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

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In summary, The conversation is about solving a linear equation and using an integrating factor. The solution is y = x^2 / cos(x) + C. The conversation also mentions the difference between the coverage of topics in single variable calculus at UCLA compared to other schools.
  • #1
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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  • #2
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.
 
  • #3
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:
 
  • #4
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:
 
  • #5
Right you are! The devil's in the details! :devil:

point noted
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 

What is the purpose of solving linear equations?

The purpose of solving linear equations is to find the value or values of the variable that make the equation true. This helps us to solve real-world problems and make predictions based on mathematical relationships.

What is a linear equation?

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. The variables in a linear equation are raised to the first power and the equation can be graphed as a straight line.

What is the difference between a linear equation and a non-linear equation?

The difference between a linear equation and a non-linear equation is that the terms in a linear equation are raised to the first power, while the terms in a non-linear equation can be raised to any power. This means that the graph of a linear equation is a straight line, while the graph of a non-linear equation can be curved.

What is the process for solving a linear equation?

The process for solving a linear equation involves isolating the variable on one side of the equation by using inverse operations. This means that whatever operation is being done to the variable, it must be done to both sides of the equation. Once the variable is isolated, its value can be found by substituting in the other values and simplifying the equation.

How can we solve the given linear equation: y' - ytanx = 2x/cosx?

To solve the given linear equation, we first need to isolate the variable, y. This can be done by adding ytanx to both sides of the equation. Then, we can divide both sides by cosx to get y by itself. Finally, we can substitute in the given values for x and solve for y. The resulting value of y will be the solution to the equation.

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