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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS3 Marks
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} $ $(\text{x} \neq 0 \ \text{and} \ \text{x} > \text{y} \ \text{or} \ \text{x} < – \text{y})$
✓

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. $(ii) = xy' = x (x \cos x + \sin x) = x^2 \cos x + x \sin x$
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)]
$= \text{x}\ \text{sin}\ \text{x}+ \text{x} \sqrt{\text{x}^2(1-\text{sin}^2\text{x})} = \text{x}\ \text{sin}\ \text{x}+\text{x}\sqrt{\text{x}^2\text{cos}^2\text{x}}$
$= x \sin x + x.x \cos x = x \sin x + x^2 \cos x$
$= x^2 \cos x + x \sin x$
$\therefore$ 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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