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

       y = x\sin x\qquad :\ xy' = y + x\sqrt{x^2 - y^2}\ (x\neq 0\ \textup{and} \ x > y\ or \ x < -y)

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Given,

y = x\sin x

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

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(xsinx) = sinx + xcosx

Substituting the values of y' in LHS,

xy' = x(sinx + xcosx)

Substituting the values of y in RHS.

\\xsinx + x\sqrt{x^2 - x^2sin^2x} = xsinx + x^2\sqrt{1-sinx^2} = x(sinx+xcosx) = LHS

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

Posted by

HARSH KANKARIA

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