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Thread: 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:
    $\displaystyle
    \frac{dy}{dx} = \frac{x^2 + y^2}{2xy}
    $
    The answer is:
    $\displaystyle
    y^2 = x(x + k)
    $

    Now, i differentiate y = xu:
    $\displaystyle \frac{dy}{dx} = u + x\frac{dy}{dx}$
    After substituting in and canceling i get:
    $\displaystyle
    u + x\frac{dy}{dx} = \frac{1 + u^2}{2u}
    $
    I then take u to the otherside and put it into the fraction, multiply and integrate to get:
    $\displaystyle -ln(1 - u^2) = lnx + c$
    Taking the exponential:
    $\displaystyle \frac{1}{1 - u^2} = e^lnx + c$
    $\displaystyle \frac{1}{1 - u^2} = kx$
    Then i substitute in y = xu ( u = y / x) :
    $\displaystyle
    \frac{x^2}{x^2 - y^2} = kx
    $
    Re-arranging:
    $\displaystyle x^2 = kx(x^2 - y^2)$
    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!

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

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

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

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

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

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

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

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

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

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

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


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

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

    I continued this way . . .

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

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

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

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


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

    Thankyou very much for your reply
    A question though, how did you get
    $\displaystyle
    \ln(C)
    $

    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 $\displaystyle \ln(C)$

    i don't see how $\displaystyle c$ is equivalent to $\displaystyle \ln c$.
    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, $\displaystyle c.$

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


    Now why do we need something like that?

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

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

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

    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:

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


    Now we can exponentiate: .$\displaystyle y \;=\;Cx^2$ . . . 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, $\displaystyle c.$

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


    Now why do we need something like that?

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

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

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

    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:

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


    Now we can exponentiate: .$\displaystyle y \;=\;Cx^2$ . . . 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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