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प्रश्न
In the following example verify that the given function is a solution of the differential equation.
`"xy" = "ae"^"x" + "be"^-"x" + "x"^2; "x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" + "x"^2 = "xy" + 2`
उत्तर
`"xy" = "ae"^"x" + "be"^-"x" + "x"^2`
∴ `"xy" - "x"^2 = "ae"^"x" + "be"^-"x"` ....(1)
Differentiating both sides w.r.t. x, we get
`"x" "dy"/"dx" + "y" * "d"/"dx" ("x") - "2x" = "ae"^"x" + "be"^-"x" xx (- 1)`
∴ `"x" "dy"/"dx" + "y" - 2"x" = "ae"^"x" - "be"^-"x"`
Differentiating again w.r.t. x, we get
`"x" * "d"/"dx" ("dy"/"dx") + "dy"/"dx" * "d"/"dx" ("x") + "dy"/"dx" - 2 xx 1 = "ae"^"x" - "be"^-"x" (- 1)`
∴ `"x" ("d"^2"y")/"dx"^2 + "dy"/"dx" xx 1 + "dy"/"dx" - 2 = "ae"^"x" + "be"^-"x"`
∴ `"x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" - 2 = "xy" - "x"^2` ....[By (1)]
∴ `"x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" + "x"^2 = "xy" + 2`
Hence, xy = `"ae"^"x" - "be"^-"x" + "x"^2` is a solution of the D.E.
`"x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" + "x"^2 = "xy" + 2`
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