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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:
$\text{y}=\sqrt{\text{a}^2-\text{x}^2}\text{x} \in(-\text{a, a}) :\ \text{x+y}\frac{\text{dy}}{\text{dx}}=\ 0 (\text{y}\neq0)$
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Answer

Given: $\text{y}=\sqrt{\text{a}^2-\text{x}^2},\ \text{x}\in(-\text{a, a}) \ ....(\text{i})$

To prove: y given by eq. (i) is a solution of differential equation $\text{x+y}\frac{\text{dy}}{\text{dx}} = 0 \ ....(\text{ii})$

Proof: From eq. (i), $\frac{\text{dy}}{\text{dx}}=\frac{1}{2}(\text{a}^2-\text{x}^2)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{a}^2-\text{x}^2)=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}(-2\text{x})$

$= \frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\ .....(\text{iii})$

Putting the values of y and $\frac{\text{dy}}{\text{dx}}$ from eq. (i) and (iii) in L.H.S. of eq. (ii),

$\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}+\sqrt{\text{a}^2-\text{x}^2}\bigg(\frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\bigg)$ $=\text{x}-\text{x}=0=\text{R.H.S. of eq.(ii)}$

Hence, Function given by eq. (i) is a solution of $\text{x+y}\frac{\text{dy}}{\text{dx}}=0.$

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