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# 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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- mlam18Lv 61 decade agoFavorite 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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