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Differential Equation Involving Trigonometric Functions

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation: [itex]\frac{dy}{dx} = cos^2 (x) cos^2 (2y)[/itex]

    3. The attempt at a solution
    I rewrote the equation

    [itex]\frac{dy}{cos^2 (2y)} = sec^2 (2y) = cos^2 (x) dx[/itex]. Then I integrated,
    [itex]\frac{tan(2y)}{2} = \frac{1}{2} (x + sin(x)cos(x)) + c[/itex]. Then I solved for y,
    [itex]y = \frac{tan^{-1} (x + sin(x)cos(x) + c)}{2}[/itex]

    But this isn't the answer my book gives (or at least it doesn't look very similar). Where did I go wrong?
     
  2. jcsd
  3. Mar 27, 2012 #2
    This is another problem I'd like my work checked on (there's no book answer for this):

    Solve the differential equation: [itex]y' = \frac{x^2}{y(1+x^3)}[/itex]

    Rearranging the equation,

    [itex]y dy = \frac{x^2 dx}{1+x^3}[/itex]. Integrating, let [itex]u= 1+x^3[/itex],
    [itex]\frac{y^2}{2} = \frac{1}{3} \int \frac{1}{u} du = \frac{1}{3} ln|1+x^3| + c[/itex], which implies,
    [itex]y = ± \sqrt{ \frac{2}{3} ln|1+x^3| + c }[/itex]
     
  4. Mar 27, 2012 #3

    Mark44

    Staff: Mentor

    Looks OK to me except that you left off the "dy" that should go with sec^2(2y) up above. How does your answer differ from the one in the book? They might have written sin(x)cos(x) as (1/2)sin(2x).

    You can always check that what you have is actually a solution. Start with the equation tan(2y) = x + sin(x)cos(x) and differentiate to find dy/dx. Do the same with the book's answer. If one of the solutions doesn't get back to the differential equation, that solution is wrong.
     
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