SOLUTION: MATH 281 University of Bridgeport Differential Equations Math Project

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Attached.

QUESTION 1
63𝑒 𝑑 βˆ’ 59𝑦𝑦 β€² = 0
63𝑒 𝑑 βˆ’ 59𝑦𝑦 β€² = 0
𝑁(𝑦)𝑦 β€² = 𝑀(π‘₯); 𝐴 π‘“π‘–π‘Ÿπ‘ π‘‘ π‘œπ‘Ÿπ‘‘π‘’π‘Ÿ π‘ π‘’π‘π‘Žπ‘Ÿπ‘Žπ‘π‘™π‘’ 𝑂𝐷𝐸 β„Žπ‘Žπ‘  π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘š π‘œπ‘“ 𝑁(𝑦)𝑦 β€² = 𝑀(π‘₯)
𝑁(𝑦) = 59𝑦; 𝑀(𝑑) = 63𝑒 𝑑 (set the equations)
59𝑦𝑦 β€² = 63𝑒 𝑑 (transmute in the constitute of 𝑁(𝑦)𝑦 β€² = 𝑀(π‘₯)) -> Integrate
59𝑦 2
2

= 63𝑒 𝑑 + 𝑐1 (Solve for y)
126𝑒 𝑑 +2𝑐1
,
59

𝑦=√

126𝑒 𝑑 +2𝑐1
59

𝑦 = βˆ’βˆš

(clear-up for 𝑐1 )

Substitute 𝑦(2) = 2 to get 𝑐1
126𝑒 2 + 2𝑐1
2=√
59
𝑐1 =

236 βˆ’ 126𝑒 2
2

𝑐1 = βˆ’695.02
126𝑒 𝑑 βˆ’695.02
59

𝑦= √

(π‘Ÿπ‘’π‘π‘™π‘Žπ‘π‘’ 𝑐1 π‘€π‘–π‘‘β„Ž π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘π‘œπ‘šπ‘π‘’π‘‘π‘’π‘‘ π‘Žπ‘π‘œπ‘£π‘’)

QUESTION 2
𝑦 β€²β€² βˆ’ 35𝑦 β€² + 63𝑦 = 0
Second regulate straight, homogeneous ODE π‘Žπ‘¦ β€²β€² + 𝑏𝑦 β€² + 𝑐𝑦 = 0
Assume a disruption of the constitute 𝑦 = 𝑒 𝛾𝑑
𝑦 β€² = 𝛾𝑒 𝛾𝑑
𝑦′′ = 𝛾 2 𝑒 𝛾𝑑
The new equation:
𝑒 𝛾𝑑 (𝛾 2 βˆ’ 35𝛾 + 63) = 0
(𝛾 2 βˆ’ 35𝛾 + 63) = 0
(𝛾) =

35+√973
2

; (𝛾) =

35βˆ’βˆš973
2

(There were 2 roots that are illmatched)

𝑦1 = 𝑒

35+√973
2

𝑑

; 𝑦2 = 𝑒

35βˆ’βˆš973
2

𝑑

𝑦(𝑑) = 𝑐1 𝑦1 (𝑑) + 𝑐2 𝑦2 (𝑑) (General Equation)
𝑦(𝑑) = 𝑐1 𝑒

35+√973
𝑑
2

+ 𝑐2 𝑒

35βˆ’βˆš973
𝑑
2

𝑦′(𝑑) =

35+√973
35βˆ’βˆš973
35 + √973
35 βˆ’ √973
𝑑
𝑐1 𝑒 2
+
𝑐2 𝑒 2
2
2

𝑦 β€²(0) =

35+√973
35βˆ’βˆš973
35 + √973
35 βˆ’ √973
𝑑
𝑐1 𝑒 2
+
𝑐2 𝑒 2
2
2

𝑑

𝑑

𝑦 β€² (0) = 𝐻 = 1.60
1.6 = 33.10𝑐1 𝑒

35+√973
(0)
2

+ βˆ’1.90𝑐2 𝑒

35...

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