Question

The function $y = f\left( x \right)$ is the solution of the differential equation $\dfrac{{dy}}{{dx}} + \dfrac{{xy}}{{{x^2} - 1}} = \dfrac{{{x^4} + 2x}}{{\sqrt {1 - {x^2}} }}$ in (-1, 1) satisfying $f\left( 0 \right) = 0$ then $\int\limits_{ - \dfrac{{\sqrt 3 }}{2}}^{\dfrac{{\sqrt 3 }}{2}} {f\left( x \right)dx} $ is
A. $\dfrac{\pi }{3} - \dfrac{{\sqrt 3 }}{2}$
B. $\dfrac{\pi }{3} - \dfrac{{\sqrt 3 }}{4}$
C. $\dfrac{\pi }{6} + \dfrac{{\sqrt 3 }}{4}$
D. $\dfrac{\pi }{6} - \dfrac{{\sqrt 3 }}{4}$

Answer 1

Hint: To solve this question, we will use the concept of linear differential equation. We have to follow the following steps to solve a linear differential equation.
Step 1: write the differential equation in the form $\dfrac{{dy}}{{dx}} + Py = Q$ and obtain P and Q.
Step 2: find integration factor (I.F.) given by $I.F. = {e^{\int {Pdx} }}$
Step 3: multiply both sides of the equation in step 1 by I.F.
Step 4: integrate both sides of the equation obtained in step 3 with respect to x to obtain $y\left( {I.F.} \right) = \int {Q\left( {I.F.} \right)dx + C} $, which gives the required solution.

