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Question
Solve the differential equation `"dy"/"dx" + 1` = ex + y.
Solution
Given that: `"dy"/"dx" + 1` = ex + y
Put x + y = t
∴ `1 + "dy"/"dx" = "dt"/"dx"`
∴ `"dt"/"dx"` = et
⇒ `"dt"/"e"^"t"` = dx
⇒ `"e"^-"t" "dt"` = dx
Integrating both sides, we have
`int"e"^-1 "dt" = int "d"x`
⇒ `-"e"^"t"` = x + c
⇒ `-"e"^(-(x + y)` = x + c
⇒`(-1)/"e"^(x + y)` = x + c
⇒ (x + c)ex + y = –1
Hence, the required solution is (x + c).ex + y + 1 = 0.
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