Question

In Problems 29 and 30 verify that the given differential equation is not exact. Multiply the given differential equation by the indicated integrating factor and verify that the new equation is exact. Solve.

$\mathbf{(}\mathbf{-}\mathit{x}\mathit{y}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{)}\mathit{d}\mathit{x}\mathbf{+}\mathbf{2}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{\mu }\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}\mathit{x}\mathit{y}$

Short answer

The equation is$x^2{y}^{2}cosx+c=0$

Step-by-step solution

To Find the equation is not exact

The given equation is, $\left(-xy\sin x+2y\mathrm{cosx}\right)dx+2x\cos xdy=0$.

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

$$\begin{gathered} M(x,y)=-xysinx+2ycosx \\ N(x,y)=2xcosx \end{gathered}$$

Condition for exactness

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

$$\begin{gathered} \frac{\partial M(x,y)}{\partial y}=-xsinx+2cosx \\ \frac{\partial N(x,y)}{\partial x}=2cosx-2xsinx \end{gathered}$$

As, $\frac{\partial M(x,y)}{\partial y}\ne \frac{\partial N(x,y)}{\partial x}$. The equation is not exact.

Then we should multiply both sides in integrating factor .

$\left(-x^2{y}^{2}sinx+2x{y}^{2}cosx\right)dx+2x^{2}ycosxdy=0$

Where (n) refers to word new Then,$M_n(x,y)=-x^2{y}^{2}sinx+2x{y}^{2}cosx$ and $N_n(x,y)=2x^{2}ycosx$

$\begin{array}{c}\frac{\partial M_n(x,y)}{\partial y}=-2x^{2}ysinx+4xycosx\\ \frac{\partial N_n(x,y)}{\partial x}=2y\left(2x\cos x-x^2\sin x\right)\\ =4xy\cos x-2x^{2}y\sin x\\ Then,\frac{\partial f}{\partial y}=2x^{2}ycosxwhere\frac{\partial f}{\partial y}=N_n(x,y)\end{array}$

Find the new equation is exact or not

Checking whether the new equation is exact or not.

As, $\frac{\partial M_n(x,y)}{\partial y}=\frac{\partial N_n(x,y)}{\partial x}$The equation is exact.

Then we have $df=\left(2x^{2}ycosx\right)dy$.

Do integration.

$F(x,y)=x^2{y}^{2}cosx+g\left(x\right)$

Do differentiation for x

We get $-x^2{y}^{2}sinx+2x{y}^{2}cosx$, solve it as,$2x{y}^{2}cosx-x^2{y}^{2}sinx+g^{¢}\left(x\right)$

$g^{¢}\left(x\right)=0$

By integrating for $g\left(x\right)=c$

By substituting with $g\left(x\right)$in $F(x,y)=x^2{y}^{2}cosx+g\left(x\right)$.

Hence, the required solution is $x^2{y}^{2}cosx+c=0$.