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Sum

**In the following example verify that the given expression is a solution of the corresponding differential equation:**

y = x^{m}; `"x"^2 ("d"^2"y")/"dx"^2 - "mx" "dy"/"dx" + "my" = 0`

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#### Solution

y = x^{m }

Differentiating twice w.r.t. x, we get

`"dy"/"dx" = "d"/"dx" ("x"^"m") = "mx"^("m - 1")`

and `("d"^2"y")/"dx"^2 = "d"/"dx" ("mx"^("m - 1")) = "m" "d"/"dx" ("x"^("m - 1")) = "m"("m" - 1) "x"^("m - 2")`

∴ `"x"^2 ("d"^2"y")/"dx"^2 - "mx" "dy"/"dx" + "my"`

`= "x"^2 * "m"("m" - 1) "x"^("m - 2") - "mx" * "mx"^("m" - 1) + "m" * "x"^"m"`

`= "m"("m - 1") "x"^"m" - "m"^2 "x"^"m" + "mx"^"m"`

`= ("m"^2 - "m" - "m"^2 + "m")"x"^"m" = 0`

This shows that y = x^{m} is a solution of the D.E.

`"x"^2 ("d"^2"y")/"dx"^2 - "mx" "dy"/"dx" + "my" = 0`.

Concept: Formation of Differential Equations

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