IIT JAM Follow
July 26, 2020 3:00 pm 30 pts
The given differential equation, dy dy - 2y 3e dx dx can be written in the form of auxiliary equation i.e m2 m- 2 0 m =-1, 2 C.F C.e- +Ce2 -D-2 se = xex 1 and P.A. general solution, y C.F. + P.I y C.ex + C.ea+ xe Now y(0) = 0 and y(0) -2 Putting x =0 in (1) y(0) = C, +C, +0 = 0 i.e. and puttingx= 0 in (3), y(0) =-C, +2C, + 1 + C, +0 = 0 C,+C, = 0 y=-C.er+2C.e2+2xe+ e ea [differentiating (1) fdifferentiating (1 - 2 =- C, + 2C, +1 - C, + 2C, =- 3 Solving (2) and (4), C = 1, C,-1 -1(1) becomes, y= e-e+ xex tact Useose
• 0 Likes
• Shares
• Vaishali sharma
for P.I. if we directly put D=2 then denominator become 0 which is not acceptable so we differentiate denominator and put extra x in numerator and it's a method
Chandra dhawan
👍
• Ruby negi
complete solution...
Ruby negi
for any doubt
• Deepak singh
see attached
Deepak singh
it's simple rule..
• Chandra dhawan
see attached file
Chandra dhawan
see
• Pranav kumar
see the attachment.
Pranav kumar
• Mahak
see in the given pi when you will put x=2 in denominator the denominator will become zero so apply L hospital rule in denomitor then denominator will be 2D-1 and the solve pi