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Thread: first order equations

  1. #1
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    first order equations

    hi there
    i need help in trying to solve the following first order equations .


    dy/dx = (x-y) / (x+y) im unsure how to seperatre the variables in this


    dy/dx - xy = e^(-x^2)*y^3

    (im using ^ as "to the power of ...")

    any help appreciated ,thankyou!
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  2. #2
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    Hello, philyc86!

    The first one is homogeneous . . .


    $\displaystyle \frac{dy}{dx} \:=\: \frac{x-y}{x+y}$
    Divide top and bottom of the fraction by $\displaystyle x\!: \;\; \frac{dy}{dx} \:=\:\frac{1 - \frac{y}{x}}{1 + \frac{y}{z}} $

    Let: .$\displaystyle v \:=\: \frac{y}{x} \quad\Rightarrow\quad y \:=\: vx \quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx} $

    Substitute: .$\displaystyle v + x\frac{dv}{dx} \:=\:\frac{1-v}{1+v} \quad\Rightarrow\quad x\frac{dv}{dx} \:=\:\frac{1-v}{1+v} - v$

    . . $\displaystyle x\frac{dv}{dx} \:=\:\frac{1-2v-v^2}{1+v} \quad\Rightarrow\quad x\frac{dv}{dx} \:=\:-\frac{v^2+2v-1}{v+1}dx$

    Separate variables: .$\displaystyle \frac{v+1}{v^2+2x-1}dv \:=\:-\frac{dx}{x}$


    Integrate: .$\displaystyle \tfrac{1}{2}\ln(v^2+2v-1) \:=\:-\ln x + c_1 \quad\Rightarrow\quad \tfrac{1}{2}\ln(v^2+2v-1) \:=\:-\ln x + \ln c_2$

    . . $\displaystyle \ln(v^2+2v-1) \;\;=\;\;-2\ln x + \ln C \;\;=\;\;\ln(x^{-2}) + \ln C \;\;=\;\;\ln(Cx^{-2})$

    Then: .$\displaystyle v^2 + 2v - 1 \:=\:Cx^{-2}$


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

    . . Multiply by $\displaystyle x^2\!:\;\;\boxed{y^2 + 2xy - x^2 \:=\:C}$

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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by philyc86 View Post
    dy/dx = (x-y) / (x+y)
    Another approach :

    Substitute $\displaystyle u=x+y$ (so $\displaystyle y=u-x$ and $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u} {\mathrm{d}x}-1$) :

    $\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}-1=\frac{x-(u-x)}{u}\implies\frac{\mathrm{d}u}{\mathrm{d}x}=\fra c{2x-u+u}{u}\implies\frac{\mathrm{d}u}{\mathrm{d}x}=\fr ac{2x}{u}\implies u\,\mathrm{d}u=2x\,\mathrm{d}x$

    Now integrate both sides to get $\displaystyle \frac{u^2}{2}=x^2+C\implies u^2=2x^2+C'$. Then back-substitute and you'll find Soroban's answer : $\displaystyle y^2+2xy-x^2=C'$.
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