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Math Help - Separable Differential Equations Problems

  1. #1
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    Separable Differential Equations Problems

    Hi
    The following questions i need help on:
    Solve the Differential Equations
    1) \frac{x}{y} * \frac{dy}{dx} + \frac{3x^2}{(1+x^2)}=0

    \frac{dy}{dx} = \frac{-3x^2}{(1+x^2)}*\frac{x}{y}

    \frac{dy}{dx} = \frac{-3x^2}{x(1+x^2)}

    \frac{dy}{dx} = \frac{-3x^2y}{x+x^3)}

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

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

    \frac{1}{-3y} * \frac{dy}{dx} = \frac{x}{(1+x^2)}

    u=1+x^2
    du=2xdx

    \frac{-1}{3}ln|y| = \int \frac{1}{u} * \frac{du}{2}

    \frac{-1}{3}ln|y| =  \frac{1}{2} \int \frac{1}{u}

    \frac{-1}{3}ln|y| =  \frac{1}{2}ln|x^2+1|

    -ln|y| = ln|(x^2+1)^\frac{3}{2}|

    y = -e^{ln|(x^2+1)^\frac{3}{2}|+C}

    A=-e^C

    y = Ae^{ln|(x^2+1)^\frac{3}{2}|}

    2) \frac{dy}{dx} = y^2cos2x

    \int \frac{1}{y^2} dy = \int cos2xdx

    u=2x du=2 dx

    \frac{-1}{y} = \frac{1}{2} \int cos(u) du

    \frac{-1}{y} = \frac{1}{2}sin(u) + C

    \frac{-2}{y} = sin(u) + C

    y = \frac{-2}{sin2x + C}

    3) \frac{dy}{dx} =\frac{y^2-1}{x}

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

    what should i do next?

    i tried coth^{-1} but that didn't work.

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    The following questions i need help on:
    Solve the Differential Equations
    1) \frac{x}{y} * \frac{dy}{dx} + \frac{3x^2}{(1+x^2)}=0

    \frac{dy}{dx} = \frac{-3x^2}{(1+x^2)}*\frac{x}{y}

    \frac{dy}{dx} = \frac{-3x^2}{x(1+x^2)}

    \frac{dy}{dx} = \frac{-3x^2y}{x+x^3)}

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

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

    \frac{1}{-3y} * \frac{dy}{dx} = \frac{x}{(1+x^2)}

    u=1+x^2
    du=2xdx

    \frac{-1}{3}ln|y| = \int \frac{1}{u} * \frac{du}{2}

    \frac{-1}{3}ln|y| =  \frac{1}{2} \int \frac{1}{u}

    \frac{-1}{3}ln|y| =  \frac{1}{2}ln|x^2+1|

    -ln|y| = ln|(x^2+1)^\frac{3}{2}|

    y = -e^{ln|(x^2+1)^\frac{3}{2}|+C}

    A=-e^C

    y = Ae^{ln|(x^2+1)^\frac{3}{2}|}

    2) \frac{dy}{dx} = y^2cos2x

    \int \frac{1}{y^2} dy = \int cos2xdx

    u=2x du=2 dx

    \frac{-1}{y} = \frac{1}{2} \int cos(u) du

    \frac{-1}{y} = \frac{1}{2}sin(u) + C

    \frac{-2}{y} = sin(u) + C

    y = \frac{-2}{sin2x + C}

    3) \frac{dy}{dx} =\frac{y^2-1}{x}

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

    what should i do next?

    i tried coth^{-1} but that didn't work.

    P.S
    For(3), let y=\sin \theta

    Are you having problems with the rest of the questions or you want to verify the answers?
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  3. #3
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    question 1) and 2) are incorrect according to the book's answers.

    The book's answers say the following:

    1) y= \frac{C}{(x+x^2)^{\frac{3}{2}}}

    2) y=\frac{2}{C-sin(2x)}
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  4. #4
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    can anyone help me??
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  5. #5
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    \int(\frac{dy}{y^2 -1}) = \int(\frac{dx}{x})

    \int(\frac{\frac{y + 1}{2} - \frac{(y - 1)}{2}}{(y + 1)(y - 1)})dy = log|x| + log|c|

    \frac{log(y - 1) - log(y + 1)}{2} = log|cx|\\

    log|\frac{y - 1}{y + 1}| = log|x^2.k|

    so

    \frac{k(y-1)}{(y + 1)} = x^2
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  6. #6
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    Your algebra for (1) is incorrect. As far as I can tell, your mistake was in going from step 1 to step 2. Try this:

    \displaystyle{\frac{x}{y} * \frac{dy}{dx} + \frac{3x^2}{(1+x^2)}=0}

    \displaystyle{\frac{x}{y} * \frac{dy}{dx} =- \frac{3x^2}{(1+x^2)}}

    \displaystyle{\frac{dy}{y}=-\frac{3x}{(1+x^2)}\,dx}.

