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  • Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y = root of 1 + x^2 : y' = xy / 1 + x^2

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

        y = \sqrt{1 + x^2}\qquad :\ y' = \frac{xy}{1 + x^2}

Answers (1)

best_answer

Given,

y = \sqrt{1 + x^2}

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

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{1 + x^2}) = \frac{2x}{2\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}}

Substituting the values of y in RHS,

\frac{x\sqrt{1+x^2}}{1 + x^2} = \frac{x}{\sqrt{1+x^2}} = LHS.

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

Posted by

HARSH KANKARIA

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