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Question
In the following example verify that the given expression is a solution of the corresponding differential equation:
y = `"a" + "b"/"x"; "x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" = 0`
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Solution
y = `"a" + "b"/"x"`
Differentiating w.r.t. x, we get
`"dy"/"dx" = 0 + "b"(- 1/"x"^2) = - "b"/"x"^2`
∴ `"x"^2 "dy"/"dx" = - "b"`
Differentiating again w.r.t. x, we get
`"x"^2 * "d"/"dx" ("dy"/"dx") + "dy"/"dx" * "d"/"dx" ("x"^2) = 0`
∴ `"x"^2 ("d"^2"y")/"dx"^2 + "dy"/"dx" xx "2x" = 0`
∴ `"x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" = 0`
Hence, y = `"a" + "b"/"x"` is a solution of the D.E.
`"x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" = 0`
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