3.3.7.8 Example 8
dydt=y23tyt2+yty(2)=1

Let y=ut or u=yt, hence dydt=tdudt+u and the above ode becomes

tdudt+u=u2t23t2ut2+ut2tdudt+u=u23u1+utdudt=u23u1+uu=u23uu(1+u)1+u=u23uuu21+u=2u24u1+u

Which is separable.

(1+u2u2+4u)du=1tdt12(1+uu2+2u)du=1tdt12ln(2u+u2)=2lnt+c1ln(2u+u2)=4lnt+c2

Or

2u+u2=c31t4

But u=yt. Hence the above becomes

(1)2yt+(yt)2=c31t4

Applying IC y(2)=1 the above becomes

212+(12)2=c31241+14=c316c3=54(16)=20

Hence (1) becomes

2yt+(yt)2=20t4

Or

y1=t2+t4+20ty2=t2t4+20t

Whenever we get more than one solution, we should verify each solution satisfies the ode and IC as some can be extraneous When we do this, we will find both solutions satisfy the ode itself, but y2 does not satisfy the IC. Hence it is now removed. The final solution is therefore

y1=t2+t4+20t