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# initial value problem

what is the solution of this initial value problem?

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

y=1/4 when x=0

please help asap!

### 1 Answer

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- lou hLv 71 decade agoFavorite Answer
1/y^(3/2) · dy = 1/2·e^(-x/4) dx

-2· 1/√y = -2· e^(-x/4) - 2· C

1/√y = 1/e^(x/4) + C

√y= e^(x/4) / [ C·e^(x/4) + 1]

with √(1/4) = 1/(C+1) --> 1/2= 1/(C+1) --> C+1=2 --> C=1

==> √y= e^(x/4) / [ e^(x/4) + 1] ==> y= e^(x/2)/ [ e^(x/4) + 1]²

Saludos

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