Question

The integrating factor of differential equation $\cos \left( x \right)dy=y\left( \sin x-y \right)dx$ is

Answer 1

Hint: -Now, first of all, we must know the standard form of Linear Differential Equation (commonly known as L.D.E.) which is given as follows
$\dfrac{dy}{dx}+Py=Q$
(Where P and Q are functions in x)
Now, for finding the integrating factor, every linear differential equation must be converted to make it similar the above mentioned equation. Then for finding the integrating factor we write as follows
$integerating\ factor\ ={{e}^{\int{Pdx}}}$
(Where p is a function in x)

Complete step-by-step answer:
As mentioned in the question, we have to find the integrating factor for the given differential equation.
Now, firstly, we will convert the given differential equation into a form which is as follows
$\dfrac{dy}{dx}+Py=Q$
So, we can make the following changes to get the required form of the equation which is as follows

& \Rightarrow \cos \left( x \right)dy=y\left( \sin x-y \right)dx \\\ \end{aligned}$$ (As we divided both sides with dx) $$\begin{aligned} & \Rightarrow \cos \left( x \right)\dfrac{dy}{dx}=y\left( \sin x-y \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{y\left( \sin x-y \right)}{\cos \left( x \right)} \\\ \end{aligned}$$ (As we divided both sides with cosx) Now, on simplification, we can get the following expression $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=y\tan x-\dfrac{{{y}^{2}}}{\cos x} \\\ & on\ dividing\ {{y}^{2}}with\ every\ term,\ we\ get \\\ & \Rightarrow \dfrac{1}{{{y}^{2}}}\cdot \dfrac{dy}{dx}=\dfrac{1}{y}\tan x-\dfrac{1}{\cos x} \\\ \end{aligned}$$ Now, it is not matching the standard form that is given to us, so, we can do the following Let $$\dfrac{1}{y}\ be\ t$$ . So, we can write as following $$\begin{aligned} & \Rightarrow \dfrac{1}{y}=t \\\ & \Rightarrow d\left( \dfrac{1}{y} \right)=dt \\\ & \Rightarrow \left( \dfrac{-1}{{{y}^{2}}} \right)dy=dt \\\ \end{aligned}$$ Now, on putting $$\dfrac{1}{y}\ as\ t$$ in the equation, we get the following results $$\begin{aligned} & \Rightarrow -\left\\{ \dfrac{-1}{{{y}^{2}}}\cdot \dfrac{dy}{dx} \right\\}=\dfrac{1}{y}\tan x-\dfrac{1}{\cos x} \\\ & \Rightarrow -\dfrac{dt}{dx}=t\tan x-\dfrac{1}{\cos x} \\\ & \Rightarrow \dfrac{dt}{dx}+t\tan x=\dfrac{1}{\cos x} \\\ \end{aligned}$$ Now, this equation is comparable to the required form and on comparing, we get that $$P(x)=\tan x\ and\ Q(x)=\dfrac{1}{\cos x}$$ So, as given in the hint, we can get that the integrating factor is as follows $$\begin{aligned} & ={{e}^{\int{P(x)dx}}} \\\ & ={{e}^{\int{\left( \tan x \right)dx}}} \\\ & \left[ \int{\left( \tan x \right)dx}=\log (\sec x) \right] \\\ & ={{e}^{\log (\sec x)}} \\\ & =\sec x \\\ \end{aligned}$$ (As we know that $${{e}^{\log \left( a \right)}}=a$$ ) Note:-For solving this question, the student must know that $${{e}^{\log \left( a \right)}}=a$$ because towards the end of the solution, knowing this fact is very essential to solve the question further. Also, knowing the integral of tanx is also very important which is as follows $$\int{\left( \tan x \right)dx}=\log (\sec x)$$ As without knowing this, one can never get to the correct solution.