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Math Help - Initial value problem

  1. #1
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    Initial value problem

    Hi, I don't understand how I am meant to solve this:

    x\frac{dy}{dx} + y = x^4, y(1) = 1

    Any help appreciated
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  2. #2
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    Hello, Beard!

    x\frac{dy}{dx} + y \:=\: x^4,\quad y(1) = 1

    Divide by x\!:\quad \frac{dy}{dx} + \frac{1}{x}\,y \:=\:x^3

    Integrating Factor: . I \:=\:e^{\int\frac{dx}{x}} \:=\:e^{\ln x} \:=\:x

    Multiply by x\!:\quad x\,\frac{dy}{dx} + y \:=\:x^4

    We have: . \frac{d}{dx}(xy) \:=\:x^4

    Integrate: . xy \:=\:\frac{1}{5}x^5 + C \quad\Rightarrow\quad y \:=\:\frac{1}{5}x^4 + \frac{C}{x}


    From y(1) = 1\!:\quad 1 \:=\:\tfrac{1}{5}(1^4) + \frac{C}{1} \quad\Rightarrow\quad C \:=\:\frac{4}{5}


    Therefore: . y \;=\;\frac{1}{5}x^4 + \frac{4}{5x}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Beard!


    Divide by x\!:\quad \frac{dy}{dx} + \frac{1}{x}\,y \:=\:x^3

    Integrating Factor: . I \:=\:e^{\int\frac{dx}{x}} \:=\:e^{\ln x} \:=\:x

    Multiply by x\!:\quad x\,\frac{dy}{dx} + y \:=\:x^4

    We have: . \frac{d}{dx}(xy) \:=\:x^4

    Integrate: . xy \:=\:\frac{1}{5}x^5 + C \quad\Rightarrow\quad y \:=\:\frac{1}{5}x^4 + \frac{C}{x}


    From y(1) = 1\!:\quad 1 \:=\:\tfrac{1}{5}(1^4) + \frac{C}{1} \quad\Rightarrow\quad C \:=\:\frac{4}{5}


    Therefore: . y \;=\;\frac{1}{5}x^4 + \frac{4}{5x}

    Thankyou but I am confused by the step "integrating factor" I don't understand how it affects the equation
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  4. #4
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    Quote Originally Posted by Beard View Post
    Thankyou but I am confused by the step "integrating factor" I don't understand how it affects the equation
    It makes the left-hand side look like the result of a product-rule differentiation. Just in case a picture helps... suppose



    is the product rule - straight continuous lines differentiating downwards (integrating up) with respect to x. Then we're trying to fit the left-hand side...

    \frac{dy}{dx} + y\ \frac{1}{x}

    ... along the bottom row. As it comes, it won't fit (and satisfy the rule). But if we accomodate the chain rule...



    ... inside the (legs-uncrossed version of the) product rule...



    ... we can see how to fix it...



    Spoiler:

    Hope that helps.

    Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    __________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus: standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; November 29th 2009 at 08:47 AM. Reason: typo
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  5. #5
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    Quote Originally Posted by tom@ballooncalculus View Post
    It makes the left-hand side look like the result of a product-rule differentiation. Just in case a picture helps... suppose



    is the product rule - straight continuous lines differentiating downwards (integrating up) with respect to x. Then we're trying to fit the left-hand side...

    \frac{dy}{dx} + y\ \frac{1}{x}

    ... along the bottom row. As it comes, it won't fit (and satisfy the rule). But if we accomodate the chain rule...



    ... inside the (legs-uncrossed version of the) product rule...



    ... we can see how to fix it...



    Spoiler:

    Hope that helps.

    Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    __________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus: standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Thanks for the attempt but that has confused me more
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  6. #6
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    Fair enough! Try to focus on the product rule, one way or another... or, maybe not - you could simply practice applying the formula. Whatever works for you...

    PS: are you clear, at least, that the equation is 'multiplied through' by the I.F.?

    i.e. given...

    A + B = C (1)

    ... it follows that

    A * I.F. + B * I.F. = C * I.F. (2)

    ... and (2) is easier than (1)
    Last edited by tom@ballooncalculus; November 30th 2009 at 05:09 AM. Reason: ps
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