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Question
In the following example verify that the given function is a solution of the differential equation.
`"x"^2 = "2y"^2 log "y", "x"^2 + "y"^2 = "xy" "dx"/"dy"`
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Solution
`"x"^2 = "2y"^2 log "y"` .....(1)
Differentiating both sides w.r.t. y, we get
`"2x" "dx"/"dy" = 2 "d"/"dy" ("y"^2 log "y")`
`= 2 ["y"^2 "d"/"dy" (log "y") + (log "y") * "d"/"dy"("y"^2)]`
`= 2 ["y"^2 xx 1/"y" + (log "y") xx "2y"]`
∴ `"x" "dx"/"dy" = "y" + 2"y" log "y"`
∴ `"xy" "dx"/"dy" = "y"^2 + 2"y"^2 log "y"`
= y2 + x2 ....[By (1)]
∴ `"x"^2 + "y"^2 = "xy" "dx"/"dy"`
Hence, x2 = 2y2 log y is a solution of the D.E.
`"x"^2 + "y"^2 = "xy" "dx"/"dy"`
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