' - ytanx = 2x/cosx

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Discussion Overview

The discussion revolves around solving a linear differential equation of the form y' - ytanx = 2x/cosx. Participants explore the process of finding an integrating factor and discuss the steps involved in solving the equation, including potential mistakes and clarifications.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in solving the equation and seeks assistance in identifying where they went wrong.
  • Another participant suggests an alternative approach by rearranging the equation instead of dividing by cos(x), indicating that the left-hand side can be expressed as a derivative.
  • A participant claims to have found a solution, y = x^2 / cos(x) + C, but expresses uncertainty about its correctness.
  • There is a correction regarding the placement of the constant "+C" in the solution, with a suggestion that it should be over cos(x).
  • Participants engage in a light-hearted exchange about the details of the solution and the learning process.
  • Discussion shifts to the educational context, with participants sharing their experiences regarding course structures and content coverage in calculus and differential equations.

Areas of Agreement / Disagreement

Participants generally agree on the steps to solve the equation, but there are minor disagreements regarding the placement of the constant in the solution and the educational content covered in their respective courses.

Contextual Notes

Participants mention variations in course content and structure across different institutions, highlighting differences in the timing and depth of topics such as polar coordinates and differential equations.

Who May Find This Useful

Students studying differential equations, calculus, or those interested in the pedagogical approaches to teaching these subjects may find this discussion relevant.

Math Is Hard
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Hi, I am having a little trouble solving a linear equation:

[tex]y' cosx = ysinx + 2x[/tex]

I translated it into standard y' + P(x)y= Q(x) format as

[tex]y' - ytanx = 2x/cosx[/tex]

then I needed an integrating factor, so I used

[tex]I(x)=e^{-\int tanx\:dx} = cosx[/tex]

when I multiply this to both sides of [tex]y' - ytanx = 2x/cosx[/tex]
it doesn't seem to do much good. I just get back to where I started, and I am not seeing anything that will wrap into a nice neat little derivative on the LHS.

Can you show me where I am going wrong? Thanks!
 
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Go back up to your first line. Instead of dividing by cos(x), bring the ysin(x) term to the LHS. Then you have:

y'cos(x)-ysin(x)=2x.

The LHS is identical to d(ycos(x))/dx.
 
oh, wow! I knew it wasn't supposed to be that hard!

so my solution is

[tex]y = x^2 / cos(x) + C[/tex]

I think that's right..

Thanks, Tom!
:smile: :smile: :smile: :smile: :smile:
 
Math Is Hard said:
[tex]y = x^2 / cos(x) + C[/tex]

The "+C" should be over the cos(x), with the x2.

Thanks, Tom!
:smile: :smile: :smile: :smile: :smile:

No problemo. :approve:
 
Right you are! The devil's in the details! :devil:

point noted
 
By the way, what course is this for? It looks like Differential Equations, but I could have sworn that not too long ago you were asking questions about subject matter from Calculus II.
 
I am still wrapping up the second part of single variable calculus - I am done in two weeks - hooray!
UCLA is on a quarter system, so for engineering series there are two quarters of single variable, two quarters of multi-variable, and then a quarter of linear algebra and a quarter of ordinary diff. equations. (I think that's how it goes, anyway)
Toward the end of this class they give us a little taste of differential equations, Taylors, and some other things we might come across if we continue on in math. What's kinda weird though is that polar coordinates aren't covered until multivariable calculus at UCLA, and I had heard that it was standard to cover those in single variable calc.
 
Math Is Hard said:
Toward the end of this class they give us a little taste of differential equations, Taylors, and some other things we might come across if we continue on in math.

A-ha. Where I work, we teach a little taste of Diff Eq in Calculus I (but not even as heavy as what you have presented here), and Taylor series are done in great detail in Calculus II.

What's kinda weird though is that polar coordinates aren't covered until multivariable calculus at UCLA, and I had heard that it was standard to cover those in single variable calc.

Yes, we cover polar coordinates extensively in Calculus II (single variable), and we do it again in Calculus III (multivariable). But you can do polar coordinates from scratch in a multivariable setting.
 

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