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प्रश्न
Solve the following differential equation:
(D2 + D – 2)y = e3x + e–3x
उत्तर
The auxiliary equation is
m2 + m – 6 = 0
(m + 3)(m – 2) = 0
Roots are real and different
The complementary function is
C.F = Aem1x + Bem2x
C.F = Ae–3x + Be2x
P.I = `1/("d"^2 + "D" - 6) "e"^(3x)`
= `1/((3)^2 + 3 - 6) "e"^(3x)`
P.I = `"e"^(3x)/(9 + 3 - 6)`
P.I(1) = `"e"^(3x)/6`
P.I(2) = `1/("D"^2 + "D" - 6) "e"^(-3x)`
Replace D by – 3
D2 + D – 6 = 0
When D = – 3
∴ P.I2 = `x * 1/(2"D" + 1) "e"^(-3x)`
= `x * 1/(2(-3) + 1) "e"^(-3x)`
= `x * 1/(-5) "e"^(-3x)`
P.I2 = `(-x)/5 "e"^(-3x)`
The general solution is y = C.F = P.I1 + P.I2
y = `"Ae"^(-3x) + "Be"^(2x) + "e"^(3x)/6 - x/5 "e"^(-3x)`
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