%%

Let y = y(x) be the solution of the differential equation xdy = (y + x3 cosx)dx with y($$\pi$$) = 0, then $$y\left( {{\pi \over 2}} \right)$$ is equal to :

JEE · Math · previous-year question

  1. A.$${{{\pi ^2}} \over 4} + {\pi \over 2}$$correct
  2. B.$${{{\pi ^2}} \over 2} + {\pi \over 4}$$
  3. C.$${{{\pi ^2}} \over 2} - {\pi \over 4}$$
  4. D.$${{{\pi ^4}} \over 4} - {\pi \over 2}$$

Answer

A. $${{{\pi ^2}} \over 4} + {\pi \over 2}$$

Explanation

$$xdy = (y + {x^3}\cos x)dx$$ $$ \Rightarrow $$ $$xdy = ydx + {x^3}\cos xdx$$ $$ \Rightarrow $$ $${{xdy - ydx} \over {{x^2}}} = {{{x^3}coxdx} \over {{x^2}}}$$ $$ \Rightarrow $$ $${d \over {dx}}\left( {{y \over x}} \right) = \int {x\cos xdx} $$ $$ \Rightarrow {y \over x} = x\sin x - \int {1.\sin xdx} $$ $$ \Rightarrow $$ $${y \over x} = x\sin x + \cos x + C$$ $$ \Rightarrow 0 = - 1 + C \Rightarrow C = 1,x = \pi ,y = 0$$ so, $${y \over x} = x\sin x + \cos x + 1$$ $$y = {x^2}\sin x + x\cos x + x$$ $$x = {\pi \over 2}$$ $$y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}$$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions