हिंदी

For the given below, verify that the given function (implicit or explicit) is a solution to the corresponding differential equation. x2=2y2logy:(x2 +y2)dydx-xy=0

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प्रश्न

For the given below, verify that the given function (implicit or explicit) is a solution to the corresponding differential equation.

`x^2 = 2y^2 log y : (x^2  + y^2) dy/dx - xy = 0`

योग
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उत्तर

We have, `x^2 = 2y^2 log y`           ....(1)

Differentiating (1) w.r.t. x, we get

`2x = 2 [2y log y + y^2 xx 1/y] dy/dx`

`= 2 [2y log y + y] dy/dx`

⇒ `dy/dx = x/ (2y log y + y) = x/ (y (2 log y + 1))`

From (1), 2 log `y = x^2/y^2`

∴ `dy/dx = x/(y [x^2/y^2 + 1])`

`= (xy/(x^2 + y^2))`

⇒ `(x^2 + y^2) dy/dx - xy = 0`

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अध्याय 9: Differential Equations - Exercise 9.7 [पृष्ठ ४२०]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 9 Differential Equations
Exercise 9.7 | Q 2.4 | पृष्ठ ४२०

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