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Misc 11 - Find particular solution: (x - y) (dx + dy) = dx - dy - Solving homogeneous differential equation

Misc 11 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 11 - Chapter 9 Class 12 Differential Equations - Part 3 Misc 11 - Chapter 9 Class 12 Differential Equations - Part 4


Transcript

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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.