# SOLUTION: MATH 281 University of Bridgeport Differential Equations Math Project Attached.

QUESTION 1
63𝑒 𝑡 − 59𝑦𝑦 ′ = 0
63𝑒 𝑡 − 59𝑦𝑦 ′ = 0
𝑁(𝑦)𝑦 ′ = 𝑀(𝑥); 𝐴 𝑓𝑖𝑟𝑠𝑡 𝑜𝑟𝑑𝑒𝑟 𝑠𝑒𝑝𝑎𝑟𝑎𝑏𝑙𝑒 𝑂𝐷𝐸 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑜𝑓 𝑁(𝑦)𝑦 ′ = 𝑀(𝑥)
𝑁(𝑦) = 59𝑦; 𝑀(𝑡) = 63𝑒 𝑡 (set the equations)
59𝑦𝑦 ′ = 63𝑒 𝑡 (transmute in the constitute of 𝑁(𝑦)𝑦 ′ = 𝑀(𝑥)) -> Integrate
59𝑦 2
2

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

𝑦=√

126𝑒 𝑡 +2𝑐1
59

𝑦 = −√

(clear-up for 𝑐1 )

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

236 − 126𝑒 2
2

𝑐1 = −695.02
126𝑒 𝑡 −695.02
59

𝑦= √

(𝑟𝑒𝑝𝑙𝑎𝑐𝑒 𝑐1 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝑎𝑏𝑜𝑣𝑒)

QUESTION 2
𝑦 ′′ − 35𝑦 ′ + 63𝑦 = 0
Second regulate straight, homogeneous ODE 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0
Assume a disruption of the constitute 𝑦 = 𝑒 𝛾𝑡
𝑦 ′ = 𝛾𝑒 𝛾𝑡
𝑦′′ = 𝛾 2 𝑒 𝛾𝑡
The new equation:
𝑒 𝛾𝑡 (𝛾 2 − 35𝛾 + 63) = 0
(𝛾 2 − 35𝛾 + 63) = 0
(𝛾) =

35+√973
2

; (𝛾) =

35−√973
2

(There were 2 roots that are illmatched)

𝑦1 = 𝑒

35+√973
2

𝑡

; 𝑦2 = 𝑒

35−√973
2

𝑡

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

35+√973
𝑡
2

+ 𝑐2 𝑒

35−√973
𝑡
2

𝑦′(𝑡) =

35+√973
35−√973
35 + √973
35 − √973
𝑡
𝑐1 𝑒 2
+
𝑐2 𝑒 2
2
2

𝑦 ′(0) =

35+√973
35−√973
35 + √973
35 − √973
𝑡
𝑐1 𝑒 2
+
𝑐2 𝑒 2
2
2

𝑡

𝑡

𝑦 ′ (0) = 𝐻 = 1.60
1.6 = 33.10𝑐1 𝑒

35+√973
(0)
2

+ −1.90𝑐2 𝑒

35...

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