Question

Discuss how the functions $\mathit{M}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$and $\mathit{N}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$can be found so that each differential equation is exact. Carry out your ideas.

(a) $\mathit{M}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathit{d}\mathit{x}\mathbf{+}\mathbf{(}\mathbf{x}{\mathbf{e}}^{\mathbf{x}\mathbf{y}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{)}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

(b)$\left(x^{-1/2}{y}^{1/2}+\frac{x}{x^2+y}\right)\mathit{d}\mathit{x}\mathbf{+}\mathit{N}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

(a) The required value is $M(x,y)=y{e}^{xy}+y^2-\frac{y}{x^2}$.

(b) The required value is $N(x,y)=x^{\frac{1}{2}}{y}^{-\frac{1}{2}}+\frac{1}{2}\frac{1}{x^2+y}$.

Step-by-step solution

To Find the Differential Equation (a)

A standard differential equation is of the form $M(x,y)dx+N(x,y)dy=0$.

Based on the above and the given differential equation,

$N(x,y)=x{e}^{xy}+2xy+\frac{1}{x}$

The condition for the exact differential equation is

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

There exists a function such that

$\frac{\partial f}{\partial y}=N(x,y)and\frac{\partial f}{\partial x}=M(x,y)$

Then,

$\frac{\partial f}{\partial y}=x{e}^{xy}+2xy+\frac{1}{x}$

Integrate the obtained function

$\begin{array}{c}f=\int \left(x{e}^{xy}+2xy+\frac{1}{x}\right)\mathrm{d}y\\ =x\int e^{xy}\mathrm{d}y+2x\int y\mathrm{d}y+\frac{1}{x}\int \mathrm{d}y\\ =x\frac{e^{xy}}{x}+2x\frac{y^2}{2}+\frac{y}{x}\\ =e^{xy}+x{y}^2+\frac{y}{x}\end{array}$

Find the required value of M

Then, find the required value of M as,

$\begin{array}{c}M(x,y)=\frac{\partial f}{\partial x}\\ =\frac{\partial }{\partial x}\left[e^{xy}+x{y}^2+\frac{y}{x}\right]\\ =y{e}^{xy}+y^2-\frac{y}{x^2}\end{array}$

Now, let's check that the obtained value is correct or not:

$\begin{array}{c}\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left[y{e}^{xy}+y^2-\frac{y}{x^2}\right]=e^{xy}+xy{e}^{xy}+2y-\frac{1}{x^2}\\ \frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left[x{e}^{xy}+2xy+\frac{1}{x}\right]=e^{xy}+xy{e}^{xy}+2y-\frac{1}{x^2}\end{array}$

Which implies that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, hence $M(x,y)=y{e}^{xy}+y^2-\frac{y}{x^2}$.

The condition for exact differential (b)

A standard differential equation is of the form .

$M(x,y)dx+N(x,y)dy=0$

Based the above and the given differential equation, we get

$M(x,y)=x^{-\frac{1}{2}}{y}^{\frac{1}{2}}+\frac{x}{x^2+y}$

The condition for exact differential equation is

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

There exists a function such that

$\frac{\partial f}{\partial y}=N(x,y)and\frac{\partial f}{\partial x}=M(x,y)$

Then,

$\frac{\partial f}{\partial x}=x^{-\frac{1}{2}}{y}^{\frac{1}{2}}+\frac{x}{x^2+y}$

Integrate the obtained function.

$\begin{array}{c}f=\int \left(x^{-\frac{1}{2}}{y}^{\frac{1}{2}}+\frac{x}{x^2+y}\right)dx\\ =y^{\frac{1}{2}}\int x^{-\frac{1}{2}}dx+\int \frac{x}{x^2+y}dx\\ =2y^{\frac{1}{2}}{x}^{\frac{1}{2}}+\frac{1}{2}\ln \left(x^2+y\right)\end{array}$

Find the required value of N

Now, let's check that the obtained value is correct or not:

$\begin{array}{c}\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left[x^{-\frac{1}{2}}{y}^{\frac{1}{2}}+\frac{x}{x^2+y}\right]=\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{-\frac{1}{2}}-\frac{x}{{\left(x^2+y\right)}^2}\\ \frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left[x^{\frac{1}{2}}{y}^{-\frac{1}{2}}+\frac{1}{2}\frac{1}{x^2+y}\right]=\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{-\frac{1}{2}}-\frac{x}{{\left(x^2+y\right)}^2}\end{array}$

Which implies that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, hence $N(x,y)=x^{\frac{1}{2}}{y}^{-\frac{1}{2}}+\frac{1}{2}\frac{1}{x^2+y}$.