Question

In Problems 31-36 solve the given differential equation by finding as in Example 4, an appropriate integrating factor.

34.$\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathit{d}\mathit{x}\mathbf{+}\left(1+\frac{2}{y}\right)\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The resulting equation is $y+2\ln y+\ln \left|\sin x\right|=c$.

Step-by-step solution

Step 1: Identifying M and N from the differential equation  

The given equation is, $cosxdx+\left(1+\frac{2}{y}\right)\sin xdy=0$.

$\begin{array}{c}M=cosx\\ N=\left(1+\frac{2}{y}\right)sinx\end{array}$

Finding $N_x$and $M_y$.

$$\begin{gathered} M_y=0 \\ {N}_x=\left(1+\frac{2}{y}\right)\cos x \end{gathered}$$

Compute $\frac{M_y-N_x}{N}$.

$\begin{array}{c}\frac{M_y-N_x}{N}=\frac{0-\left(1+\frac{2}{y}\right)\cos x}{\left(1+\frac{2}{y}\right)\sin x}\\ =-\cot x\end{array}$

Then, the integrating factor is,

$\begin{array}{c}\mu \left(x\right)=e^{\int -\cot xdx}\\ =e^{-\ln \left|\sin x\right|}\\ =e^{\ln \left|\csc x\right|}\\ =\csc x\end{array}$

Since $\frac{M_y-N_x}{N}$only depends on x, the integrating factor is $\csc x$. Multiply the given equation by $\csc x$.

$\begin{array}{c}\csc x\cos xdx+\csc x\left(1+\frac{2}{y}\right)\sin xdy=0\\ \frac{\cos x}{\sin x}dx+\frac{1}{\sin x}\left(a+\frac{2}{y}\right)\sin xdy=0\\ \cot xdx+\left(1+\frac{2}{y}\right)dy=0\end{array}$

Then we have,

$\begin{array}{c}M(x,y)=\cot x\\ N(x,y)=1+\frac{2}{y}\end{array}$

Now, checking whether the differential equation is exact.

$\begin{array}{c}\frac{\partial M}{\partial y}=0\\ \frac{\partial N}{\partial x}=0\end{array}$

As, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

Therefore, the equation is exact.

Find the integrate

Now, the function is $f(x,y)=y+2\ln y+g\left(x\right)\sqrt{b^2-4ac}$.

Integrate$N(x,y)=\frac{\partial f}{\partial y}=1+\frac{2}{y}$.

$\frac{\partial f}{\partial x}=g\text{'}\left(x\right)$

Take$\frac{\partial f}{\partial x}$.

$\begin{array}{c}g\text{'}\left(x\right)=\cot x\\ g\left(x\right)=\ln \left|\sin x\right|\end{array}$

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

$g\text{'}\left(x\right)=0$

Therefore, the required solution is $y+2lny+ln\left|sinx\right|=c$.