How do you solve $y'=-xy+sqrty$ given y(0)=1?

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Answer 1 · 2017-10-09T11:39:46

See below.

Making $y = z^2$ we obtain

$z (x z + 2 z'-1) = 0$ or $z=0$ and $x z + 2 z'-1=0$

We discard $z=0$ due to the initial conditions so we follow with

$2 z'+x z-1=0$ which is a linear non-homogeneous differential equation with solution

$z = C e^(-(x^2/4))+ e^(-(x^2/4)) int_0^(x/2)e^(-xi^2) d xi$ and then

$y = pm sqrt( e^(-(x^2/4))(C+ int_0^(x/2)e^(-xi^2) d xi))$

and

$y(0) = pm sqrt(C) = 1$ then $C = 1$ and

$y = pm sqrt( e^(-(x^2/4))(1+ int_0^(x/2)e^(-xi^2) d xi))$

How do you solve y'=-xy+sqrtyy'=-xy+sqrty given y(0)=1?