Question

If $y(x)$ satisfies the differential equation $y^{′}−y\tan x=2x\sec x$ and $y(0)=0$ then

• A. $y\left(\frac{Ï€}{4}\right)=\frac{{Ï€}^2}{8\sqrt{2}}$
  
• B. $y^{′}\left(\frac{Ï€}{4}\right)=\frac{{Ï€}^2}{18}$
  
• C. $y\left(\frac{Ï€}{3}\right)=\frac{{Ï€}^2}{9}$
  
• D. $y^{′}\left(\frac{Ï€}{3}\right)=\frac{4Ï€}{3}+\frac{2{Ï€}^2}{3\sqrt{3}}$
  

Answer 1

Answer: (A), (D)

PLAN Linear differential equation under one variable.
$\frac{dy}{dx}+Py=Q;IF=e^{∫Pdx}$
$∴$ Solution is, $y(IF)=∫Q.(IF)dx+C$
$y^{′}−y\tan x=2x\sec x$ and $y(0)=0$
$⇒\frac{dy}{dx}−y\tan x=2x\sec x$
$∴IF=∫e^{−\tan x}dx=e^{\log âˆ\pounds \cos xâˆ\pounds }=\cos x$
Solution is $y.\cos x=∫2x\sec x.\cos xdx+C$
$⇒y.\cos x=x^2+C$
As $y(0)=0$
$⇒C=0$
$∴y=x^2\sec x$
Now, $y\left(\frac{Ï€}{4}\right)=\frac{{Ï€}^2}{8\sqrt{2}}$
$⇒y^{′}\left(\frac{π}{4}\right)=\frac{π}{\sqrt{2}}+\frac{{π}^2}{8\sqrt{2}}$
$y\left(\frac{Ï€}{3}\right)=\frac{2{Ï€}^2}{9}$
$⇒y^{′}\left(\frac{π}{3}\right)=\frac{4π}{3}+\frac{2{π}^2}{3\sqrt{3}}$