# Math Help - Initial value problem

1. ## Initial value problem

Having no luck with this either

Find the solution of the initial value problem

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

2. Originally Posted by offahengaway and chips
Having no luck with this either

Find the solution of the initial value problem

dy/dx=(1/2e^-x/4).(y sqrt y) (y>0) y=1/4 when x =0
do you mean $y' = \frac 12e^{-x/4} \cdot y \sqrt{y}$ ?

if so, this si separable, divide both sides by $y \sqrt{y} = y^{3/2}$ (we can do this since $y \ne 0$). then, multiply both sides by $dx$, we get

$\Rightarrow y^{-3/2}~dy = \frac 12e^{-x/4}~dx$

now integrate both sides and continue