# Misc 11 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Nov. 14, 2019 by Teachoo

Miscellaneous

Misc 1 (i)

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)

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

Last updated at Nov. 14, 2019 by Teachoo

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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