Question

For,$\mathit{u}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{y}$

(a) Verify that ${\mathbf{\partial }}^{\mathbf{2}}\mathit{u}\mathbf{/}\mathbf{\partial }\mathit{x}\mathbf{\partial }\mathit{y}\mathbf{=}{\mathbf{\partial }}^{\mathbf{2}}\mathit{u}\mathbf{/}\mathbf{\partial }\mathit{y}\mathbf{\partial }\mathit{x}$;

(b) Verify that${\mathbf{\partial }}^{\mathbf{2}}\mathit{u}\mathbf{/}\mathbf{\partial }{\mathit{x}}^{\mathbf{2}}\mathbf{=}{\mathbf{\partial }}^{\mathbf{2}}\mathit{u}\mathbf{/}\mathbf{\partial }{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

Short answer

The value of provide equation is $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$.

Step-by-step solution

Explanation of solution

Introduction:

In this question it is provide equation $u=e^x\cos y$and to verify that

${\partial }^{2}u/\partial x\partial y={\partial }^{2}u/\partial y\partial x$ and ${\partial }^{2}u/\partial x^2={\partial }^{2}u/\partial y^2=0$.

Partial differentiation

The procedure of calculating the partial derivative of a function is called partial differentiation. This method is used to get the partial derivative of a function with respect to one variable while keeping the other constant.

Calculation

(a)

Consider that,

$u=e^x\cos y$

Implies that,

role="math" localid="1659092296034" $\frac{\partial u}{\partial x}=e^x\cos y$

Implies that,

$\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}\right)=-e^x\sin y$

Implies that,

$\frac{{\partial }^{2}u}{\partial x\partial y}=-e^x\sin y\_\_\_\_\_\_\left(1\right)$

Now,

$\frac{\partial u}{\partial y}=-e^x\sin y$

Since,

$\frac{{\partial }^{2}u}{\partial y\partial x}=-e^x\sin y\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\_\left(2\right)$

(b)

Consider $u=e^x\cos y$

Impleis that,

$\frac{\partial u}{\partial x}=-e^x\cos y$

Implies that,

$\frac{{\partial }^{2}u}{\partial x^2}=-e^x\cos y\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\left(3\right)$

And,

role="math" localid="1659094574208" $\frac{{\partial }^{2}u}{\partial y\partial x}=-e^x\cos y\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\left(4\right)$

Add equation (3) and (4) ,

$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=-e^x\cos y-e^x\cos y\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\_\hspace{0.17em}\left(4\right)$

Implies that,

$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$