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प्रश्न
Obtain the differential equation by eliminating the arbitrary constants from the following equation:
y = `"Ae"^(3"x" + 1) + "Be"^(- 3"x" + 1)`
उत्तर
y = `"Ae"^(3"x" + 1) + "Be"^(- 3"x" + 1)` ....(1)
Differentiating twice w.r.t. x, we get
`"dy"/"dx" = "Ae"^(3"x" + 1) * "d"/"dx" (3"x" + 1) + "Be"^(- 3"x" + 1) * "d"/"dx" (- 3"x" + 1)`
`= "Ae"^(3"x" + 1) xx (3 + 1 + 0) + "Be"^(- 3"x" + 1) xx (- 3 xx 1 + 0)`
`= 3"Ae"^(3"x" + 1) - 3 "Be"^(- 3"x" + 1)`
and `("d"^2 "y")/"dx"^2 = 3 "Ae"^(3"x" + 1) * "d"/"dx" ("3x" + 1) - 3 "Be"^(- 3"x" + 1) * "d"/"dx" (- 3"x" + 1)`
`= 3"Ae"^(3"x" + 1) xx (3 xx 1 + 0) - 3"Be"^(- 3"x" + 1) xx (- 3 xx 1 + 0)`
`= 9"Ae"^(3"x" + 1) - 9"Be"^(- 3"x" + 1)`
`= 9 ("Ae"^(3"x" + 1) + "Be"^(- 3"x" + 1))`
= 9y ....[By (1)]
∴ `("d"^2"y")/"dx"^2 - 9"y" = 0`
This is the required D.E.
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