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Math Help - Integration need help

  1. #1
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    Integration need help

    Solve (x^2+1)y' + y^2 + 1 =0, y(3)=2. Find y in terms of x only. Simplify so that your answer does not contain any trigonometric or inverse trig functions.


    (x^2+1)y' + y^2 + 1 =0, y(3)=2
    dy/dx + y^2 = -\frac{1}{x^2+1}

    I'm stuck here, how do I get it so I can have y^2 multiplied by dy and have the other side multiplied by dx so I can integrate?
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  2. #2
    Junior Member F.A.P's Avatar
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    You should end up here...

    dy/dx= -\frac{y^2+1}{x^2+1}

    since (x^2+1) is only factor for y' and not y^2.

    Go on from there...
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  3. #3
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    ok so when you integrate, on the left side u get y, but how do u deal with the right side since there's both x and y?
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    You have a separable DE.

    \left( {x^2 + 1} \right)y' + \left( {y^2 + 1} \right) = 0 \implies \left( {x^2 + 1} \right)\,\frac{{dy}}<br />
{{dx}} = - \left( {y^2 + 1} \right)

    so we have  <br />
- \frac{1}<br />
{{y^2 + 1}}\,dy = \frac{1}<br />
{{x^2 + 1}}\,dx<br />

    Now it remains to integrate both sides.
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  5. #5
    Junior Member F.A.P's Avatar
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    Quote Originally Posted by circuscircus View Post
    ok so when you integrate, on the left side u get y, but how do u deal with the right side since there's both x and y?
    I only intended to make you aware of a mistake during your first steps in rearranging the factors...

    ....follow Krizalid's post...
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