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Q10 E

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Found in: Page 14

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.${\mathbf{y}}{\mathbf{-}}{{\mathbf{log}}}_{{\mathbf{e}}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{1}}$, $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{=}}\frac{\mathbf{2}\mathbf{xy}}{\mathbf{y}\mathbf{-}\mathbf{1}}$

The given relation is an implicit solution to the given differential equation.

See the step by step solution

## Step1: Differentiating the given relation

As, in the given relation $\mathrm{y}-{\mathrm{log}}_{\mathrm{e}}\mathrm{y}={\mathrm{x}}^{2}+1$, y is defined implicitly as the function of x, so by using implicit differentiation, we will differentiate the given relation concerning x,

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}-{\mathrm{log}}_{\mathrm{e}}\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}+1\right)\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}$

## Step 2: Simplification of the differential equation obtained in step 1

$\left(1-\frac{1}{\mathrm{y}}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}\phantom{\rule{0ex}{0ex}}\left(\frac{\mathrm{y}-1}{\mathrm{y}}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{xy}}{\mathrm{y}-1}$

Which is identical to the given differential equation.

Thus, the relation ${\mathbf{y}}{\mathbf{-}}{{\mathbf{log}}}_{{\mathbf{e}}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{1}}$ is an implicit solution to the differential equation $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{=}}\frac{\mathbf{2}\mathbf{xy}}{\mathbf{y}\mathbf{-}\mathbf{1}}$.

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