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**In the following example verify that the given expression is a solution of the corresponding differential equation:**

y = `"e"^"ax"; "x" "dy"/"dx" = "y" log "y"`

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#### Solution

y = `"e"^"ax"`

∴ log y = log `"e"^"ax"` = ax log e

∴ log y = ax .....(1) .....[∵ log e = 1]

Differentiating w.r.t. x, we get

`1/"y" * "dy"/"dx" = "a" xx 1`

∴ `"dy"/"dx" = "ay"`

∴ `"x""dy"/"dx" = ("ax")"y"`

∴ `"x" "dy"/"dx" = "y" log "y"` ....[By (1)]

Hence, y = `"e"^"ax"` is a solution of the D.E. `"x" "dy"/"dx" = "y" log "y"`.

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