    Integrate, etc.

    As for the second problem, your answer and the book's answer are equivalent. You've just re-defined your constant of integration once or twice. Your constant is the negative of the book's constant. That is an unimportant difference.

    As for #3, raj007's solution is a bit bizarre, in my opinion. I would go with a straight-forward trig substitution, or look the integral up in a table if you're allowed to do that.
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  7. #7
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    Concerning Problem #3: My idea of trig substitution is not, I think, useful. Try partial fractions after factoring the denominator on the LHS.
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  8. #8
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    y=sin\theta dy=cos\theta d\theta

    \int \frac{1}{sin^2(\theta)-1} * cos\theta d\theta = ln|x|+C

    \int \frac{cos\theta}{sin^2(\theta)-1} = ln|x|+C

    \int \frac{1}{cos^2(\theta)} = ln|x|+C

    \int sec^2(\theta)=ln|x|+C

    What should i do next?
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  9. #9
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    The cosine in your numerator dissappeared. A minus in the denominator has also gone astray.

    And a shorter way will be to factor y^2 -1 into (y+1)(y-1) and then do partial fractions.
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  10. #10
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    \int \frac{1}{(y-1)(y+1)} dy = \int \frac{1}{x} dx

    \frac{-1}{2}\int\frac{1}{y-1} + \frac{1}{2} \int \frac{1}{y+1}dy = ln|x|+C

    \frac{-1}{2}ln|y-1 + \frac{1}{2} ln|y+1| = ln|x|+C

    \frac{-1}{2}(ln|y-1|-ln|y+1|) = ln|x|+C

    \frac{-1}{2}ln|\frac{y-1}{y+1}| = ln|x|+C

    what should be done next??
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  11. #11
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    Quote Originally Posted by Paymemoney View Post
    \int \frac{1}{(y-1)(y+1)} dy = \int \frac{1}{x} dx

    \frac{-1}{2}\int\frac{1}{y-1} + \frac{1}{2} \int \frac{1}{y+1}dy = ln|x|+C

    \frac{-1}{2}ln|y-1 + \frac{1}{2} ln|y+1| = ln|x|+C

    \frac{-1}{2}(ln|y-1|-ln|y+1|) = ln|x|+C

    \frac{-1}{2}ln|\frac{y-1}{y+1}| = ln|x|+C

    what should be done next??
    From step 3 to step 4, how did the + becomes - ?

    Next, put the equation in exponents form.
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  12. #12
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    oops my mistake it's meant to be:
    <br />
\frac{1}{2}(ln|y+1|-ln|y-1|) = ln|x|+C

    ln|\frac{y+1}{y-1}|=2ln|x|+2C

    \frac{y+1}{y-1}=e^{ln|x^2|}*e^{2C}

    y+1=x^2*e^{2C}
    y+1=(x^2*e^{2C})(y-1)
    y+1=x^2*e^{2C}y-x^2*e^{2C}
    1+x^2*e^{2C}=x^2*e^{2C}y-y
    1+x^2*e^{2C}=y(x^2*e^{2C}-1)

    y=\frac{1+x^2*e^{2C}}{-1+x^2*e^{2C}}
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  13. #13
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    Quote Originally Posted by Paymemoney View Post
    y=\frac{1+x^2*e^{2C}}{-1+x^2*e^{2C}}
    However answers are \frac{1+e^{2C}x^2}{1-e^{2C}x^2}
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  14. #14
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    That looks good, except that in going from step 3 to 4, you lost your y-1 factor. And then in going from step 4 to step 5, you regained it! I wouldn't recommend dropping terms willy-nilly, unless you have a good reason for doing it (approximation, for example).

    Recommendation: rename your constant of integration so it looks a bit cleaner. That is, define, say, K = exp(2C), and use K in both places there. This is not a fundamentally important aspect of the problem, however. It is merely cosmetic.
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  15. #15
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    To fix the sign error, I'd go back and re-visit your partial fraction decomposition. I think you might have those signs flipped.
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