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