Question

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

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

So, the correct answer is “Option B”.

Note: when the differential equation is in the form ${\displaystyle \frac{dx}{dy}}+Rx=S$, then
Step 1: write the differential equation in the form ${\displaystyle \frac{dx}{dy}}+Rx=S$ 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.