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

        y = \sqrt{a^2 - x^2}\ x\in (-a,a)\qquad : \ x + y \frac{dy}{dx} = 0\ (y\neq 0)

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Given,

y = \sqrt{a^2 - x^2}

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{a^2 - x^2}) = \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}

Substituting the values of y and y' in LHS,

x + y \frac{dy}{dx} = x + (\sqrt{a^2 - x^2})(\frac{-x}{\sqrt{a^2 - x^2}}) = 0 = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Posted by

HARSH KANKARIA

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