Question

Determine whether the given differential equation is exact. If it is exact, solve it.

$\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{lnx}\mathbf{+}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{)}\mathit{d}\mathit{x}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{lnx}\mathbf{)}\mathit{d}\mathit{y}$

Short answer

The given differential equation is exact and the solution is$\left(ylnx-y\right)+xlnx=c$

Step-by-step solution

Step 1: Define exact differential equation 

The given equation is$\left(1+lnx+\frac{y}{x}\right)dx=\left(1-lnx\right)dy$. An equation of the form $Mdx+Ndy=f\left(\text{x}\right)$is exact only if it satisfieslocalid="1663835605091" $\frac{\partial \text{M}}{\partial \text{y}}=\frac{\partial \text{N}}{\partial \text{x}}$

Determining the exactness of the differential equation

Find M and N using given equation

$\begin{array}{l}M\left(x,y\right)=\left(1+lnx+\frac{y}{x}\right)\\ N\left(x,y\right)=-\left(1-lnx\right)\end{array}$

Find localid="1663835615485" $\frac{\partial M}{\partial y}$and localid="1663835621211" $\frac{\partial N}{\partial x}$

$\begin{array}{l}\frac{\partial M}{\partial y}=\frac{1}{x}\\ \frac{\partial N}{\partial x}=\frac{1}{x}\end{array}$

As,localid="1663835634640" $\frac{\partial \text{M}}{\partial \text{y}}=\frac{\partial \text{N}}{\partial \text{x}}$the equation is exact.

Find the solution of differential equation

The solution of the differential equation is a function such that $\frac{\partial f}{\partial x}=\text{M}$and $\frac{\partial f}{\partial x}=N$

Here $\frac{\partial f}{\partial x}=\left(1+lnx+\frac{y}{x}\right)=M\left(x,y\right)$and $\frac{\partial f}{\partial y}=-\left(1-lnx\right)=N\left(x,y\right)$

$M\left(x,y\right)$is obtained by taking the partial derivative of the previous expression with respect to y and setting the result to $M\left(x,y\right)$

$\begin{array}{c}\frac{\partial f}{\partial x}=\frac{y}{x}+h\text{'}\left(x\right)\\ M\left(x,y\right)=\left(1+lnx+\frac{y}{x}\right)\end{array}$

On comparing these two equations we get

$\begin{array}{l}h\text{'}\left(x\right)=1+lnx\\ h\left(x\right)=xlnx-x+x\\ h\left(x\right)=xlnx\end{array}$

Hence the solution is$\left(ylnx-y\right)+xlnx=c$