Question

If $y\left(x\right)$ satisfies the differential equation $y’-y\tan x=2xsecx$ and $y\left(0\right)=0$, then:

• A. $y\left(\frac{\mathrm{\pi }}{4}\right)=\frac{{\mathrm{\pi }}^2}{8\sqrt{2}}$
• B. $y\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=\frac{{\mathrm{\pi }}^2}{18}$
• C. $y\left(\frac{\mathrm{\pi }}{3}\right)=\frac{{\mathrm{\pi }}^2}{9}$
• D. $y\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{4{\mathrm{\pi }}^2}{3}+\frac{2{\mathrm{\pi }}^2}{3\sqrt{2}}$

Answer 1

Answer: (A), (D)

$y\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{4{\mathrm{\pi }}^2}{3}+\frac{2{\mathrm{\pi }}^2}{3\sqrt{2}}$

Explanation for correct option:

Step 1: Differentiate the Equation

The given expression is,

$y’-y\tan x=2xsecx$ $\left(1\right)$

$\Rightarrow$$\frac{dy}{dx}-y\frac{\sin x}{\cos x}=\frac{2x}{\cos x}$

$\Rightarrow$$\frac{dy}{dx}\cos x-y\sin x=2x$

$\Rightarrow$$\frac{d}{dx}\left(y\cos x\right)=2x$ $\left[\because \frac{d}{\mathrm{dx}}\left(\mathrm{uv}\right)=u\frac{d}{\mathrm{dx}}\left(v\right)+v\frac{d}{\mathrm{dx}}\left(u\right)\right]$

Step 2: Integrate the Equation

Integrating on both sides we get,

$\int d\left(y\cos x\right)=2\int xdx$

$\Rightarrow$$y\cos x=x^2+c$ $\left[\because \int \mathrm{dx}=x;\int x^n=\frac{x^{n+1}}{n+1}\right]$

Given that $y\left(0\right)=0$, hence $c=0$

$\therefore y\cos x=x^2$ $\left(2\right)$

Option (A):

Substituting $x=\frac{\mathrm{\pi }}{4}$ in equation $\left(2\right)$

$y\left(\frac{\mathrm{\pi }}{4}\right).\frac{1}{\sqrt{2}}={\left(\frac{\mathrm{\pi }}{4}\right)}^2$ $\left[\because \cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\right]$

$\Rightarrow$$y\left(\frac{\mathrm{\pi }}{4}\right)=\frac{{\mathrm{\pi }}^2}{8\sqrt{2}}$

Option (D):

Substituting $x=\frac{\mathrm{\pi }}{3}$ in equation $\left(1\right)$

$y\text{'}\left(\frac{\mathrm{\pi }}{3}\right)-\frac{2{\mathrm{\pi }}^2}{9}\tan \left(\frac{\mathrm{\pi }}{3}\right)=2.\frac{\mathrm{\pi }}{3}.2$ $\left[\because \tan \left(\frac{\pi }{3}\right)=\sqrt{3};\sec \left(\frac{\pi }{3}\right)=2\right]$

$\Rightarrow$$y\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=\sqrt{3.}\frac{2{\mathrm{\pi }}^2}{9}+\frac{4\mathrm{\pi }}{3}$

$\Rightarrow$$y\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{2{\mathrm{\pi }}^2}{3\sqrt{3}}+\frac{4\mathrm{\pi }}{3}$

Explanation for Incorrect Options:

Option (B):

Substituting $x=\frac{\mathrm{\pi }}{3}$ in equation $\left(2\right)$

$y\left(\frac{\mathrm{\pi }}{3}\right).\frac{1}{2}={\left(\frac{\mathrm{\pi }}{3}\right)}^2$ $\left[\because \cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\right]$

$\Rightarrow$$y\left(\frac{\mathrm{\pi }}{3}\right)=\frac{2{\mathrm{\pi }}^2}{9}$

Option (C):

Substituting $x=\frac{\mathrm{\pi }}{4}$ in equation $\left(1\right)$

$y\text{'}\left(\frac{\mathrm{\pi }}{4}\right)-\frac{{\mathrm{\pi }}^2}{8\sqrt{2}}\tan \left(\frac{\mathrm{\pi }}{4}\right)=2.\frac{\mathrm{\pi }}{4}.\sqrt{2}$ $\left[\because \tan \left(\frac{\pi }{4}\right)=1;\sec \left(\frac{\pi }{4}\right)=\sqrt{2}\right]$

$\Rightarrow$$y\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{{\mathrm{\pi }}^2}{8\sqrt{2}}+\frac{\mathrm{\pi }}{\sqrt{2}}$

Hence, the correct options are (A) and (D).