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Thread: How to solve for the particular solution of the separable Differential equation

  1. #1
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    Red face How to solve for the particular solution of the separable Differential equation

    1. 2xydx+(1+y)dy=0 ; y(2)=1

    2. y'+y^2 sinx=0 ; y(0)=0

    3. 2xdx-dy=x(xdy-2ydx) ; y(-3)=1

    4. dy=x(2ydx-xdy) ; x=1 , y=4

  2. #2
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    You should start by separating, since you know they are separable.

    Let's see what you get.

  3. #3
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    Hello, tops!

    You've never done one of these? . . . Ever?
    I'll do the first one.


    [tex](1)\;\;2xy\,dy + (1+y)dy \;=\; 0 \quad y(2) = 1 [/tex]

    Separate variables: .[[tex]\frac{y+1}{y}\,dy \;=\;2x \,dx[/tex]

    Integrate: .[tex]\displaystyle\int\left(1 + \tfrac{1}{y}\right)\,dy \;=\;\int 2x\,dx[/tex]

    . . . . . . . . . . . . .[tex]y + \ln y \;=\;x^2 + C[/tex]


    [tex]y(2) = 1\!:\;\; 1 + \ln 1 \:=\:2^2+C \quad\Rightarrow\quad C \,=\,-3[/tex]


    [tex]\text{Therefore: }\;y + \ln y \;=\;x^2 - 3[/tex]

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