MATH 3001 Lecture Notes - Lecture 3: Linear Map, Integrating FactorExam
Course CodeMATH 3001
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APPM 2360 Exam 1, Review 1 Solutions
IVPs: Consider the diﬀerential equation dy
(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
y dy =t
t2+ 1 dt,
Zy dy =Zt
t2+ 1 dt.
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
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Picard’s Theorem: T/F + Explanation: Both conditions of Picard’s Theorem hold for the following IVP:
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.
∂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
(In other words, we can draw a rectangle that doesn’t intersect v= 0).
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