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Answer: we have given the differential eqrationy^('')+4y=0" (1) "We have to transform (1) into Bessel diff. eqn, to do this we make an transform let y=ux^(1//2)- (2) where u is a function of x.by (2), (dy)/(dx)=x^(1//2)(dy)/(dx)+(1)/(2)x^(-1//2)yand (d^(2)y)/(delx^(2))=(1)/(2)x^(-1//2)(del y)/(dx)+x^(1//2)(d^(2)y)/(delx^(2))+(1)/(2)x^((1)/(2))(dy)/(del x)(-1)/(4)x^(-2)y=>(d^(2y))/(dx^(2))=x^(1//2)(d^(2)y)/(dx^(2))+x^(-1//2)(dy)/(dx)-(1)/(4)x^(-3//2u)Now put the value of (d^(2)y)/(dx^(2)) and y into the enn^(n) (1) we get{:[x^(1//2)(d^(2)y)/(delx^(2))+x^(-1//2)(del y)/(del x)-(1)/(4)x^(-3//2)y+4x^(1//2)u=0],[=>x^(2)(d^(2)y)/(delx^(2))+x(dy)/(del x)+(-(1)/(4))y^(')+4x^(2)y=0],[=>x^(2)(d^(2)u)/(dx^(2))+x(dy)/(del x)+(4x^(2)-(1)/(4))u=0]:}equ (3) is still not the Bessel diff. eq&qu ... See the full answer