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