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Question: The solution of \[\dfrac{dy}{dx}-y\tan x={{e}^{x}}\sec x\] is: (a) \[y{{e}^{x}}=\cos x+c\] (b) \...

The solution of dydx−ytan⁡x=exsec⁡x\dfrac{dy}{dx}-y\tan x={{e}^{x}}\sec x is:
(a) yex=cos⁡x+cy{{e}^{x}}=\cos x+c
(b) ycos⁡x=ex+cy\cos x={{e}^{x}}+c
(c) ysin⁡x=ex+cy\sin x={{e}^{x}}+c
(d) yex=sin⁡x+cy{{e}^{x}}=\sin x+c

Explanation

Solution

Hint: First of all consider the given equation and convert it in the form dydx+P(x)y=Q(x)\dfrac{dy}{dx}+{{P}_{\left( x \right)}}y=Q\left( x \right). Solve it by solving y(I.F)=∫Q(I.F)dxy\left( I.F \right)=\int{Q\left( I.F \right)dx} where I.F=e∫PdxI.F={{e}^{\int{Pdx}}}. So, find I.F and substitute in the above equation and solve it.

Complete step-by-step answer:
In this question, we have to solve the differential equation
dydx−ytan⁡x=exsec⁡x\dfrac{dy}{dx}-y\tan x={{e}^{x}}\sec x
Let us consider the differential equation given in the question
dydx−ytan⁡x=exsec⁡x\dfrac{dy}{dx}-y\tan x={{e}^{x}}\sec x
We can also write the above equation as
dydx+(−tan⁡x)y=exsec⁡x\dfrac{dy}{dx}+\left( -\tan x \right)y={{e}^{x}}\sec x
Now, the above equation is of the form dydx+Py=Q\dfrac{dy}{dx}+Py=Q where P and Q are functions of x. By comparison, we get, P = – tan x and Q=exsec⁡xQ={{e}^{x}}\sec x.
To solve this equation, we will first find the integration factor that is,
I.F=e∫PdxI.F={{e}^{\int{Pdx}}}
By substituting the value of P = – tan x, we get,
I.F=e∫−tan⁡x dxI.F={{e}^{\int{-\tan x\text{ }dx}}}
We know that ∫tan⁡θdθ=−mcos⁡θ\int{\tan \theta d\theta =-m\cos \theta }, so by using this, we get,
I.F=eln⁡cos⁡xI.F={{e}^{\ln \cos x}}
We know that alog⁡b.blog⁡a{{a}^{\log b}}.{{b}^{\log a}}, by using this, we get,
I.F=(cos⁡x)ln⁡eI.F={{\left( \cos x \right)}^{\ln e}}
We know that ln e = 1. So, we get,
I.F=cos⁡xI.F=\cos x
Now, the equation is solved by using
y(I.F)=∫Q(I.F)dxy\left( I.F \right)=\int{Q\left( I.F \right)dx}
By substituting the value of I.F = cos x and Q=exsec⁡xQ={{e}^{x}}\sec x, we get,
y(cos⁡x)=∫exsec⁡xcos⁡xdxy\left( \cos x \right)=\int{{{e}^{x}}\sec x\cos xdx}
We know that, cos⁡θ.sec⁡θ=1\cos \theta .\sec \theta =1. So, we get,
y(cos⁡x)=∫exdxy\left( \cos x \right)=\int{{{e}^{x}}dx}
We know that ∫exdx=ex+c\int{{{e}^{x}}dx}={{e}^{x}}+c. By using this, we get,
y(cos⁡x)=ex+cy\left( \cos x \right)={{e}^{x}}+c
ycos⁡x=ex+cy\cos x={{e}^{x}}+c
Hence, option (b) is the right answer.

Note: In this type of question, first of all, it is very important to identify the form and convert it into the standard linear differential equation of the first order that is dydx+Py=Q\dfrac{dy}{dx}+Py=Q. Also, remember that I.F or integrating factor is e∫Pdx{{e}^{\int{Pdx}}} and not e∫Qdx{{e}^{\int{Qdx}}}. So, always compare the values properly and then only substitute them to get the correct answer.