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प्रश्न
Obtain the differential equation by eliminating the arbitrary constants from the following equation:
c1x3 + c2y2 = 5
उत्तर
c1x3 + c2y2 = 5 .....(1)
Differentiating twice w.r.t. x, we get
`"c"_1 xx "3x"^2 + "c"_2 xx "2y" "dy"/"dx" = 0`
∴ `3"c"_1"x"^2 + 2"c"_2"y" "dy"/"dx" = 0` ....(2)
Differentiating again w.r.t. x, we get
`3"c"_1 xx "2x" + 2"c"_2 ["y"."d"/"dx"("dy"/"dx") + "dy"/"dx" * "dy"/"dx"] = 0`
∴ `6"c"_1"x" + 2"c"_2 ["y" ("d"^2"y")/"dx"^2 + ("dy"/"dx")^2] = 0`
The equations (1), (2) and (3) in c1, c2 are consistent.
∴ determinant of their consistency is zero.
∴ `|("x"^3, "y"^2, 5),("3x"^2, "2y""dy"/"dx", 0),(6"x", "2y" ("d"^2"y")/"dx"^2 + 2("dy"/"dx")^2, 0)|`
∴ `"x"^3 (0 - 0) - "y"^2(0 - 0) + 5["6x"^2"y" ("d"^2"y")/"dx"^2 + "6x"^2("dy"/"dx")^2 - 12"xy" "dy"/"dx"] = 0`
∴ `"6x"^2"y" ("d"^2"y")/"dx"^2 + "6x"^2("dy"/"dx")^2 - 12"xy" "dy"/"dx" = 0`
∴ `"xy" ("d"^2"y")/"dx"^2 + "x"("dy"/"dx")^2 - "2y" "dy"/"dx" = 0`
This is the required D.E.
Notes
The answer in the textbook is incorrect.
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