First order differentials: separating variables

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Homework Help Overview

The discussion revolves around finding the general solution to a first-order differential equation by separating variables, specifically the equation 3tan(y) - dy/dx(sec(x)) = 0. Participants are exploring the steps involved in separating the variables and integrating both sides.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial setup of the equation and the process of separating variables. There are attempts to clarify integration steps and the correct handling of variables during the process. Questions arise regarding potential mistakes in the setup and the interpretation of the results.

Discussion Status

The discussion is ongoing, with participants providing corrections and guidance on the integration process. Some participants suggest that the original poster's steps were mostly correct but point out specific errors in variable handling and the need for a constant of integration. There is no explicit consensus on the final form of the solution, but there is a recognition of the general solution's nature.

Contextual Notes

Participants note the importance of correctly identifying variables and constants during integration. There is also a mention of the limitations in solving for y explicitly in the context of the differential equation.

raincheck
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"Find the general solution to the differential equation by separating variables:
3tany - dy/dx(secx) = 0"

This is what I set up:

3tany dx = secx dy
1/secx dx = 1/3tany dy
cosx dx = 1/3tany dy
[int] cosx dx = [int] 1/3tany dy
sinx = (1/3)ln|sinx|

I'm stuck as to what to do next, how to solve the equation.. did I mess up in the beginning or what am I doing wrong?
 
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you have to separate the same variables. So:

[tex]3\tan y \; dx = \sec x \; dy[/tex]

Multiply by [tex]\frac{1}{dy\cdot dx}[/tex]. Then you get:

[tex]\frac{dx}{\sec x} = \frac{dy}{3\tan y}[/tex]

[tex]\int \cos x \; dx = \frac{1}{3}\int \cot y + C[/tex]

[tex]\sin x = \frac{1}{3}\ln|\sin y|[/tex]

[tex]3\sin x = \ln|\sin y|[/tex]

[tex]\sin y = e^{3\sin x}[/tex]

[tex]y = \arcsin(e^{3\sin x}) + C[/tex]
 
Last edited:
raincheck said:
[int] cosx dx = [int] 1/3tany dy
sinx = (1/3)ln|sinx|

It looks like the y's became x's between these two steps :)
 
raincheck said:
"Find the general solution to the differential equation by separating variables:
3tany - dy/dx(secx) = 0"

This is what I set up:

3tany dx = secx dy
1/secx dx = 1/3tany dy
cosx dx = 1/3tany dy
[int] cosx dx = [int] 1/3tany dy
Every thing exactly right up to here.

sinx = (1/3)ln|sinx|

No, sin x= (1/3) ln |cos(y)|+ C. You accidently wrote x instead of y and forgot the constant of integeration.

[/quote]I'm stuck as to what to do next, how to solve the equation.. did I mess up in the beginning or what am I doing wrong?[/QUOTE]
What exactly do you mean by "solve the equation". In general you cannot solve for y. What you have, with the corrections, is the general solution.
 

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