In each of the verify that the given function (explicit or implic… - Vidyadip

Question

In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y} = \text{x} \ \text{sin} \ \text{x}\ : \ \text{xy}' = \text{y}+\text{x}\sqrt{\text{x}^2-\text{y}^2} $ $(\text{x} \neq 0 \ \text{and} \ \text{x} > \text{y} \ \text{or} \ \text{x} < – \text{y})$

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Answer

Given: y = x sin x .....(i)

To prove: y given by eq. (i) is a solution of differential equation $\text{xy}'=\text{y}+\text{x} \sqrt{\text{x}^2-\text{y}^2}\ ....(\text{ii})$

Proof: From eq. (i), $\frac{\text{dy}}{\text{dx}}(=\text{y}') = \text{x} \frac{\text{d}}{\text{dx}}\text{sin}\ \text{x}+ \text{sin}\ \text{x} \frac{\text{d}}{\text{dx}} \text{x} = \text{x}\ \text{cos}\ \text{x}\ + {\text{sin}} \ \text{x}$

L.H.S. of eq. (ii) = xy' = x (x cos x + sin x) = x² cos x + x sin x

R.H.S. of eq. (ii) $= \text{y} + \text{x}\sqrt{\text{x}^2-\text{y}^2} = \text{x}\ \text{sin} \ \text{x}+\text{x} \sqrt{\text{x}^2-\text{x}^2\ \text{sin}^2\text{x}}$ [From eq. (i)]

$= \text{x}\ \text{sin}\ \text{x}+ \text{x} \sqrt{\text{x}^2(1-\text{sin}^2\text{x})} = \text{x}\ \text{sin}\ \text{x}+\text{x}\sqrt{\text{x}^2\text{cos}^2\text{x}}$

= x sin x + x.x cos x = x sin x + x² cos x

= x² cos x + x sin x

$\therefore$ L.H.S. = R.H.S

Hence, y given by eq. (i) is a solution of $\text{xy}'=\text{y}+\text{x}\sqrt{\text{x}^2-\text{y}^2}.$

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