The particular solution of the differential equation ² y d x d y +x= y when x=0 and y= 3 4 is

Step-by-step solution

We have, ² y d x dy + x = y d x dy + ( cosec ² y ) x = y cosec ² y I.F. = e ^ cosec ² ydy = e ^ - y xe ^ - y = e ^ - y y ec ² y dy In RHS, put - y = t cosec ² y dy = dt xe ^ - y = e ^ t (- t ) dt =- t e ^ t dt =- [ t e ^ t - e ^ t dt ]=- t e ^ t + e ^ t = e ^ t (1- t ) x e ^ - y = e ^ - y (1+ y)+C ...(2) At x=0, y= 3 4 , we get 0= e (1-1)+C From equation (2) , required solution is x e ^ - y = e ^ - y (1+ y) x =1+ yx=1+ y