Question Solved1 Answer Solve the following ordinary differential equation in terms of Bessel functions. You may initially transform the given equation into the Bessel differential equation. Show the steps of transformation and find the transformed equation. Then, write the solution in terms of Bessel functions. y" + 4y = 0
Get 4 Free Unlocks by registration
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