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Question
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"`
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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