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Math Help - Substitution D.E

  1. #1
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    Substitution D.E

    Ok, so i do everything to one point and can't get the right answer.
    The question is:
    By using the substitution y = xu solve:
    <br />
\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}<br />
    The answer is:
    <br />
y^2 = x(x + k)<br />

    Now, i differentiate y = xu:
    \frac{dy}{dx} = u + x\frac{dy}{dx}
    After substituting in and canceling i get:
    <br />
u + x\frac{dy}{dx} = \frac{1 + u^2}{2u}<br />
    I then take u to the otherside and put it into the fraction, multiply and integrate to get:
    -ln(1 - u^2) = lnx + c
    Taking the exponential:
    \frac{1}{1 - u^2} = e^lnx + c
    \frac{1}{1 - u^2} = kx
    Then i substitute in y = xu ( u = y / x) :
    <br />
\frac{x^2}{x^2 - y^2} = kx<br />
    Re-arranging:
    x^2 = kx(x^2 - y^2)
    Making y the subject does not give the anwser.

    Any help would be greatly appriciated. Thank-you
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  2. #2
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    Hello, Ashley!

    \frac{dy}{dx} \:=\: \frac{x^2 + y^2}{2xy}

    The answer is: . y^2 \:= \:x(x + k)

    Now, i differentiate y = xu\!:\;\;\frac{dy}{dx} \:=\: u + x\frac{du}{dx}

    After substituting and canceling, i get: . u + x\frac{dy}{dx} \:=\: \frac{1 + u^2}{2u}

    I then take u to the other side and put it into the fraction:

    . . x\frac{du}{dx} \:=\:\frac{1+u^2}{2u} - u \:=\:\frac{1-u^2}{2u} \quad\Longrightarrow\quad \frac{2u}{1-u^2}\,du \:=\:\frac{dx}{x}

    Integrate to get: . -\ln(1 - u^2) \:=\: \ln x + c
    I continued this way . . .

    -\ln(1-u^2) \:=\:\ln x + c \quad\Longrightarrow\quad \ln(1-u^2) \:=\:-\ln x + \ln(C) \:=\:\ln\left(\frac{C}{x}\right)

    Exponentiate: . 1-u^2 \:=\:\frac{C}{x}

    Back-substitute: . 1 - \left(\frac{y}{x}\right)^2 \:=\:\frac{C}{x} \quad\Longrightarrow\quad \frac{x^2-y^2}{x^2}\:=\:\frac{C}{x}

    . . . . . . x^2-y^2 \;=\;Cx \quad\Longrightarrow\quad y^2 \;=\;x^2 - Cx \quad\Longrightarrow\quad y^2 \;=\;x^2 + kx


    Therefore: . y^2 \;=\;x(x+k)

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

    I continued this way . . .

    -\ln(1-u^2) \:=\:\ln x + c \quad\Longrightarrow\quad \ln(1-u^2) \:=\:-\ln x + \ln(C) \:=\:\ln\left(\frac{C}{x}\right)

    Exponentiate: . 1-u^2 \:=\:\frac{C}{x}

    Back-substitute: . 1 - \left(\frac{y}{x}\right)^2 \:=\:\frac{C}{x} \quad\Longrightarrow\quad \frac{x^2-y^2}{x^2}\:=\:\frac{C}{x}

    . . . . . . x^2-y^2 \;=\;Cx \quad\Longrightarrow\quad y^2 \;=\;x^2 - Cx \quad\Longrightarrow\quad y^2 \;=\;x^2 + kx


    Therefore: . y^2 \;=\;x(x+k)

    Thankyou very much for your reply
    A question though, how did you get
    <br />
\ln(C)<br />

    because i don't see how 'c' is equivalent to ln c ?
    Or am i missing something :S?
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  4. #4
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    Hello, Ashley!

    How did you get \ln(C)

    i don't see how c is equivalent to \ln c.
    Or am i missing something?
    You should be familiar with this procedure by now . . .


    First of all, I used two different C's.

    I had an arbitrary constant, c.

    Since c is the log of some other number, C,
    . . I let: . c \:=\:\ln(C)


    Now why do we need something like that?

    Suppose our equation came out to: . \ln y \:=\:2\ln x + c

    We can exponentiate both sides: . e^{\ln y} \;=\;e^{2\ln x + c}

    . . and get: . y \;=\;e^{2\ln x + c}

    Technically, we're done . . . We have the answer.
    But we should not be surprised that the book has simplified the answer.


    Before we exponentiate, we have:

    . . \ln y \;=\; \underbrace{2\ln x}_{\downarrow} + \underbrace{c}_{\downarrow}
    . . \ln y= \;\underbrace{\ln x^2 + \ln C}_{\downarrow}
    . . \ln y \;=\;\;\;\overbrace{\ln(Cx^2)}


    Now we can exponentiate: . y \;=\;Cx^2 . . . see?

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Ashley!

    You should be familiar with this procedure by now . . .


    First of all, I used two different C's.

    I had an arbitrary constant, c.

    Since c is the log of some other number, C,
    . . I let: . c \:=\:\ln(C)


    Now why do we need something like that?

    Suppose our equation came out to: . \ln y \:=\:2\ln x + c

    We can exponentiate both sides: . e^{\ln y} \;=\;e^{2\ln x + c}

    . . and get: . y \;=\;e^{2\ln x + c}

    Technically, we're done . . . We have the answer.
    But we should not be surprised that the book has simplified the answer.


    Before we exponentiate, we have:

    . . \ln y \;=\; \underbrace{2\ln x}_{\downarrow} + \underbrace{c}_{\downarrow}
    . . \ln y= \;\underbrace{\ln x^2 + \ln C}_{\downarrow}
    . . \ln y \;=\;\;\;\overbrace{\ln(Cx^2)}


    Now we can exponentiate: . y \;=\;Cx^2 . . . see?

    I see . We were taught it differently and not explained to us properlyy. So thankyou very much for clearing that up, much appreciated.
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