Complete step-by-step answer:
Given that,
 Differential equation is:
$\dfrac{{dy}}{{dx}} + \dfrac{{xy}}{{{x^2} - 1}} = \dfrac{{{x^4} + 2x}}{{\sqrt {1 - {x^2}} }}$ …….. (i)
Comparing this with the general form of differential equation, i.e. $\dfrac{{dy}}{{dx}} + Py = Q$
We get,
$ \Rightarrow P = \dfrac{x}{{{x^2} - 1}}$ and $Q = \dfrac{{{x^4} + 2x}}{{\sqrt {1 - {x^2}} }}$
We know that,
Integration factor (I.F.) is given by,
$I.F. = {e^{\int {Pdx} }}$
Putting the value of P, we will get
$ \Rightarrow I.F. = {e^{\int {\dfrac{x}{{{x^2} - 1}}dx} }}$ ……… (ii)
First, we will find $\int {\dfrac{x}{{{x^2} - 1}}} dx$
So, let $t = {x^2} - 1$
Differentiate both sides,
$dt = 2xdx$
$\dfrac{{dt}}{2} = xdx$
Using this, we can write the above integration as:
$ \Rightarrow \int {\dfrac{1}{t}} \dfrac{{dt}}{2}$
Integrating this, we will get
$ \Rightarrow \dfrac{1}{2}\ln \left| t \right| + C$
Replace $t = {x^2} - 1$,
$ \Rightarrow \dfrac{1}{2}\ln \left| {{x^2} - 1} \right| + C$
Putting this value in equation (ii), we will get
$ \Rightarrow I.F. = {e^{\dfrac{1}{2}\ln \left| {{x^2} - 1} \right| + C}}$
According to the question,
The differential equation satisfies (-1, 1)
So, we can say that,
$\left| {{x^2} - 1} \right| = 1 - {x^2}$
Hence,
$ \Rightarrow I.F. = {e^{\ln \left( {\sqrt {1 - {x^2}} } \right) + C}}$
Solving this, we will get
$ \Rightarrow I.F. = \sqrt {1 - {x^2}} $ [$\therefore {e^{\ln x}} = x$]
Now,
The required solution will be,
$y\left( {I.F.} \right) = \int {Q\left( {I.F.} \right)dx + C} $
Putting the required values, we will get
\[ \Rightarrow y\sqrt {1 - {x^2}} = \int {\dfrac{{{x^4} + 2x}}{{\sqrt {1 - {x^2}} }}\sqrt {1 - {x^2}} dx + C} \]
\[ \Rightarrow y\sqrt {1 - {x^2}} = \int {\left( {{x^4} + 2x} \right)dx + C} \]
Solving this, we will get
\[ \Rightarrow y\sqrt {1 - {x^2}} = \dfrac{{{x^5}}}{5} + {x^2} + C\]
we know that,
$y = f\left( x \right)$
So,
\[ \Rightarrow f\left( x \right)\sqrt {1 - {x^2}} = \dfrac{{{x^5}}}{5} + {x^2} + C\] ……. (iii)
Putting x = 0, we will get
\[ \Rightarrow f\left( 0 \right)\sqrt {1 - {0^2}} = \dfrac{{{0^5}}}{5} + {0^2} + C\]
We have given $f\left( 0 \right) = 0$
Hence, we get
$ \Rightarrow C = 0$
Therefore, equation (iii) will become,
\[ \Rightarrow f\left( x \right)\sqrt {1 - {x^2}} = \dfrac{{{x^5}}}{5} + {x^2}\]
\[ \Rightarrow f\left( x \right) = \dfrac{{\left( {\dfrac{{{x^5}}}{5} + {x^2}} \right)}}{{\sqrt {1 - {x^2}} }}\]
According to the question, we have to find $\int\limits_{ - \dfrac{{\sqrt 3 }}{2}}^{\dfrac{{\sqrt 3 }}{2}} {f\left( x \right)dx} $
Putting the value of f(x),
\[ \Rightarrow \int\limits_{ - \dfrac{{\sqrt 3 }}{2}}^{\dfrac{{\sqrt 3 }}{2}} {f\left( x \right)dx} = \int\limits_{ - \dfrac{{\sqrt 3 }}{2}}^{\dfrac{{\sqrt 3 }}{2}} {\dfrac{{\left( {\dfrac{{{x^5}}}{5} + {x^2}} \right)}}{{\sqrt {1 - {x^2}} }}dx} \]
We can write this as,
\[ \Rightarrow \int\limits_{ - \dfrac{{\sqrt 3 }}{2}}^{\dfrac{{\sqrt 3 }}{2}} {\left( {\dfrac{{{x^5}}}{{5\sqrt {1 - {x^2}} }}} \right)dx} + \int\limits_{ - \dfrac{{\sqrt 3 }}{2}}^{\dfrac{{\sqrt 3 }}{2}} {\left( {\dfrac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} \right)dx} \] ………. (iv)
Using the identity, we know that
$\int\limits_{ - a}^a {f\left( x \right)dx} = 0$, if f is an odd function and,
$\int\limits_{ - a}^a {f\left( x \right)dx} = 2\int\limits_0^a {f\left( x \right)dx} $, if f is an even function.
Here we can see that,
\[\dfrac{{{x^5}}}{{5\sqrt {1 - {x^2}} }}\] is an odd function.
So,
\[ \Rightarrow \int\limits_{ - \dfrac{{\sqrt 3 }}{2}}^{\dfrac{{\sqrt 3 }}{2}} {\left( {\dfrac{{{x^5}}}{{5\sqrt {1 - {x^2}} }}} \right)dx} = 0\]
Then, equation (iv) will become,
\[ \Rightarrow 2\int\limits_0^{\dfrac{{\sqrt 3 }}{2}} {\left( {\dfrac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} \right)dx} \]
Put $x = \sin \theta $
Differentiate both sides,
$dx = \cos \theta d\theta $
When $x = 0$, $\theta = {\sin ^{ - 1}}0 = 0$
And when $x = \dfrac{{\sqrt 3 }}{2}$, $\theta = {\sin ^{ - 1}}\dfrac{{\sqrt 3 }}{2} = \dfrac{\pi }{3}$
Using this, we will get
\[ \Rightarrow 2\int\limits_0^{\dfrac{\pi }{3}} {\left( {\dfrac{{{{\sin }^2}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}} \right)\cos \theta d\theta } \]
\[ \Rightarrow 2\int\limits_0^{\dfrac{\pi }{3}} {\left( {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right)\cos \theta d\theta } \] [$\therefore \sqrt {1 - {{\sin }^2}\theta } = \cos \theta $]
\[ \Rightarrow 2\int\limits_0^{\dfrac{\pi }{3}} {{{\sin }^2}\theta d\theta } \] ….. (v)
We know that,
$ \Rightarrow \cos 2x = 1 - 2{\sin ^2}x$
So,
$ \Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}$
Hence equation (v) will become,
\[ \Rightarrow 2\int\limits_0^{\dfrac{\pi }{3}} {\dfrac{{1 - \cos 2x}}{2}d\theta } \]
Integrating this, we will get
\[ \Rightarrow 2\left[ {\dfrac{\theta }{2} - \dfrac{{\sin 2\theta }}{4}} \right]_0^{\dfrac{\pi }{3}}\]
\[ \Rightarrow 2\left[ {\left( {\dfrac{\pi }{6} - \dfrac{{\sin \dfrac{{2\pi }}{3}}}{4}} \right) - \left( {\dfrac{0}{2} - \dfrac{{\sin 20}}{4}} \right)} \right]\]
\[ \Rightarrow 2\left[ {\dfrac{\pi }{6} - \dfrac{{\sqrt 3 }}{8} - 0} \right]\]
\[ \Rightarrow \dfrac{\pi }{3} - \dfrac{{\sqrt 3 }}{4}\]
Hence, we can say that the value of $\int\limits_{ - \dfrac{{\sqrt 3 }}{2}}^{\dfrac{{\sqrt 3 }}{2}} {f\left( x \right)dx} $ is \[\dfrac{\pi }{3} - \dfrac{{\sqrt 3 }}{4}\]

So, the correct answer is “Option B”.

Note: when the differential equation is in the form $\dfrac{{dx}}{{dy}} + Rx = S$, then
Step 1: write the differential equation in the form $\dfrac{{dx}}{{dy}} + Rx = S$ and obtain R and S.
Step 2: find integration factor (I.F.) given by $I.F. = {e^{\int {Rdy} }}$
Step 3: multiply both sides of the equation in step 1 by I.F.
Step 4: integrate both sides of the equation obtained in step 3 with respect to y to obtain $x\left( {I.F.} \right) = \int {S\left( {I.F.} \right)dy + C} $, which gives the required solution.