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Question
Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.
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Solution
`"dy"/"dx"` = yex
⇒ `int "dy"/y = int "e"^x "d"x`
⇒ logy = ex + c
Substituting x = 0 and y = e
We get loge = e0+ c
i.e., c = 0 ....(∵ loge = 1)
Therefore, log y = ex.
Now, substituting x = 1 in the above
We get log y = e
⇒ y = ex.
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