


Miscellaneous
Misc 1 (ii)
Misc 1 (iii) Important
Misc 2 (i)
Misc 2 (ii) Important
Misc 2 (iii)
Misc 2 (iv) Important
Misc 3 Deleted for CBSE Board 2022 Exams
Misc 4 Important
Misc 5 Important Deleted for CBSE Board 2022 Exams
Misc 6
Misc 7 Important
Misc 8
Misc 9 Important
Misc 10 Important
Misc 11 You are here
Misc 12 Important
Misc 13
Misc 14 Important
Misc 15 Important
Misc 16 (MCQ)
Misc 17 (MCQ) Important Deleted for CBSE Board 2022 Exams
Misc 18 (MCQ)
Last updated at Nov. 14, 2019 by Teachoo
Misc 11 Find a particular solution of the differential equation ( )( + )= , given that = 1 , when =0 ( : = ) ( )( + )= + y dx y dy = dx dy x dx y dx dx = xdy + y dy dy (x y 1) dx = ( x + y 1)dy / = ( 1)/( + 1) / = (( 1))/( ( + 1)) Put x y = t Diff w.r.t.x 1 / = / Putting value of t & dt in (1) 1 / = (( 1))/( + 1) / = 1 + (( 1))/( + 1) / = ( + 1 + 1)/( + 1) / = 2 /( 1) ( 1)/2 dt = Integrating both sides 1 ( + 1 )/2 = 1 1 ( /2 +1/2 ) = x + c 1 (1/2+1/2 ) = x + c /2 + 1/2 log | | = x + c Putting value of t = x y ( )/2 + 1/2 log | | = x + C Given y = 1 when x = 0 Put x = 0 & y = 1 in (2) (0 + 1)/2 + 1/2 log 1 = 0 + C 1/2 = C C = 1/2 Putting value in (2) ( )/2+1/2 log | |= +1/2 ( )/2+1/2 log | |=(2 + 1)/2 x y + log | |=2 +1 log | | = 2x + 1 x + y log | | = x + y + 1