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प्रश्न
Obtain the differential equation by eliminating the arbitrary constants from the following equation:
y = `sqrt("a" cos (log "x") + "b" sin (log "x"))`
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उत्तर
y = `sqrt("a" cos (log "x") + "b" sin (log "x"))`
∴ y2 = a cos (log x) + b sin (log x) ....(1)
Differentiating both sides w.r.t. x, we get
`"2y" "dy"/"dx" = "a" "d"/"dx" [cos (log "x")] + "b" "d"/"dx" [sin (log "x")]`
`= "a" [ - sin (log "x")] * "d"/"dx" (log "x") + "b" cos (log "x") * "d"/"dx" (log "x")`
`= - "a" sin (log "x") xx 1/"x" + "b" cos (log "x") xx 1/"x"`
∴ `"2xy" "dy"/"dx" = - "a" sin (log "x") + "b" cos (log "x")`
Differentiating again w.r.t. x, we get
`2 ["xy" * "d"/"dx" ("dy"/"dx") + "dy"/"dx" * "d"/"dx" ("xy")]`
`= - "a" "d"/"dx" [sin (log "x")] + "b" "d"/"dx" [cos (log "x")]`
∴ `2 ["xy" ("d"^2"y")/"dx"^2 + "dy"/"dx" ("x" "dy"/"dx" + "y" xx 1)]`
`= - "a" cos (log "x") * "d"/"dx" (log "x") + "b"[- sin (log "x")] * "d"/"dx" (log "x")`
∴ `2"xy" ("d"^2"y")/"dx"^2 + 2"x" ("dy"/"dx")^2 + "2y" "dy"/"dx"
`= - "a" cos (log "x") xx 1/"x" - "b" sin (log "x") xx 1/"x"`
∴ `2"x"^2"y" ("d"^2"y")/"dx"^2 + 2"x"^2("dy"/"dx")^2 + 2"xy" "dy"/"dx"`
`= -["a" cos (log "x") + "b" sin (log "x")] = - "y"^2` ......[By (1)]
∴ `2"x"^2"y" ("d"^2"y")/"dx"^2 + 2"x"^2 ("dy"/"dx")^2 + 2"xy" "dy"/"dx" + "y"^2 = 0`
This is the required D.E.
Notes
The answer in the textbook is incorrect.
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