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Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: y = x sin x : xy' = y+x x2-y2 (x ≠ 0 and x > y or x < -y) - Mathematics

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Question

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = x sin x : xy' = `y + x  sqrt (x^2 - y^2)`  (x ≠ 0 and x > y or x < -y)

Sum
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Solution

y = x sin x …(i)

y’ = x cos x + sin x

and cos x + `sqrt(1 - sin^2 x) = sqrt((1 - y^2)/x^2)`

`= sqrt(x^2 - y^2)/x`

Hence,  y' = x. `sqrt(x^2 - y^2)/x +y/x`

`xy' = x sqrt(x^2 - y^2) + y`

The given function is the solution of the given differential equation.

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Chapter 9: Differential Equations - Exercise 9.2 [Page 385]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 9 Differential Equations
Exercise 9.2 | Q 6 | Page 385

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