Separable Differential Equation

  • Thread starter hpayandah
  • Start date
  • #1
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Homework Statement


Can someone please verify if I am solving this equation right.


Homework Equations


Please refer to attachment.


The Attempt at a Solution


Please refer to attachment.
 

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Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
how about checking by differentiating
[tex]sin(y) = \frac{A}{\sqrt{x^2+1}} [/tex]

[tex]\frac{d}{dx}sin(y) = \frac{d}{dx}\frac{A}{\sqrt{x^2+1}} [/tex]

[tex]cos(y)\frac{dy}{dx} = 2A(x^2+1)^{-3/2}\frac{-1}{2}2x [/tex]

[tex]cos(y) dy \frac{A}{\sqrt{x^2+1}} = -A(x^2+1)^{-3/2}2x.dx.sin(y) [/tex]

[tex]cos(y)dy(x^2+1)= -sin(y)2x.dx [/tex]

which is looking ok, you probably want to use your initial condition though
 

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