Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

initial-value problem?

dy/dx = 1/2e^(-x/4)y(sqrt y) (y>0)

y= 1/4 when x=0

please help!!

1 Answer

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  • mlam18
    Lv 6
    1 decade ago
    Favorite Answer

    dy/dx = 1/2e^(-x/4)y(sqrt y)

    dy/y(sqrt y) = 1/2 e^(-x/4) dx

    ∫ dy/y(sqrt y) = ∫1/2 e^(-x/4) dx

    ∫ y^(-3/2) dy = ∫1/2 e^(-x/4) dx

    -2 y^(-1/2) = (1/2)(-4)e^(-x/4) + c

    y= 1/4 when x=0

    -2 (1/4)^(-1/2) = (1/2)(-4)e^(-0/4) + c

    -4 = -2 + c

    c = -2

    -2 y^(-1/2) = (1/2)(-4)e^(-x/4) -2

    -2 y^(-1/2) = -2e^(-x/4) -2

    y^(-1/2) = e^(-x/4) +1

    You can finish the rest.

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