Q30 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Implicit Function Theorem. Let $\mathbf{G}\left(x,y\right)$ have continuous first partial derivatives in the rectangle $\mathbf{R}\mathbf{=}\left\{\left(x,y\right):a<x<b,c<y<d\right\}$ containing the point $(x_0,y_0)$. If $\mathbf{G}\left(x_0,y_0\right)\mathbf{=}\mathbf{0}$ and the partial derivative ${\mathbf{G}}_{\mathbf{y}}\left(x_0,y_0\right)\mathbf{\ne }\mathbf{0}$, then there exists a differentiable function $\mathbf{y}\mathbf{=}\mathbf{\varphi }\left(x\right)$, defined in some interval $\mathbf{I}\mathbf{=}\left(x_0-\delta ,y_0+\delta \right)$, that satisfies G for all for $\mathbf{G}\left(x,\varphi \left(x\right)\right)$ all $\mathbf{x}\mathbf{\in }\mathbf{I}$.

The implicit function theorem gives conditions under which the relationship $\mathbf{G}\left(x,y\right)\mathbf{=}\mathbf{0}$ implicitly defines y as a function of x. Use the implicit function theorem to show that the relationship $\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{xy}}\mathbf{=}\mathbf{0}$ given in Example 4, defines y implicitly as a function of x near the point $\left(0,-1\right)$.

Step-by-Step Solution

Answer

The given relationship implicitly defines y as a function of x near the point $\left(0,-1\right)$.

Step 1: Finding the partial derivative of the given relation concerning y

$\frac{\partial }{\partial \mathrm{y}}\left(\mathrm{x}+\mathrm{y}+{\mathrm{e}}^{\mathrm{xy}}\right)=1+\mathrm{x} {\mathrm{e}}^{\mathrm{xy}}$

Step 2: Determining whether the given relation defines y implicitly as a function of x or not

As one sees in Step 1, the partial derivative of the given relation is not equal to zero for all x in the interval $\left(0.-1\right)$. 

Hence, the given relationship $\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{xy}}\mathbf{=}\mathbf{0}$ defines y implicitly as a function of x near the point $\left(0,-1\right)$.