Question

An integrating factor of the differential equation $(1+y+x^{2}y)dx+(x+x^3)dy=0$ is

• A. $\log x$
• B. $x$
• C. $e^x$
• D. $\frac{1}{x}$
• E. $-\frac{1}{x}$

Answer 1

The correct option is B

$x$

Explanation for the correct option:

Step 1. Given that $(1+y+x^{2}y)dx+(x+x^3)dy=0$

Dividing the entire equation by $\text{'}dx\text{'}$

$(1+y+x^{2}y)+(x+x^3)\frac{dy}{dx}=0$

Step 2. Differentiate it with respect to $x$:

$\Rightarrow$ $\frac{dy}{dx}=-\frac{\left(1+y\left(1+x^2\right)\right)}{\left(x+x^3\right)}$

$\Rightarrow$ $\frac{dy}{dx}=-\left[\frac{1}{\left(x+x^3\right)}+\frac{y\left(1+x^2\right)}{\left(x+x^3\right)}\right]$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-\left[1+y(1+x^2)\right]}{x\left(1+x^2\right)}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-1-y\left(1+x^2\right)}{x\left(1+x^2\right)}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-1}{x\left(1+x^2\right)}-\frac{y\left(1+x^2\right)}{x\left(1+x^2\right)}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-1}{x\left(1+x^2\right)}-\frac{y}{x}$

$\Rightarrow$ $\frac{dy}{dx}+\frac{y}{x}=\frac{-1}{x\left(1+x^2\right)}$

$\Rightarrow$ $\begin{array}{rcl}\frac{dy}{dx}+\frac{1}{x}.y& =& -\frac{1}{x(1+x^2)}\to \left(ii\right)\end{array}$

Step 3. Compare equation $\left(i\right)\&\left(ii\right)$

It is a Linear Differential equation

$\frac{dy}{dx}+yP\left(x\right)=Q\left(x\right)$

$\begin{array}{rcl}I.F& =& e^{\int \mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\end{array}$

$\begin{array}{rcl}P\left(x\right)& =& \frac{1}{x}\end{array}$

$\begin{array}{rcl}& =& e^{\int \frac{1}{\mathrm{x}}\mathrm{dx}}\end{array}$

$\begin{array}{rcl}& =& e^{\log \left(\mathrm{x}\right)}\end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\mathit{I}\mathbf{.}\mathbf{}\mathit{F}\mathbf{}& \mathbf{=}& \mathit{x}\end{array}$

Hence, option $\left(B\right)$ is correct.