First order differentials: separating variables

  • Thread starter raincheck
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  • #1
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?
 

Answers and Replies

  • #2
courtrigrad
1,236
1
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:
  • #3
slearch
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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 :)
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
43,021
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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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