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IVPs: Consider the diﬀerential equation dy

dt =t

t2y+y.

(a) Find the general solution for this DE.

(b) Find the solution for this DE that passes though the point (1,2).

(a) First, we may separate variables and integrate

dy

dt =t

t2y+y,

dy

dt =t

y(t2+ 1),

y dy =t

t2+ 1 dt,

Zy dy =Zt

t2+ 1 dt.

Integrating yields

1

2y2=1

2ln |t2+ 1|+c,

y(t) = ±pC+ ln(t2+ 1), C = 2c.

(b) We have an initial condition y(1) = 2, so we take the positive solution from above:

2 = pC+ ln [(1)2+ 1],

2 = √C+ ln 2,

C= 4 −ln 2.

Substituting back into the general solution gives

y(t) = p4−ln 2 + ln(t2+ 1),

=s4 + ln 1

2(t2+ 1)

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Picard’s Theorem: T/F + Explanation: Both conditions of Picard’s Theorem hold for the following IVP:

dv

dt =kv2/3,

v(3) = 4.

If we treat this DE as v′=f(t, v), we have f(t, v) = kv2/3. This function is continuous for all values of tand

v, and so Picard’s Theorem guarantees that a solution exists for any initial condition.

Considering ∂f

∂v =fv(t, v) = 2k

3v1/3, we see that this function is discontinuous at v= 0. However, given our

initial condition v(3) = 4, it is possible to draw a rectangle around the IC such that fyis continous everywhere in

the rectangle.

(In other words, we can draw a rectangle that doesn’t intersect v= 0).

t

v

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