Separable Differential Equation

Click For Summary
SUMMARY

The discussion focuses on solving a separable differential equation involving the function sin(y) and its relationship with the variable x. The user attempts to differentiate the equation sin(y) = A/√(x² + 1) and applies the chain rule to find dy/dx. The solution process appears correct, but the user is advised to incorporate initial conditions for a complete solution. The mathematical manipulations and differentiation steps are crucial for verifying the solution's accuracy.

PREREQUISITES
  • Understanding of separable differential equations
  • Familiarity with differentiation techniques, particularly the chain rule
  • Knowledge of trigonometric functions and their derivatives
  • Ability to apply initial conditions in solving differential equations
NEXT STEPS
  • Study the method of solving separable differential equations in detail
  • Learn about the application of initial conditions in differential equations
  • Explore the implications of trigonometric identities in differential equations
  • Practice differentiating complex functions involving trigonometric and algebraic expressions
USEFUL FOR

Students studying calculus, particularly those focusing on differential equations, as well as educators looking for examples of solving separable equations.

hpayandah
Messages
18
Reaction score
0

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.
 

Attachments

  • IMG-20120124-00015.jpg
    IMG-20120124-00015.jpg
    38.7 KB · Views: 411
Physics news on Phys.org
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
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
1K
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K