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Initial-Value Problem Integration Question

  1. Aug 29, 2011 #1
    Hi there,

    I'm doing a maths uni pre course with some questions on differentiation and integration.

    I'm looking for a few pointers on how to proceed.

    I need to solve:

    dy/dx=(cos(3x))/(2-sin(3x)) when y=2 and x=0

    Can anyone help with how I'm meant to approach this and how I go about actually integrating the equation. Its been some time since I've done things like this and the information I have is not very easy to understand.

    Many thanks

    Bob
     
  2. jcsd
  3. Aug 29, 2011 #2
    you got:

    [tex]\frac{dy}{dx}=\frac{\cos(3x)}{2-\sin(3x)}[/tex]

    You can now separate variables and integrate:

    [tex]\int_{y_0}^y dy=\int_{x_0}^x \frac{\cos(3x)}{2-\sin(3x)}dx[/tex]

    The left side is just y-y_0. The other side you can do right? Just make a u-substitution. For example, what is the derivative of the denominator? Isn't that real close to what's in the numerator? Can you adjust it to be just right then integrate?
     
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