SOLUTION: MATH 281 University of Bridgeport Differential Equations Math Project

Attached.

QUESTION 1
63π π‘ β 59π¦π¦ β² = 0
63π π‘ β 59π¦π¦ β² = 0
π(π¦)π¦ β² = π(π₯); π΄ ππππ π‘ πππππ π ππππππππ ππ·πΈ βππ  π‘βπ ππππ ππ π(π¦)π¦ β² = π(π₯)
π(π¦) = 59π¦; π(π‘) = 63π π‘ (set the equations)
59π¦π¦ β² = 63π π‘ (transmute in the constitute of π(π¦)π¦ β² = π(π₯)) -> Integrate
59π¦ 2
2

= 63π π‘ + π1 (Solve for y)
126π π‘ +2π1
,
59

π¦=β

126π π‘ +2π1
59

π¦ = ββ

(clear-up for π1 )

Substitute π¦(2) = 2 to get π1
126π 2 + 2π1
2=β
59
π1 =

236 β 126π 2
2

π1 = β695.02
126π π‘ β695.02
59

π¦= β

(πππππππ π1 π€ππ‘β π‘βπ π£πππ’π πππππ’π‘ππ ππππ£π)

QUESTION 2
π¦ β²β² β 35π¦ β² + 63π¦ = 0
Second regulate straight, homogeneous ODE ππ¦ β²β² + ππ¦ β² + ππ¦ = 0
Assume a disruption of the constitute π¦ = π πΎπ‘
π¦ β² = πΎπ πΎπ‘
π¦β²β² = πΎ 2 π πΎπ‘
The new equation:
π πΎπ‘ (πΎ 2 β 35πΎ + 63) = 0
(πΎ 2 β 35πΎ + 63) = 0
(πΎ) =

35+β973
2

; (πΎ) =

35ββ973
2

(There were 2 roots that are illmatched)

π¦1 = π

35+β973
2

π‘

; π¦2 = π

35ββ973
2

π‘

π¦(π‘) = π1 π¦1 (π‘) + π2 π¦2 (π‘) (General Equation)
π¦(π‘) = π1 π

35+β973
π‘
2

+ π2 π

35ββ973
π‘
2

π¦β²(π‘) =

35+β973
35ββ973
35 + β973
35 β β973
π‘
π1 π 2
+
π2 π 2
2
2

π¦ β²(0) =

35+β973
35ββ973
35 + β973
35 β β973
π‘
π1 π 2
+
π2 π 2
2
2

π‘

π‘

π¦ β² (0) = π» = 1.60
1.6 = 33.10π1 π

35+β973
(0)
2

+ β1.90π2 π

35...

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