Question

The Integrating factor of the differential equation $\dfrac{dy}{dx}-y\tan x=\cos x$ is:

Answer 1

Hint: Convert the equation to the form $\dfrac{dy}{dx}+p(x)y(x)=q(x)$ and solve it. The Integrating Factor is used to solve the differential equations of such kind and is given as:
Integrating factor(I.F) $={{e}^{\int{\tan xdx}}}$

Complete step by step answer:
Integrating factors are useful for solving ordinary differential equations that can be expressed in the form.
$\dfrac{dy}{dx}+p(x)y(x)=q(x)$
where $p$ and $q$ are given continuous functions, can be made integral by letting $v(x)$ be a function such that
So $v(x)=\int{p(x)dx}$ and $\dfrac{dv(x)}{dx}=p(x)$
So integrating factor,
An integrating factor is a function by which an ordinary differential equation can be multiplied in order to make it integrable.
An integrating factor is a function that is chosen to facilitate the solving of a given equation involving differentials. It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus when multiplying through by an integrating factor allows an inexact differential to be made into an exact differential.
This is especially useful in thermodynamics where temperature becomes the integrating factor that makes entropy an exact differential.
Integrating factors are useful for solving ordinary differential equations
In proper words I.F $={{e}^{\int{p(x)dx}}}={{e}^{v(x)}}$
Then ${{e}^{v(x)}}$ would be the integrating factor.
So integrating factor for $\dfrac{dy}{dx}-y\tan x=\cos x$
Here $p(x)=\tan x$
Integrating factor (I.F) $={{e}^{\int{\tan xdx}}}$ ………. (1)
So we have to integrate $\tan x$ ,
$\int{\tan xdx}$
So we know $\tan x=\dfrac{\sin x}{\cos x}$ ,
So writing the above in integration of$\tan x$we get,
$\int{\dfrac{\sin x}{\cos x}dx}$
Now let $\cos x=u$
So differentiating $\cos x$ w.r.t $x$ we get,
$-\sin xdx=du$
So we get $\sin xdx=-du$
Now substituting $\sin xdx=-du$ and in $\int{\dfrac{\sin x}{\cos x}dx}$ we get,
$-\int{\dfrac{du}{u}}$
So we know $\int{\dfrac{1}{x}dx}=\log (x)+c$
So we get,
$-\int{\dfrac{du}{u}}=-\log (u)+c$
So we know $\cos x=u$ ,
So we get,
$-\log (u)+c=-\log (\cos x)+c$
We know the property that $-\log (a)=\log ({{(a)}^{-1}})=\log \left( \dfrac{1}{a} \right)$ ,
So using the property we get,
$\log ({{(\cos x)}^{-1}})+c=\log \left( \dfrac{1}{\cos x} \right)+c=\log (\sec x)+c$
So we get final integration of $\tan x$ as,
$\int{\tan xdx=\log \sec x+c}$
So substituting $\int{\tan xdx=\log \sec x+c}$ in (1) we get,
So I.F. $={{e}^{\log (\sec x)}}$
So using property ${{e}^{\log a}}=a$ We get it as,
I.F $=\sec x$
So the Integrating factor of $\dfrac{dy}{dx}-y\tan x=\cos x$ is $\sec x$.

Note: The main form should be known $\dfrac{dy}{dx}+p(x)y(x)=q(x)$ and so I.F $={{e}^{\int{p(x)dx}}}={{e}^{v(x)}}$ so you can solve it. It may be any form you have to convert it in first order form. While finding the integrating factor you should know what is $p(x)$ so you can easily calculate the Integrating factor. Here it was direct $p(x)=\tan x$ Sometimes it is different one you have to simplify and write in this form$\dfrac{dy}{dx}+p(x)y(x)=q(x)$ .