Question

Which of the following is the correct solution of $x\cos x\left(\frac{dy}{dx}\right)+y(x\sin x+\cos x)=1$?

• A. $yxsecy=c+\tan x$
• B. $yx\cos y=c+\tan x$
• C. $yxsecx=c+\tan x$
• D. None of these

Answer 1

The correct option is C

$yxsecx=c+\tan x$

Explanation for the correct options:

Step 1: Simplifying the given equation.

Given $x\cos x\left(\frac{dy}{dx}\right)+y(x\sin x+\cos x)=1$

Divide Both sides of the equation by $x\cos x$,

$$\begin{gathered} \frac{xcosx\left(\frac{dy}{dx}\right)}{xcosx}+\frac{y(xsinx+cosx)}{xcosx}=\frac{1}{xcosx} \\ \Rightarrow \left(\frac{dy}{dx}\right)+\frac{y(xsinx+cosx)}{xcosx}=\frac{1}{xcosx} \end{gathered}$$

We can consider $P=\frac{(xsinx+cosx)}{xcosx}$ and $Q=\frac{1}{xcosx}$

Therefore,

$\Rightarrow \frac{dy}{dx}+Py=Q$

Step 2: Finding the Integrating factor

The integrating factor is $=e^{\int Pdx}$

$$\begin{gathered} \Rightarrow I.F.=e^{\int \frac{(xsinx+cosx)}{xcosx}dx} \\ \Rightarrow I.F.=e^{\int (tanx+\frac{1}{x})dx} \\ \Rightarrow I.F.=e^{\int tanxdx+\int \frac{1}{x}dx} \\ \Rightarrow I.F.=e^{logsecx+logx} \end{gathered}$$

$\Rightarrow I.F.=e^{log(xsecx)}$ $\left[\because \log A+\log B=\log (A\times B)\right]$

$\Rightarrow I.F.=secx\times x$ $\left[\because e^{logx}=x\right]$

Step 3: Apply integration

$$\begin{gathered} y\times I.F.=\int (Q\times I.F.)dx \\ \Rightarrow y\times xsecx=\int \left(\frac{1}{x\cos x}\times xsecx\right)dx \end{gathered}$$

$\Rightarrow y\times xsecx=\int se{c}^{2}xdx+c$ $\left[\because \int se{c}^{2}xdx=tanx+c\right]$

$\Rightarrow yx\mathrm{secx}=tanx+c$
Hence, the correct option is option (c).