differential equation Lecture#10

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Transcript of differential equation Lecture#10

DIFFERENTIAL EQUATION (MT-202) SYED AZEEM INAM

DIFFERENTIAL EQUATION (MT-202)

LECTURE #10

HIGHER ORDER DIFFERENTIAL EQUATIONS:

HOMOGENOUS LINEAR DIFFERENTIAL EQUATIONS:

Consider the equations

π‘Ž0

𝑑𝑛𝑦

𝑑π‘₯𝑛+ π‘Ž1

π‘‘π‘›βˆ’1𝑦

𝑑π‘₯π‘›βˆ’1+ β‹― + π‘Žπ‘›βˆ’1

𝑑𝑦

𝑑π‘₯+ π‘Žπ‘›π‘¦ = 0 (1)

Where π‘Ž0, π‘Ž1, … , π‘Žπ‘›βˆ’1,π‘Žπ‘› are real constants.

To find the solution consider exponential function. Her we have for 𝑦 = π‘’π‘šπ‘₯

𝑑𝑦

𝑑π‘₯= π‘šπ‘’π‘šπ‘₯,

𝑑2𝑦

𝑑π‘₯2= π‘š2π‘’π‘šπ‘₯, . . . ,

π‘‘π‘›βˆ’1𝑦

𝑑π‘₯π‘›βˆ’1= π‘šπ‘›βˆ’1π‘’π‘šπ‘₯,

𝑑𝑛𝑦

𝑑π‘₯𝑛= π‘šπ‘›π‘’π‘šπ‘₯

Substituting in (1)

π‘Ž0π‘šπ‘›π‘’π‘šπ‘₯ , π‘Ž1π‘šπ‘›βˆ’1π‘’π‘šπ‘₯ + β‹― + π‘Žπ‘›βˆ’1π‘šπ‘’π‘šπ‘₯ + π‘Žπ‘›π‘’π‘šπ‘₯ = 0

π‘’π‘šπ‘₯(π‘Ž0π‘šπ‘›, π‘Ž1π‘šπ‘›βˆ’1 + β‹― + π‘Žπ‘›βˆ’1π‘š + π‘Žπ‘›) = 0

Since π‘’π‘šπ‘₯ β‰  0

π‘Ž0π‘šπ‘›, π‘Ž1π‘šπ‘›βˆ’1 + β‹― + π‘Žπ‘›βˆ’1π‘š + π‘Žπ‘› = 0 (2)

Thus π‘š is a solution of (1)if and only if π‘š is the solution (2).

Eq→ (2) is called the characteristics (or auxiliary) equation of the given differential

equation (1)

Three cases arise as the roots (2) are

(i) Real and distinct

(ii) Real and repeated

(iii) Complex

CASE I: DISTINCT REAL ROOTS

Let π‘š1, π‘š2, π‘š3, … , π‘šπ‘› be the ′𝑛′ real distinct roots then

𝑦 = 𝑐1π‘’π‘š1π‘₯ + 𝑐2π‘’π‘š2π‘₯ + β‹― + π‘π‘›βˆ’1π‘’π‘šπ‘›βˆ’1π‘₯ + π‘π‘›π‘’π‘šπ‘›π‘₯

DIFFERENTIAL EQUATION (MT-202) SYED AZEEM INAM

DIFFERENTIAL EQUATION (MT-202)

CASE# II: REPEATED REAL ROOTS

Let π‘š1, π‘š2, π‘š3, … , π‘šπ‘› be the ′𝑛′ real distinct roots where π‘š1 = π‘š2then

𝑦 = (𝑐1 + 𝑐2)π‘’π‘š1π‘₯ + β‹― + π‘π‘›βˆ’1π‘’π‘šπ‘›βˆ’1π‘₯ + π‘π‘›π‘’π‘šπ‘›π‘₯

CASE III: COMPLEX ROOTS:

Let π‘š be a complex roots then

𝑦 = π‘’π‘Žπ‘₯(𝑐1𝑠𝑖𝑛𝑏π‘₯ + 𝑐2π‘π‘œπ‘ π‘π‘₯)

EXAMPLE #1: (𝐷2 + 4𝐷 + 3)𝑦 = 0

EXAMPLE #2: (𝐷3 βˆ’ 5𝐷2 + 7𝐷 βˆ’ 3)𝑦 = 0

EXAMPLE #3: (𝐷3 βˆ’ 𝐷2 + 𝐷 βˆ’ 1)𝑦 = 0

EXAMPLE #4: (𝐷2 + 𝐷 βˆ’ 12)𝑦 = 0

EXAMPLE #5: (𝐷2 + 4𝐷 + 5)𝑦 = 0

EXAMPLE #6: (𝐷3 βˆ’ 3𝐷2 + 4)𝑦 = 0

EXAMPLE#7: (9𝐷2 βˆ’ 12𝐷 + 4)𝑦 = 0

EXAMPLE #8: (75𝐷2 + 50𝐷 + 12)𝑦 = 0

EXAMPLE #9: (𝐷3 βˆ’ 4𝐷2 + 𝐷 + 6)𝑦 = 0

EXAMPLE #10: (𝐷3 βˆ’ 6𝐷2 + 12𝐷 βˆ’ 8)𝑦 = 0

EXAMPLE #11: (𝐷3 βˆ’ 27)𝑦 = 0

EXAMPLE #12: (4𝐷4 βˆ’ 4𝐷3 βˆ’ 3𝐷2 + 4𝐷 βˆ’ 1)𝑦 = 0