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))#