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Sum

**Obtain the differential equation by eliminating the arbitrary constants from the following equation:**

y = Ae^{5x} + Be^{-5x}

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#### Solution

y = Ae^{5x} + Be^{-5x} ...(1)

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

`"dy"/"dx" = "Ae"^"5x" xx 5 + "Be"^(- "5x") xx (- 5)`

∴ `"dy"/"dx" = 5"Ae"^"5x" - 5"Be"^(- "5x")`

and `("d"^2"y")/"dx"^2 = "Ae"^"5x" xx 5 + "Be"^(- "5x") xx (- 5)`

`= 25"Ae"^"5x" + 25"Be"^(- "5x")`

`= 25("Ae"^"5x" + "Be"^(- "5x")) = 25"y"` ....[By(1)]

∴ `("d"^2"y")/"dx"^2 - 25"y" = 0`

This is the required D.E.

Concept: Formation of Differential Equations

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