InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Let `y=g(x)` be the solution of the differential equation `sin (dy)/(dx)+y cos x=4x, x in (0,pi)` If y(pi/2)=0`, then `y(pi/6)` is equal toA. (a) `4/(9sqrt3)pi^(2)`B. (b) `(-8)/(9sqrt3)pi^(2)`C. (c) `-8/(9)pi^(2)`D. (d) `-4/(9)pi^(2)` |
|
Answer» Correct Answer - (c) Wh have, `sin x dy/dx +y cos x = 4x rArr dy/dx+ y cot x = 4 x cosec x` This is a linear differential equation of form `dy/dx+ Py=Q` where `P=cot x,Q=4x cosec x` Now, `IF=e^(int pdx)=e^(int cot xdx)=e^(logsinx)=sinx` Solution of the differential equation is `y cdot sin x = int 4x cosec x sin x dx +C` `rArr ysin x = int 4xdx +C = 2x^(2)+C` Put `x=pi/2,y=0,` we get `C= -pi^(2)/2 rArr y sin x = 2x^(2) - pi^(2)/2` Put `x =pi/6` `therefore y(1/2)=2(pi^(2)/36)-pi^(2)/2` `rArr y=pi^(2)/9-pi^(2) rArr y=-(8pi^(2))/9` Alternate Method We have , `sin x dy/dx+y cos x = 4x,` which can be written as `d/dx(sin x cdoty) =4x` On integrating both sides, we get `int d/dx (sin x cdoty)cdot dx =int 4x cdot dx` `rArr y cdot sin x =(4x^(2))/2 + C rArr y cdot sin x =2x^(2)+C` Now, as y=0 when `x = pi/2` `therefore C=-pi^(2)/2` `rArr y cdot sin x = 2x^(2) -pi^(2)/2` Now, putting `x=pi/6,` we get `y(1/2)=2((pi^(2))/36)-pi^(2)/2 rArr y = pi^(2)/9-pi^(2)=-(8pi^(2))/9` |
|