Question

In Problems 1-20 determine whether the given differential equation is exact. If it is exact, solve it.

$\left(1-\frac{3}{y}+x\right)\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{y}\mathbf{=}\frac{\mathbf{3}}{\mathbf{x}}\mathbf{-}\mathbf{1}$

Short answer

The solution is$x+y+xy-3\ln xy=C$

Step-by-step solution

Given information

The given equation is $\left(1-\frac{3}{y}+x\right)\frac{dy}{dx}+y=\frac{3}{x}-1$

Multiply both side by $dx$

$\left(1-\frac{3}{y}+x\right)dy+\left(1-\frac{3}{y}+y\right)dx=0$

Compare the given equation with $Mdx+Ndy=0$

$\begin{array}{l}M\left(x,y\right)=1-\frac{3}{y}+y\\ N\left(x,y\right)=1-\frac{3}{y}+x\end{array}$

Condition for exactness

An equation of the form $Mdx+Ndy=0$, is said to be exact, if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Determining the exactness of the differential equation

Checking whether the differential equation is exact by finding $\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$

$\frac{\partial M}{\partial y}=1$

$\frac{\partial N}{\partial x}=1$

It can be observed that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. So, the given equation is exact. Next, we have to find the solution of the given equation. Where,$f(x,y)=y-3lny+xy+g\left(x\right)$

Find the solution

Integrate$N(x,y)=\frac{\partial f}{\partial y}=1-\frac{3}{y}+x$

$\begin{array}{c}\frac{\partial f}{\partial x}=y+g\text{'}\left(x\right)\\ g\text{'}\left(x\right)=1-\frac{3}{x}\\ g\left(x\right)=x-3lnx\end{array}$

Since $M(x,y)=\frac{\partial f}{\partial x},$ we see that $g\text{'}\left(x\right)=1-\frac{3}{x}$

The solution is$x+y+xy-3\ln xy=C$