Question: Consider the initial value problem y ' - 3 2 y = 3 t + 2 e t , y ( 0 ) = y

Consider the initial value problem
y'-32y=3t+2et,y(0)=y0
Find the value of y0 that separates solutions that grow positively as t from those that grow negatively. How does the solution that corresponds to this critical value of y0 behave as t?
Consider the initial value problem y ' - 3 2 y =

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