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Math Help - Simple Differential Equation

  1. #1
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    Simple Differential Equation

    \frac{dy}{dx} = \frac{4 x + 20 y}{5 x}

    y(1)=-1

    What did I do wrong here?
    y'=\frac{4x}{5x}+\frac{20y}{5x}

    y'-\frac{20}{5}x^{-1}y=\frac{4x}{5x} < Is this proper first order linear?

    let \eta=e^{\int\frac{20}{5x}}= e^{-4ln(x)}

    Mulitplay eta on both sides, and integrate, I get:

    <br />
yx^{-4}=\frac{4}{5}\int x^{-4} dx<br />

    <br />
y=\frac{4x}{15}+x^4C<br />

    To solve for C:
    <br />
-1=\frac{4}{15}+C, C= -\frac{4}{3}<br />

    My answer is:
    <br />
y=\frac{4x}{15}-\frac{4x^4}{3}<br />

    which is totally wrong. What am I missing?

    Thank you!
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  2. #2
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    Quote Originally Posted by Vamz View Post
    \frac{dy}{dx} = \frac{4 x + 20 y}{5 x}

    y(1)=-1

    What did I do wrong here?
    y'=\frac{4x}{5x}+\frac{20y}{5x}

    y'-\frac{20}{5}x^{-1}y=\frac{4x}{5x} < Is this proper first order linear?

    let \eta=e^{\int\frac{20}{5x}}= e^{-4ln(x)}

    Mulitplay eta on both sides, and integrate, I get:

    <br />
yx^{-4}=\frac{4}{5}\int x^{-4} dx<br />

    <br />
y=\frac{4x}{15}+x^4C<br />
    I don't know about totally wrong. You missed a minus sign in this last line:
    <br />
y=-\frac{4x}{15}+x^4C<br />

    -Dan
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  3. #3
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    Thanks. But it still appears to be wrong
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  4. #4
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    Quote Originally Posted by Vamz View Post
    Thanks. But it still appears to be wrong
    Have you been told what the correct answer is? If not, have you tried differentiating and substituting back into the DE to see if the DE is satisfied?
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  5. #5
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    No, I don't have the correct answer. When I derive it, it does not seem to be correct.
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  6. #6
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    Quote Originally Posted by Vamz View Post
    No, I don't have the correct answer. When I derive it, it does not seem to be correct.
    Then like I said, substitute your answer and your derivative into your DE. Find out if the solution satisfies the DE...
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  7. #7
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    Quote Originally Posted by Vamz View Post
    \frac{dy}{dx} = \frac{4 x + 20 y}{5 x}

    y(1)=-1

    What did I do wrong here?
    y'=\frac{4x}{5x}+\frac{20y}{5x}

    y'-\frac{20}{5}x^{-1}y=\frac{4x}{5x} < Is this proper first order linear?

    let \eta=e^{\int\frac{20}{5x}}= e^{-4ln(x)}

    Mulitplay eta on both sides, and integrate, I get:

    <br />
yx^{-4}=\frac{4}{5}\int x^{-4} dx<br />

    <br />
y=\frac{4x}{15}+x^4C<br />

    To solve for C:
    <br />
-1=\frac{4}{15}+C, C= -\frac{4}{3}<br />

    My answer is:
    <br />
y=\frac{4x}{15}-\frac{4x^4}{3}<br />

    which is totally wrong. What am I missing?

    Thank you!
    \displaystyle 5xy'=4+20y\Rightarrow 5xy'-20y=4\Rightarrow y'-\frac{4y}{x}=\frac{4}{5}

    \displaystyle\frac{y}{x^4}=\int \frac{4}{5}x^{-4} \ dx\Rightarrow yx^{-4}=\frac{-4x^{-3}}{15}+C

    \displaystyle y=Cx^4-\frac{4x}{15}

    \displaystyle f(1)=C-\frac{4}{15}=-1\Rightarrow C=-\frac{11}{15}

    This is what I obtained.
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