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Math Help - Separable 1st order ODE

  1. #1
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    Separable 1st order ODE

    I've already solved this problem, but the assessment system I'm supposed to insert my answer to keeps telling me it's incorrect, so I must have made a mistake somewhere and can't spot it.

    So I'll write out, step by step, what I've done:

    y''-21-7x^2+3y^2+x^2y^2=0

    => y''=(7-y^2)(x^2+3)

    =>\int{\frac{1}{7-y^2} dy}=\int{x^2+3 dx}

    Right Hand Side:

    \int{x^2+3 dx}=\frac{x^3}{3}+3x+c

    Left Hand Side:

    Rewriting LHS using partial fractions:

    \frac{1}{7-y^2}= \frac {A}{\sqrt{7}-y} + \frac{B}{\sqrt{7}+y}=\frac{A(\sqrt{7}+y)}{(\sqrt{7  }-y)(\sqrt{7}+y)}+\frac{B(\sqrt{7}-y)}{(\sqrt{7}-y)(\sqrt{7}+y)} =\frac{A(\sqrt{7}+y)+B(\sqrt{7}-y)}{7-y^2}

    A(\sqrt{7}+y)+B(\sqrt{7}-y)=1

    Let y=\sqrt{7}
    A(\sqrt{7}+\sqrt{7})+B(\sqrt{7}-\sqrt{7})=1
    => A(2\sqrt{7})=1
    => A=\frac{1}{2\sqrt{7}}

    Let y=-\sqrt{7}
    A(\sqrt{7}-\sqrt{7})+B(\sqrt{7}+\sqrt{7})=1
    =>B(2\sqrt{7})=1
    => B=\frac{1}{2\sqrt{7}}

    So then I get LHS to be:
    \frac{1}{2\sqrt{7}}(\frac {1}{\sqrt{7}-y} + \frac{1}{\sqrt{7}+y})

    Integrating the LHS with respect to y gives
    \frac{1}{14}\sqrt{7}\ln{(\sqrt{7}+y)}-\frac{1}{14}\sqrt{7}\ln{(\sqrt{7}-y)}+c

    So now I have
    \frac{1}{14}\sqrt{7}\ln{(\frac{(\sqrt{7}+y)}{\sqrt  {7}-y)})}=\frac{x^3}{3}+3x

    =>y=\sqrt{7}\frac{1-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{1+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}+c

    But apparently that's the wrong answer and I can't, for the life of me, figure out where I'm going wrong.
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  2. #2
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    Quote Originally Posted by aspotonthewall View Post
    I've already solved this problem, but the assessment system I'm supposed to insert my answer to keeps telling me it's incorrect, so I must have made a mistake somewhere and can't spot it.

    So I'll write out, step by step, what I've done:

    y''-21-7x^2+3y^2+x^2y^2=0

    => y''=(7-y^2)(x^2+3)

    =>\int{\frac{1}{7-y^2} dy}=\int{x^2+3 dx}

    Right Hand Side:

    \int{x^2+3 dx}=\frac{x^3}{3}+3x+c

    Left Hand Side:

    Rewriting LHS using partial fractions:

    \frac{1}{7-y^2}= \frac {A}{\sqrt{7}-y} + \frac{B}{\sqrt{7}+y}=\frac{A(\sqrt{7}+y)}{(\sqrt{7  }-y)(\sqrt{7}+y)}+\frac{B(\sqrt{7}-y)}{(\sqrt{7}-y)(\sqrt{7}+y)} =\frac{A(\sqrt{7}+y)+B(\sqrt{7}-y)}{7-y^2}

    A(\sqrt{7}+y)+B(\sqrt{7}-y)=1

    Let y=\sqrt{7}
    A(\sqrt{7}+\sqrt{7})+B(\sqrt{7}-\sqrt{7})=1
    => A(2\sqrt{7})=1
    => A=\frac{1}{2\sqrt{7}}

    Let y=-\sqrt{7}
    A(\sqrt{7}-\sqrt{7})+B(\sqrt{7}+\sqrt{7})=1
    =>B(2\sqrt{7})=1
    => B=\frac{1}{2\sqrt{7}}

    So then I get LHS to be:
    \frac{1}{2\sqrt{7}}(\frac {1}{\sqrt{7}-y} + \frac{1}{\sqrt{7}+y})

    Integrating the LHS with respect to y gives
    \frac{1}{14}\sqrt{7}\ln{(\sqrt{7}+y)}-\frac{1}{14}\sqrt{7}\ln{(\sqrt{7}-y)}+c

    So now I have
    \frac{1}{14}\sqrt{7}\ln{(\frac{(\sqrt{7}+y)}{\sqrt  {7}-y)})}=\frac{x^3}{3}+3x

    =>y=\sqrt{7}\frac{1-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{1+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}+c

    But apparently that's the wrong answer and I can't, for the life of me, figure out where I'm going wrong.
    You have + c. I think this is wrong. From what you have

    \frac{1}{14}\sqrt{7}\ln{(\frac{(\sqrt{7}+y)}{\sqrt  {7}-y)})}=\frac{x^3}{3}+3x + \frac{1}{14}\sqrt{7}\, \ln c (the last part is the arbitrary constant). Solving for y gives

    y=\sqrt{7}\,\frac{c-\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{c+\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}} (Note where my c's are.)
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  3. #3
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    I just tried it and entered it into the assessment system, and it's still wrong, but the error I got was:

    "Your answer does satisfy the differential equation but either does not have a constant specified or you did not use "c" (small case) as instructed. Your mark for this attempt is 0."
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  4. #4
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    Quote Originally Posted by aspotonthewall View Post
    I just tried it and entered it into the assessment system, and it's still wrong, but the error I got was:

    "Your answer does satisfy the differential equation but either does not have a constant specified or you did not use "c" (small case) as instructed. Your mark for this attempt is 0."
    How about

    <br />
y=\sqrt{7}\,\frac{1-c\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}{1+c\exp{(\frac{-2}{3}\sqrt{7}x^3-6\sqrt{7}x)}}?
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  5. #5
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    I get the exact same error

    Oh well... thanks anyway!
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  6. #6
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    Quote Originally Posted by aspotonthewall View Post
    I get the exact same error

    Oh well... thanks anyway!
    One last attempt. Try

    <br />
y=\sqrt{7}\,\frac{\exp{(\frac{2}{3}\sqrt{7}x^3+6\s  qrt{7}x + c)} - 1}{\exp{(\frac{2}{3}\sqrt{7}x^3+6\sqrt{7}x+c)}+1}.

    Also try

     <br />
y = \sqrt{7} \tanh \left( 3 \sqrt{7} x + \frac{\sqrt{7}}{3} x^3 + \sqrt{7} c_1 \right)<br />

    This is what Maple gave as the solution.
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