Question

In Problems determine whether the given differential equation is exact.

If it is exact, solve it.

$\left(siny-ysinx\right)\mathit{d}\mathit{x}\mathbf{+}\left(cosx+xcosy-y\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The given differential equation is exact and the solution is$xsiny+ycosx-\frac{1}{2}{y}^2+c=0$

Step-by-step solution

Given Information

The given equation is$\left(siny-ysinx\right)dx+\left(cosx+xcosy-y\right)dy=0$. An equation of the form $Mdx+Ndy=f\left(\text{x}\right)$is exact only if it satisfieslocalid="1663835154407" $\frac{\partial \text{M}}{\partial \text{y}}=\frac{\partial \text{N}}{\partial \text{x}}$

Determining the exactness of the differential equation

Find M and N using given equation

$\begin{array}{l}M(x,y)=siny-ysinx\\ N(x,y)=cosx+xcosy-y\end{array}$

Findlocalid="1663835166498" $\frac{\partial \text{M}}{\partial \text{y}}$andlocalid="1663835172426" $\frac{\partial \text{N}}{\partial \text{x}}$

$\begin{array}{l}\frac{\partial M(x,y)}{\partial y}=cosy-sinx\\ \frac{\partial N(x,y)}{\partial x}=-sinx+cosy\end{array}$

As ,localid="1663835186842" $\frac{\partial \text{M}}{\partial \text{y}}=\frac{\partial \text{N}}{\partial \text{x}}$so, the equation is exact.

Find the solution of differential equation

The solution of the differential equation is a function f(x,y) such that $\frac{\partial f}{\partial x}=\text{M}$and $\frac{\partial f}{\partial \text{y}}=\text{N}$. Here $\frac{\partial f}{\partial x}=siny-ysinx$and $\frac{\partial f}{\partial y}=xsiny+ycosx$. After integrating the first of these equations, we get

$\begin{array}{c}\int df=\int \left(siny-ysinx\right)dx\\ F(x,y)=xsiny+ycosx+h\left(y\right)\end{array}$

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

$\begin{array}{c}\frac{\partial f}{\partial y}=xcosy+cosx+h\text{'}\left(y\right)\\ =cosx+xcosy-y\end{array}$

On comparing these two equations we get

$\begin{array}{l}{h}^{\text{'}}\left(y\right)=-y\\ h\left(y\right)=\frac{-1}{2}{y}^2+c\end{array}$

Therefore, function $f(x,y)$ becomes $f(x,y)=xsiny+ycosx-\frac{1}{2}{y}^2$ As a result, the implicit solution of the problem is$xsiny+ycosx-\frac{1}{2}{y}^2+c=0$