If y(t)=3e^(t^2)is known to be the solution of the initial value problem

y'+p(t)y=0, y(0)=y_0,

what must the function p(t) and the constant y_0 be?

Differentiate y(t):

y'(t)=3*2te^(t^2) = 6te^(t^2)

Given
y'(t)+p(t)y=0
=>
p(t)=y'(t)/y(t)=-2t

The original DE becomes
y'(t)-2ty(t)=0

If y(0)=y_0, then
y'(0)-2(0)y_0=0
which is always true irrespective of the value of y_0, since
y'(0)=6te^(t^2)≡0, and
-2(0)y_0≡0

So p(t)=-2t, and y_0 ∈R can be any value.