Question

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

Answer 1

Hint: First of all consider the given equation and convert it in the form \[\dfrac{dy}{dx}+{{P}_{\left( x \right)}}y=Q\left( x \right)\]. Solve it by solving \[y\left( I.F \right)=\int{Q\left( I.F \right)dx}\] where \[I.F={{e}^{\int{Pdx}}}\]. 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
\[\dfrac{dy}{dx}-y\tan x={{e}^{x}}\sec x\]
Let us consider the differential equation given in the question
\[\dfrac{dy}{dx}-y\tan x={{e}^{x}}\sec x\]
We can also write the above equation as
\[\dfrac{dy}{dx}+\left( -\tan x \right)y={{e}^{x}}\sec x\]
Now, the above equation is of the form \[\dfrac{dy}{dx}+Py=Q\] where P and Q are functions of x. By comparison, we get, P = – tan x and \[Q={{e}^{x}}\sec x\].
To solve this equation, we will first find the integration factor that is,
\[I.F={{e}^{\int{Pdx}}}\]
By substituting the value of P = – tan x, we get,
\[I.F={{e}^{\int{-\tan x\text{ }dx}}}\]
We know that \[\int{\tan \theta d\theta =-m\cos \theta }\], so by using this, we get,
\[I.F={{e}^{\ln \cos x}}\]
We know that \[{{a}^{\log b}}.{{b}^{\log a}}\], by using this, we get,
\[I.F={{\left( \cos x \right)}^{\ln e}}\]
We know that ln e = 1. So, we get,
\[I.F=\cos x\]
Now, the equation is solved by using
\[y\left( I.F \right)=\int{Q\left( I.F \right)dx}\]
By substituting the value of I.F = cos x and \[Q={{e}^{x}}\sec x\], we get,
\[y\left( \cos x \right)=\int{{{e}^{x}}\sec x\cos xdx}\]
We know that, \[\cos \theta .\sec \theta =1\]. So, we get,
\[y\left( \cos x \right)=\int{{{e}^{x}}dx}\]
We know that \[\int{{{e}^{x}}dx}={{e}^{x}}+c\]. By using this, we get,
\[y\left( \cos x \right)={{e}^{x}}+c\]
\[y\cos x={{e}^{x}}+c\]
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 \[\dfrac{dy}{dx}+Py=Q\]. Also, remember that I.F or integrating factor is \[{{e}^{\int{Pdx}}}\] and not \[{{e}^{\int{Qdx}}}\]. So, always compare the values properly and then only substitute them to get the correct answer.