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Sum

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

xy = log y +c; `"dy"/"dx" = "y"^2/(1 - "xy")`

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#### Solution

xy = log y +c

Differentiating w.r.t. x, we get

`"x" * "dy"/"dx" + "y" xx 1 = 1/"y" * "dy"/"dx" + 0`

∴ `"x" "dy"/"dx" + "y" = 1/"y" * "dy"/"dx"`

`("x" - 1/"y")"dy"/"dx" = - "y"`

∴ `(("xy" - 1)/"y") "dy"/"dx" = - "y"`

∴ `"dy"/"dx" = (- "y"^2)/("xy" - 1) = "y"^2/(1 - "xy")`, if xy ≠ 1

Hence, xy = log y + c is a solution of the D.E.

`"dy"/"dx" = "y"^2/(1 - "xy")`, if xy ≠ 1.

Concept: Formation of Differential Equations

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