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Math Help - Ordinary differentiation equations (ODE)-seperable

  1. #1
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    Ordinary differentiation equations (ODE)-seperable

    hey there..
    i've got unsolved solution here..



    Solve initial value problem..

    xy’ – y = 2x2y ; y(1) = 1

    x dy/dx = y + 2x2y
    dy/dx = y/x + 2xy
    dy/dx = y[(1/x) + 2x]
    ∫1/y dy = ∫ (1/x) + 2x dx
    ln |y| = ln |x| + x2 + C
    apply exp to both side
    y = x + ex^2 + e^C

    substitute the initial value
    1 = 1 + e1 + e^C
    e^(1+C) = 0
    apply ln
    1 + C = ln 0?
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  2. #2
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    Lexington, MA (USA)
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    Hello, nameck!

    We're supposed to guess which numbers are exponents?


    xy’ - y \:=\: 2x^2y\qquad y(1) \,=\, 1

    We have: . x\,\frac{dy}{dx} \;=\;2x^2y + y

    . . . . . . . . x\,\frac{dy}{dx}\;=\;(2x^2+1)y

    . . . . . . . . . \frac{dy}{y} \;=\;\frac{2x^2+1}{x}\,dx

    . . . . . . . . . \frac{dy}{y} \;=\;\left(2x + \frac{1}{x}\right)\,dx

    . . . . . . . . \ln y \;=\;x^2 + \ln x + C

    . . . . . . . . . . y \;=\;e^{x^2+\ln x+C}

    . . . . . . . . . . y \;=\;e^{x^2}\cdot e^{\ln x}\cdot e^C

    . . . . . . . . . . y \;=\;e^{x^2}\cdot x \cdot C

    . . Hence: . . y \;=\;Cx\,e^{x^2}


    Since y(1)=1, we have: . C\cdot 1\cdot e^{1^2} \:=\:1  \quad\Rightarrow\quad C \:=\:\frac{1}{e}


    Therefore: . y \;=\;\frac{1}{e}\,x\,e^{x^2} \quad\Rightarrow\quad\boxed{ y\;=\;x\,e^{x^2-1}}

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