EqualRoots If F(r) has equal roots ri = r2, then we have one solutionyi(x)=xi* We could use reduction of order to get a second solutioninstead, we will consider an alternative methodSince F(r) has a double root ri, F(r) = (r - r), and F(r) = 0 This suggests differentiating L[x'] with respect to r and thensetting r equal to ri, as follows:L[x']=x"r2 +(α-1)r+β]-x'(r-r)oax(r-r)L[x']=ararL[x' In x]=x" In x(r -r) +2(r -r)x=y2(x)=xi ln x, x>0
Equal Roots If F(r) has equal roots r1 = r2 , then we have one solution We could use reduction of order to get a second solution; instead, we will consider an alternative method. Since F(r) has a double root r1 , F(r) = (r - r1 ) 2 , and F'(r1 ) = 0. This suggests differentiating L[x r ] with respect to r and then setting r equal to r1 , as follows: 1 ( ) 1 r y x = x ( ) ( ) ( ) ( ) ( ) ln , 0 [ ln ] ln 2 [ ] [ ] ( 1) 1 2 1 2 1 2 1 2 1 2 = = − + − − = = + − + = − y x x x x L x x x x r r r r x x r r r L x r L x x r r x r r r r r r r r r r r
Equal Roots* Thus in the case of equal roots ri = r2, we have two solutionsyi(x) = xi, y2(x)= xi ln x* Nowxi n xr'iyi y2Wyi y2/rx-l xn-(ri In x+1)= x21-(r ln x+1)-rix2r- In x= x2n-l ± 0 for all x > 0.* Thus yi and y2 are linearly independent, and the generalsolution to our differential equation isy(x)=cx +c,xi In x=(c +c, n x)x, x >0
Equal Roots Thus in the case of equal roots r1 = r2 , we have two solutions Now Thus y1 and y2 are linearly independent, and the general solution to our differential equation is y x x y x x x r r ( ) , ( ) ln 1 1 1 = 2 = ( ) ( ) 0 for all 0. ln 1 ln ln 1 ln 2 1 2 1 1 1 2 1 1 1 1 1 2 1 1 2 1 1 1 1 1 1 1 = = + − + = = − − − − − x x x r x r x x r x x r x x x x y y y y W r r r r r r r ( ) ln ( ln ) , 0 1 1 1 y x = c1 x + c2 x x = c1 + c2 x x x r r r