Question

Obtain the differential equation by eliminating the arbitrary constants from the following equation:

y = `sqrt("a" cos (log "x") + "b" sin (log "x"))`

Answer 1

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.