Q10 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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.

$\mathbf{y}\mathbf{-}{\mathbf{log}}_{\mathbf{e}}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{2}\mathbf{xy}}{\mathbf{y}\mathbf{-}\mathbf{1}}$

Step-by-Step Solution

Answer

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

Step1: Differentiating the given relation

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

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}-{\log }_{\mathrm{e}}\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2+1\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x} \end{gathered}$$

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

$$\begin{gathered} \left(1-\frac{1}{\mathrm{y}}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x} \\ \left(\frac{\mathrm{y}-1}{\mathrm{y}}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{xy}}{\mathrm{y}-1} \end{gathered}$$

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 $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{2}\mathbf{xy}}{\mathbf{y}\mathbf{-}\mathbf{1}}$.