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Math Help - <help> Differential Equations... MAPS (urgent)

  1. #1
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    <help> Differential Equations... MAPS (urgent)

    Task:
    There are some differential equations that require other techniques in order to solve them. One type, called homogenous, requires a substitution in order to solve.
    The substitution used can be of the form y = v x , or y = v / x,
    where v is a function of x.

    By using the substituion y = v x, and then the substitution
     u^{2} = 1 + v^{2} , find the general solution of the differential equation  <br />
y\frac{dy}{dx} = \left[ -x + \sqrt{x^{2} + y^{2}}\right]<br />
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jungohyj View Post
    Task:
    There are some differential equations that require other techniques in order to solve them. One type, called homogenous, requires a substitution in order to solve.
    The substitution used can be of the form y = v x , or y = v / x,
    where v is a function of x.

    By using the substituion y = v x, and then the substitution
     u^{2} = 1 + v^{2} , find the general solution of the differential equation  <br />
y\frac{dy}{dx} = \left[ -x + \sqrt{x^{2} + y^{2}}\right]<br />
    First, y = vx \implies \frac{dy}{dx} = \frac{dv}{dx}x + v, so
    y\frac{dy}{dx} = \left[ -x + \sqrt{x^{2} + y^{2}}\right]
    becomes
    (vx) \left ( \frac{dv}{dx}x + v \right ) = \left [ -x + \sqrt{x^2 + (vx)^2} \right ]

    x^2 v \frac{dv}{dx} + xv^2  = \left [ -x + \sqrt{x^2 + x^2v^2} \right ]

    x^2 v \frac{dv}{dx} + xv^2  = \left [ -x + x \sqrt{1 + v^2} \right ]

    x \left ( xv \frac{dv}{dx} + v^2 \right )  = x \left [ -1 +  \sqrt{1 + v^2} \right ]

    xv \frac{dv}{dx} + v^2  = \left [ -1 +  \sqrt{1 + v^2} \right ]

    Now let u^2 = 1 + v^2 \implies 2u \frac{du}{dx} = 2v \frac{dv}{dx} or u \frac{du}{dx} = v \frac{dv}{dx}

    Thus
    xu \frac{du}{dx} + (u^2 - 1)= \left [ -1 +  \sqrt{u^2} \right ]

    xu \frac{du}{dx} = -(u^2 - 1) + (|u| - 1)

    xu \frac{du}{dx} = -u^2  + |u|

    -\frac{u}{u^2 - |u|}du = \frac{dx}{x}

    So this is either:
    -\frac{1}{u - 1}du = \frac{dx}{x} if u > 0
    or
    -\frac{1}{u + 1}du = \frac{dx}{x} if u < 0

    I'm sure you can take the solution from here.

    Note: Earlier we divided both sides of the equation by x, so we need to exclude x = 0 from our solution.

    -Dan
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  3. #3
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    Hello, jungohyj!

    I used a slightly different approach and got a slightly different result . . .


    By using the substituion y \,= \,vx, and then the substitution  u^{2} = 1 + v^{2} ,
    find the general solution of: .  <br />
y\frac{dy}{dx} = \left[ -x + \sqrt{x^{2} + y^{2}}\right]<br />

    Divide by x\!:\;\;\frac{y}{x}\!\cdot\!\frac{dy}{dx} \;=\;-\frac{x}{x} + \frac{\sqrt{x^2+y^2}}{x}\;=\;\sqrt{1 + \left(\frac{y}{x}\right)^2} - 1 . . Note: x \neq 0

    Let: v = \frac{y}{x}\quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\!\cdot\!\frac{dv}{dx}

    Substitute: . v\left(v + x\!\cdot\!\frac{dv}{dx}\right) \;=\;\sqrt{1 + v^2} - 1\quad\Rightarrow\quad v^2 + vx\!\cdot\!\frac{dv}{dx} \;=\;\sqrt{1+v^2} - 1

    and we have: . vx\!\cdot\!\frac{dv}{dx}\;=\;\sqrt{1+v^2} - (1 + v^2)\quad\Rightarrow\quad vx\!\cdot\!\frac{dv}{dx} \;=\;\sqrt{1+v^2}\left(1 - \sqrt{1+v^2}\right)

    . . \Rightarrow\quad x(v\,dv) \;=\;\sqrt{1+v^2}\left(1 - \sqrt{1+v^2}\right)\,dx


    Let u \:=\:\sqrt{1+v^2}\quad\Rightarrow\quad v^2 \:=\:u^2-1\quad\Rightarrow\quad v\,dv \:=\:u\,du

    Substitute: . x(u\,du) \;=\;u(1 - u)\,dx

    . . and we have: . x\,du \;=\;(1 - u)\,dx\quad\Rightarrow\quad \frac{du}{1-u} \:=\:\frac{dx}{x}

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post
    Let u \:=\:\sqrt{1+v^2}\quad\Rightarrow\quad v^2 \:=\:u^2-1\quad\Rightarrow\quad v\,dv \:=\:u\,du
    But
    u^2 = 1 + v^2 \implies |u| = \sqrt{1 + v^2} in general, because \sqrt{u^2} = |u| .

    I don't see how you are getting around the absolute value.

    -Dan
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