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Separable Differential Equation dy/dx

  • Thread starter aznkid310
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1. Homework Statement

dy/dx =[cos^2(x)][cos^2(y)]


2. Homework Equations

The solution to this problem is y = +/- [(2n + 1)*pi]/4

How? Do i just plug C back into the equation? That seems a little messy

3. The Attempt at a Solution

dy/cos^2(y) = cos^2(x) dx

After integrating: (1/2)tan(2y) = (1/2)(x + cos(x) + sin(x) + C)

C = 2tan(2y) - 2x -sin(2x) for cos(2y) not = 0
 

Answers and Replies

You have not supplied enough information to make your solution unique. What point must the curve pass through?
 
tiny-tim
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dy/cos^2(y) = cos^2(x) dx

After integrating: (1/2)tan(2y) = (1/2)(x + cos(x) + sin(x) + C)

C = 2tan(2y) - 2x -sin(2x) for cos(2y) not = 0
Hi aznkid310! :smile:

No … you're getting your ^2 and your 2 mixed up … it's just tany on the left.

And on the right … you've gone all weird! :rolleyes:

Use cos²x = 1/2(1 + cos(2x)).
 
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It just saved solve the differential equation. But the solution included all of that.

As for the integration, i checked w/ an integral calculator and the left side is correct. For the right side, its actually cos^2(2x), my mistake.
 
D H
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The solution to this problem is y = +/- [(2n + 1)*pi]/4
Firstly, its [itex]y=\frac{2n+1}2\pi[/itex]. You don't need the [itex]\pm[/itex] and the denominator is 2, not 4.

Secondly, this is not "the" solution. There are infinitely many solutions you can find via integrating the differential equation. Hint: You will not get these particular solutions by integrating the differential equation. Big hint: what is the derivative of [itex]y=c[/itex] with respect to [itex]x[/itex]?

Thirdly, your integration ran afoul somewhere.
 
Last edited:

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