Question

An integrating factor for the D. E: $\left( 1+{{y}^{2}} \right)dx-\left( {{\tan }^{-1}}y-x \right)dy=0$ is
(a)${{\tan }^{-1}}y$
(b)${{e}^{{{\tan }^{-1}}y}}$
(c)$\dfrac{1}{1+{{y}^{2}}}$
(d)$\dfrac{1}{x\left( 1+{{y}^{2}} \right)}$

Answer 1

Hint: For calculation of integrating factor, we represent the given linear differential equation of first order either in the form of $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ or $\dfrac{dx}{dy}+P\left( y \right)x=Q\left( y \right)$. The integrating factor in first case will be = ${{e}^{\int{P\left( x \right)dx}}}$ and in the second case, it will be = ${{e}^{\int{P\left( y \right)dy}}}$.

Complete step-by-step answer:
Here in the question we have to find the integrating factor of the given differential equation. First, we must know what is an integrating factor. An integrating factor is an expression which when multiplied to a differential equation converts it into an exact form. In other words, an integrating factor facilitates the solving of a given equation involving differentials. So, first we will check, whether the given differential equation is exact or not. If equation is of the form $M+N\dfrac{dy}{dx}=0$ then if the differential equation is exact then it will satisfy following condition:
$\Rightarrow \dfrac{dM}{dy}=\dfrac{dN}{dx}$
In our case, $M=1+{{y}^{2}}$ and $N=x-{{\tan }^{-1}}y$.

& \Rightarrow \dfrac{dM}{dy}=\dfrac{d}{dy}\left( 1+{{y}^{2}} \right)=2y \\\ & \Rightarrow \dfrac{dN}{dx}=\dfrac{d}{dx}\left( x-{{\tan }^{-1}}y \right)=1-0=1 \\\ \end{aligned}$$ Thus, we can see that $$\dfrac{dM}{dy}\ne \dfrac{dN}{dx}$$ so it is not of the exact form. So in this case, we can calculate the integrating factor. So our differential equation can also be represented as: $$\Rightarrow \left( 1+{{y}^{2}} \right)dx=\left( {{\tan }^{-1}}y-x \right)dy$$ - (1) We have to represent this equation either in the form of $$\dfrac{dy}{dx}+Py=Q$$ or$$\dfrac{dx}{dy}+P\left( y \right)x=Q\left( y \right)$$. Thus, we will get, $$\begin{aligned} & \Rightarrow \dfrac{dx}{dy}=\dfrac{{{\tan }^{-1}}y-x}{1+{{y}^{2}}} \\\ & \Rightarrow \dfrac{dx}{dy}=\dfrac{{{\tan }^{-1}}y}{1+{{y}^{2}}}-\dfrac{x}{1+{{y}^{2}}} \\\ & \Rightarrow \dfrac{dx}{dy}+\dfrac{x}{1+{{y}^{2}}}=\dfrac{{{\tan }^{-1}}y}{1+{{y}^{2}}} \\\ \end{aligned}$$ Thus, the form of the differential equation is $$\dfrac{dx}{dy}+P\left( y \right)x=Q\left( y \right)$$. The integrating factor in this case will be given by: Integrating factor = $${{e}^{\int{P\left( y \right)dy}}}$$. Here, $$P\left( y \right)=\dfrac{1}{1+{{y}^{2}}}$$. We will first calculate the value of integral. $$\Rightarrow $$$$I=\int{P\left( y \right)dy}$$ $$\Rightarrow I=\int{\dfrac{1}{1+{{y}^{2}}}dy}$$ - (2) Let $$y=\tan u$$ then, $$dy={{\sec }^{2}}udu$$. We will put these values in (2). $$\begin{aligned} & \Rightarrow I=\int{\dfrac{{{\sec }^{2}}udu}{1+{{\tan }^{2}}u}} \\\ & \Rightarrow I=\int{\dfrac{{{\sec }^{2}}udu}{{{\sec }^{2}}u}} \\\ \end{aligned}$$ Here we have used the identity, $$1+{{\tan }^{2}}x={{\sec }^{2}}x$$. $$\begin{aligned} & \Rightarrow I=\int{du} \\\ & \Rightarrow I=u \\\ \end{aligned}$$ We know that, $$y=\tan u$$. Thus $$u={{\tan }^{-1}}y$$. Therefore, we get: $$I={{\tan }^{-1}}y$$ Thus the integrating factor will be: Integrating factor = $${{e}^{x}}$$ = $${{e}^{{{\tan }^{-1}}y}}$$. Hence, option (b) is correct. Note: During the calculation of this indefinite integral, we add a constant of integration but in the case of calculation of integrating factor, we do not add constant while calculating the integral.