Question

Solve the differential equation:  ` ("x" + 1) (d"y")/(d"x") = 2e^-"y" - 1; y(0) = 0.`

Answer 1

`("x" + 1) (d"y")/(d"x") = 2e^-"y" - 1`

⇒ `(d"y")/(2e^-"y" - 1) = (d"x")/("x" + 1)`

⇒  `(e^"y" d"y")/(2 -e^"y") = (d"x")/("x" + 1)`

Integrating both sides, we get:

`int_  (e^"y" d"y")/(2 -e^"y") = log |"x" + 1| + log "C"`  .....(1)

Let `2 -e^"y" = t.`

∴ `(d)/(d"y") (2 - e^"y") = (dt)/(d"y")`

⇒ `-e^"y" = (dt)/(d"y")`

⇒ `e^"y" dt  = -dt`

Substituting ths value in equation (1), we get:

`int_  (-dt)/(t) = log|"x" + 1| + log "C" `

⇒ `-log|"r"| = log| "C" ("x" + 1)`

⇒ `-log|2 - e^"y"| = log |"C"("x" + 1)|`

⇒ `(1)/(2 - e^"y") = "C" ("x" + 1)`

⇒ `2 - e^"y" = (1)/("C"("x" + 1)`     ....(2)

Now, at x = 0 and y = 0, equation (2) becomes:

⇒  `2 - 1 = (1)/("C")`

⇒ `"C" = 1`

Substituting C = 1 in equation (2), we get:

`2 -e^"y" = (1)/("x" + 1)`

⇒ `e^"y" = 2 -(1)/("x" + 1)`

⇒ `e^"y" = (2"x" + 2 - 1)/("x" + 1)`

⇒ `e^"y" = (2"x" + 1)/("x" +1)`

⇒ `"y" log|(2"x" + 1)/("x" + 1)|. ("x" ≠ - 1) `

This is the required particular solution of the given differential equation.