f (ax+b)e^y/x=x then show that x^3d^2y/dx^2=(xdy/dx-y)^2

Question

Arun · Answer

lhs and rhs on differentiating
by uv method
(a+ bx) e^y/x . dy/dx (-1/x^2) + e^y/x (b) =1
since x= (a+ bx)e^y/x substituting for that

x.dy/dx(-1/x^2) + be^y/x=1
-1/x dy/dx + be^y/x=1 ; ------A
again differentiating

-1/x d^2y/dx^2 + dy/dx(1/x^2) + b e^y/x. dy/dx(-1/x^2)=0
d^2y/dx^2(-1/x) + 1/x^2 dy/dx[1-be^y/x]=1
d^2y/dx^2 (-1/x) =1-1/x^2dy/dx[1-be^y/x]
d^2y/dx^2=-x+1/x dy/dx[1-be^y/x]
from the given question
e^y/x= x/a+bx
d^2y/dx^2= -x+1/xdy/dx[ 1- bx/a+bx]

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f (ax+b)e^y/x=x then show that x^3d^2y/dx^2=(xdy/dx-y)^2

f (ax+b)e^y/x=x then show that x^3d^2y/dx^2=(xdy/dx-y)^2

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