Question

2. Consider the differential equation dy
dt = t
y .
(a) Show by substitution into the equation that y = t is a solution, satisfying the
initial condition y(1) = 1. Over what range of t is this solution valid?
(b) Find the solution satisfying the intitial condition y(0) = 2.

Answer 1

im not sure what the equation is
is it dy/dt = ty ?

Answer 2

t-y

Answer 3

sorry t/y

Answer 4

dy/dt =t/y
ydy= tdt
integrate both sides
y^2/2=t^2/2+c
y^2=t^2+C
given y(1)=1
1=1+c
so c=0
y^2=t^2
y=t this is valid for t>=0
2nd part
y(0)=2
2=0+c
so
y^2=t^2+2