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First Order Differential Equation

  1. Oct 4, 2013 #1
    Find the general solution to the following differential equation.

    dy/dx = 2x( (y^2) + 1)

    2. Relevant equations

    3. The attempt at a solution

    I got all x terms on one side and all y terms on the otherside

    2x dx = 1/( (y^2) + 1`)dy


    x^2 + c = tan(y)

    y = tan-inverse(x^2 + c)

    Can somebody tell me if this is right ?

    my book says the solution is y = tan(x^2 + c)
  2. jcsd
  3. Oct 4, 2013 #2


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    Science Advisor

    It's not right. What is
    [tex] \int \frac{dy}{1+y^2}[/tex]?
  4. Oct 4, 2013 #3

    sorry I see where I went wrong now

    it should have been x^2 + c = tan-inverse(y)
  5. Oct 4, 2013 #4
    ∫dy/(1+y2) = tan-1y

    So,tan-1y = x2 + c

    or , y = tan(x2 + c)
  6. Oct 4, 2013 #5
    Could you help me with another one?

    Given the differential equation

    x(dy/dx) = √(4 - y^2) , find the general solution given that y = 0 when x = 1

    I got all the x terms to one side and all of the y terms to the opposite side.

    1/√(4 - y^2) dy = 1/x dx

    integrating gives

    sin-inverse(y/2) = lnx

    using the limits and we get

    sin-inverse(y/2) = lnx

    y/2 = sin(lnx)

    y = 2sin(lnx)

    my book says the solution is y = sin(lnx)
  7. Oct 4, 2013 #6
    y=2sin(lnx) looks to be the correct answer.You can verify that by putting the solution in the original equation and checking whether it satisfies the eq.

    Note : Even though you came up with the correct answer ,please do not ignore the constant of integration C while integrating.

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