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प्रश्न
Obtain the differential equation by eliminating the arbitrary constants from the following equation:
y = Ae5x + Be-5x
उत्तर
y = Ae5x + Be-5x ...(1)
Differentiating twice w.r.t. x, we get
`"dy"/"dx" = "Ae"^"5x" xx 5 + "Be"^(- "5x") xx (- 5)`
∴ `"dy"/"dx" = 5"Ae"^"5x" - 5"Be"^(- "5x")`
and `("d"^2"y")/"dx"^2 = "Ae"^"5x" xx 5 + "Be"^(- "5x") xx (- 5)`
`= 25"Ae"^"5x" + 25"Be"^(- "5x")`
`= 25("Ae"^"5x" + "Be"^(- "5x")) = 25"y"` ....[By(1)]
∴ `("d"^2"y")/"dx"^2 - 25"y" = 0`
This is the required D.E.
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