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प्रश्न
In the following example verify that the given function is a solution of the differential equation.
`"y" = 3 "cos" (log "x") + 4 sin (log "x"); "x"^2 ("d"^2"y")/"dx"^2 + "x" "dy"/"dx" + "y" = 0`
उत्तर
`"y" = 3 "cos" (log "x") + 4 sin (log "x")` ...(1)
Differentiating both sides w.r.t. x, we get
`"dy"/"dx" = 3 "d"/"dx" [cos (log "x")] + 4 "d"/"dx" [sin (log "x")]`
`= 3 [- sin (log "x")] "d"/"dx" (log "x") + 4 cos (log "x") "d"/"dx" (log "x")`
`= - 3 sin (log "x") xx 1/"x" + 4 cos (log "x") xx 1/"x"`
∴ `"x" "dy"/"dx" = - 3 sin (log "x") + 4 cos (log "x")`
Differentiating again w.r.t. x, we get,
`"x" "d"/"dx" ("dy"/"dx") + "dy"/"dx" * "d"/"dx" ("x") = - 3 "d"/"dx" [sin (log "x")] + 4 "d"/"dx"[cos (log "x")]`
∴ `"x" ("d"^2"y")/"dx"^2 + "dy"/"dx" xx 1 = - 3 cos (log "x") * "d"/"dx" (log "x") + 4 [- sin (log "x")]* "d"/"dx" (log "x")`
∴ `"x" ("d"^2"y")/"dx"^2 + "dy"/"dx" = - 3 cos (log "x") xx 1/"x" - 4 sin (log "x") xx 1/"x"`
∴ `"x"^2 ("d"^2"y")/"dx"^2 + "x" "dy"/"dx" = - [3 cos (log "x") + 4 sin (log "x")] = - "y"` ...[By (1)]
∴ `"x"^2 ("d"^2"y")/"dx"^2 + "x" "dy"/"dx" + "y" = 0`
Hence, y = 3 cos (log x) + 4 sin (log x) is a solution of the D.E. `"x"^2 ("d"^2"y")/"dx"^2 + "x" "dy"/"dx" + "y" = 0`
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