InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Solve the differential equation : `x(dy)/(dx)+y-x+xycotx=0`, `x ne 0`. |
|
Answer» The given differential may be written as ` x (dy)/(dx) + (1 + x cot x )y = x ` `rArr (dy)/(dx) + ((1)/(x) + cot x ) y = 1 ` This is of the form ` (dy)/(dx) + Py = Q`, where ` P = ((1)/(x) + cot x ) and Q = 1 ` Thus, the given differential equation is linear. ` IF = e ^(int Pdx) = e ^( int ((1)/(x) + cot x ) dx) = e ^(log x + log sin x) ` `" " = e ^( log (x sinx )) = x sinx ` So, the solution of the given differential equation is given by `y xx (IF) = int (Q xx IF) dx + C`, i.e. , `y (x sin x) = int ( 1 xx (x sin x ) dx +C ` ` " " = int underset("I")( x )underset("II") (sin) x dx +C` ` " " = x ( - cos x ) - int 1 *( - cos x ) dx + C ` ` " " = - xcos x + int cos x dx + C ` ` " " = - x cos x + sin x +C`. ` therefore y = (1)/(x) - cot x + (C)/( x sin x ) ` is the required solution. |
|