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Math Help - Solving Ordinary Differential Equations

  1. #1
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    Smile Solving Ordinary Differential Equations

    1. I want to solve this ordinary diff equation:
    (1-x^2)dy/dx + 2xy = (1-x^2)^3/2

    Here is my working:
    dy/dx = (1-x^2)^3/2 - 2xy / (1-x^2)
    = (1-x^2)^1/2 -2xy/(1-x^2)
    It's unseparable and I can't use substitution since I can't see a (x/y), what should I do next to solve this?

    2. I also want to solve: (y-x)dy/dx + 2x + 3y = 0
    But I get:
    dy/dx = -(2x + 3y)/(y-x)
    and I'm stuck on what to do next.

    Could someone please help me? I really want to undertand how to solve these.
    Thankyou for your time and help, I really do appreciate it.
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  2. #2
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    Re: Solving Ordinary Differential Equations

    Hello, happyfingerrs!

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

    Divide by 1-x^2:\;\;\frac{dy}{dx} + \frac{2x}{1-x^2}y \:=\:(1-x^2)^{\frac{1}{2}}

    Integrating factor:
    . . I \:=\:e^{\int\frac{2x\,dx}{1-x^2}} \:=\:e^{-\ln(1-x^2)} \:=\:e^{(1-x^2)^{-1}} \:=\:(1-x^2)^{-1} \:=\:\frac{1}{1-x^2}

    \text{Multiply by }I:\;\;\frac{1}{1-x^2}\frac{dy}{dx} + \frac{2x}{(1-x^2)^2}y \:=\:\frac{1}{(1-x^2)^{\frac{1}{2}}}

    \text{We have: }\:\frac{d}{dx}\left[\frac{1}{1-x^2}y\right] \;=\;\frac{1}{\sqrt{1-x^2}}

    \text{Integrate: }\:\frac{1}{1-x^2}y \;=\;\sin^{\text{-}1}(x) + C

    \text{Therefore: }\:y \;=\;(1-x^2)\left[\sin^{\text{-}1}(x) + C\right]




    2.\;(y-x)\tfrac{dy}{dx} + 2x + 3y \:=\: 0

    This one is homogeneous . . .

    \text{We have: }\:\frac{dy}{dx} \;=\;-\frac{3y+2x}{y-x}


    \text{Divide numerator and denominator by }x:
    . . \frac{dy}{dx} \;=\;-\frac{3\frac{y}{x}+2}{\frac{y}{x} - 1}

    \text{Let }v \,=\, \frac{y}{x} \quad\Rightarrow\quad y \,=\,vx \quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}

    \text{Substitute: }\:v + x\frac {dv}{dx} \;=\;-\frac{3v+2}{v-1}

    . . which simplifies to: . \frac{v-1}{v^2+2v+2}dv \;=\;-\frac{dx}{x}

    Got it?
    Thanks from happyfingerrs
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  3. #3
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    Re: Solving Ordinary Differential Equations

    Oops!
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  4. #4
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    Re: Solving Ordinary Differential Equations

    Quote Originally Posted by Soroban View Post
    Hello, happyfingerrs!


    Divide by 1-x^2)^{\frac{1}{2}}" alt="1-x^2:\;\;\frac{dy}{dx} + \frac{2x}{1-x^2}y \:=\1-x^2)^{\frac{1}{2}}" />

    Integrating factor:
    . . 1-x^2)^{-1} \:=\:\frac{1}{1-x^2} " alt="I \:=\:e^{\int\frac{2x\,dx}{1-x^2}} \:=\:e^{-\ln(1-x^2)} \:=\:e^{(1-x^2)^{-1}} \:=\1-x^2)^{-1} \:=\:\frac{1}{1-x^2} " />

    \text{Multiply by }I:\;\;\frac{1}{1-x^2}\frac{dy}{dx} + \frac{2x}{(1-x^2)^2}y \:=\:\frac{1}{(1-x^2)^{\frac{1}{2}}}

    \text{We have: }\:\frac{d}{dx}\left[\frac{1}{1-x^2}y\right] \;=\;\frac{1}{\sqrt{1-x^2}}

    \text{Integrate: }\:\frac{1}{1-x^2}y \;=\;\sin^{\text{-}1}(x) + C

    \text{Therefore: }\:y \;=\;(1-x^2)\left[\sin^{\text{-}1}(x) + C\right]





    This one is homogeneous . . .

    \text{We have: }\:\frac{dy}{dx} \;=\;-\frac{3y+2x}{y-x}


    \text{Divide numerator and denominator by }x:
    . . \frac{dy}{dx} \;=\;-\frac{3\frac{y}{x}+2}{\frac{y}{x} - 1}

    \text{Let }v \,=\, \frac{y}{x} \quad\Rightarrow\quad y \,=\,vx \quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}

    \text{Substitute: }\:v + x\frac {dv}{dx} \;=\;-\frac{3v+2}{v-1}

    . . which simplifies to: . \frac{v-1}{v^2+2v+2}dv \;=\;-\frac{dx}{x}

    Got it?

    Hello Soroban, thankyou very much for your help Your working was incredibly useful
    My only remaining question is, how would you integrate v-1/v^2 + 2v + 2? Do you have to complete the square or use trig identities?

    Thank you so much
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