Solving a Separable ODE: y'+ytanx=cosx with Initial Condition y(0)=1

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SUMMARY

The discussion focuses on solving the separable ordinary differential equation (ODE) given by y' + y tan(x) = cos(x) with the initial condition y(0) = 1. The user correctly identifies the integrating factor as μ = 1/cos(x) and simplifies the equation accordingly. The conversation confirms that the user is on the right track and emphasizes the importance of simplifying the integrand during the solution process. The collaborative effort highlights the significance of understanding integrating factors in solving separable ODEs.

PREREQUISITES
  • Understanding of separable ordinary differential equations (ODEs)
  • Knowledge of integrating factors in differential equations
  • Familiarity with trigonometric identities, specifically involving cos(x)
  • Basic calculus skills, including integration techniques
NEXT STEPS
  • Study the method of integrating factors in depth
  • Practice solving additional separable ODEs
  • Explore trigonometric identities and their applications in differential equations
  • Learn about the existence and uniqueness theorem for ODEs
USEFUL FOR

Students studying differential equations, mathematics educators, and anyone looking to enhance their problem-solving skills in the context of separable ODEs.

eeriana
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Homework Statement


y'+ytanx = cos x y(0)=1


Homework Equations





The Attempt at a Solution



We are studying separable ode's and integrating factor right now, I am a little confused... If someone could steer me in the right direction, it would be greatly appreciated... This is what I have so far:

P= tanx
\intP = -ln|cosx|
\mu=e^{}-lncosx
\mu= 1/cosx

(1/cosx*y)' =\int1/cosx cosx

and this is where I get stuck... am I even on the right track?

Thanks

Eeriana
 
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You are on the right track and have calculated the IF correctly and written the complete differential correctly. Note that you can simplify the integrand...
 
But isn't 1/cosx * cosx = 1 Or am I having an algebraic malfunction?
 
eeriana said:
But isn't 1/cosx * cosx = 1 Or am I having an algebraic malfunction?
Nope you are indeed correct.
 
I thought I was doing something wrong... hmmm..now I am going to see if I can finish it!

Thanks for the help
 

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