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.

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

Short answer

Yes, the given relation is an implicit solution to the given differential equation.

Step-by-step solution

Differentiate the given relation

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

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{xy}}+\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-1\right) \\ {\mathrm{e}}^{\mathrm{xy}}\left(\mathrm{y}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\right)+\frac{\mathrm{dy}}{\mathrm{dx}}=1 \\ \frac{\mathrm{dy}}{\mathrm{dx}}[\mathrm{x}{\mathrm{e}}^{\mathrm{xy}}+1]=1-\mathrm{y}{\mathrm{e}}^{\mathrm{xy}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-{\mathrm{ye}}^{\mathrm{xy}}}{1+{\mathrm{xe}}^{\mathrm{xy}}} \end{gathered}$$

Simplification of the differential equation

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-{\mathrm{ye}}^{\mathrm{xy}}}{1+{\mathrm{xe}}^{\mathrm{xy}}}\times \frac{{\mathrm{e}}^{-\mathrm{xy}}}{{\mathrm{e}}^{-\mathrm{xy}}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{e}}^{-\mathrm{xy}}-\mathrm{y}}{{\mathrm{e}}^{-\mathrm{xy}}+\mathrm{x}} \end{gathered}$$

Which is identical to the given differential equation.

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