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Math Help - Separable D.E. problem

  1. #1
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    Separable D.E. problem

    So here's my problem, I'm having trouble starting out:

    Solve the separable differential equation:
    9x-5y\sqrt{x^2+1}\frac{dy}{dx}=0
    Subject to the initial condition: y(0)=2

    So I have this to start out with.

    -5y dy=-9x+\frac{1}{\sqrt{x^2+1}} dx

    Am I headed in the right direction? Any help is greatly appreciated!
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  2. #2
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    Quote Originally Posted by ascendancy523 View Post
    So here's my problem, I'm having trouble starting out:

    Solve the separable differential equation:
    9x-5y\sqrt{x^2+1}\frac{dy}{dx}=0
    Subject to the initial condition: y(0)=2

    So I have this to start out with.

    -5y dy=-9x+\frac{1}{\sqrt{x^2+1}} dx

    Am I headed in the right direction? Any help is greatly appreciated!
    9x - 5y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0

    \frac{9x}{\sqrt{x^2 + 1}} - 5y\,\frac{dy}{dx} = 0

    5y\,\frac{dy}{dx} = \frac{9x}{\sqrt{x^2 + 1}}

    \int{5y\,\frac{dy}{dx}\,dx} = \int{\frac{9x}{\sqrt{x^2 + 1}}\,dx}

    \int{5y\,dy} = \int{9x(x^2 + 1)^{-\frac{1}{2}}\,dx}.


    You should be able to continue from here. You will need to use a u substitution to evaluate the RHS.
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  3. #3
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    Where would I use the u substitution?
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  4. #4
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    Quote Originally Posted by ascendancy523 View Post
    Where would I use the u substitution?
    Taking a wild guess, I'd say in the integral that requires it ....
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  5. #5
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    Quote Originally Posted by ascendancy523 View Post
    Where would I use the u substitution?

    Let u = x^2 + 1

    Then \frac{du}{dx} = 2x

    Substituting that into your original integral:

    \frac{9}{2} \int \frac{1}{\sqrt{u}} \ du
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