Ch5.5:EulerEquations* A relatively simple differential equation that has a regularsingular point is the Euler equation,Llyl=xy"+αxy'+βy=0where α, β are constants* Note that xo = O is a regular singular point*The solution of the Euler equation is typical of the solutionsof all differential equations with regular singular points, andhence we examine Euler equations before discussing themore general problem
Ch 5.5: Euler Equations A relatively simple differential equation that has a regular singular point is the Euler equation, where , are constants. Note that x0 = 0 is a regular singular point. The solution of the Euler equation is typical of the solutions of all differential equations with regular singular points, and hence we examine Euler equations before discussing the more general problem. [ ] 0 2 L y = x y + xy + y =
Lly]=x"y"+αxy'+βy=0Solutions of the Form y=x* In any interval not containing the origin, the general solutionof the Euler equation has the formy(x) = Cy(x)+C2y2(x)* Suppose x is in (O, co), and assume a solution of the formy= x'. Theny=x", y'=rxr-l,y"=r(r-1)x"-2* Substituting these into the differential equation, we obtainL[x']=r(r-l)x"+αrx"+βx"=0orL[x']=x'[r(r-1)+αr+β]=0orL[x']= x[r2 +(α-1)r+β]= 0
Solutions of the Form y = x r In any interval not containing the origin, the general solution of the Euler equation has the form Suppose x is in (0, ), and assume a solution of the form y = x r . Then Substituting these into the differential equation, we obtain or or ( ) ( ) ( ) 1 1 2 2 y x = c y x +c y x 1 2 , , ( 1) − − = = = − r r r y x y r x y r r x [ ] = ( −1) + + = 0 r r r r L x r r x r x x [ ] ( 1) 0 2 L x = x r + − r + = r r L[x ] = x r(r −1) + r + = 0 r r [ ] 0 2 L y = x y + xy + y =
Quadratic Equation* Thus, after substituting y = x' into our differential equation,we arrive atx(r2 +(α-1)r+β)=0, x>0and hencer=-(α-1)±V(α-1)°-4β2* Let F(r) be defined byF(r)=r2 +(α-1)r +β=(r -r)(r-r)* We now examine the different cases for the roots ri, r2
Quadratic Equation Thus, after substituting y = x r into our differential equation, we arrive at and hence Let F(r) be defined by We now examine the different cases for the roots r1 , r2 . ( ( 1) ) 0, 0 2 x r + − r + = x r ( ) ( 1) ( )( ) 1 2 2 F r = r + − r + = r − r r − r 2 ( 1) ( 1) 4 2 − − − − r =
Real. Distinct Roots* If F(r) has real roots r; ± r2, thenyi(x) =xi, y(x)= xare solutions to the Euler equation. Note thatxxiyiy2Wrr3-1[y y2rxi-1=rxitn-l -rxi+h-!=(rz -r)xi+n-l #0 for all x >0.* Thus yi and y2 are linearly independent, and the generalsolution to our differential equation isy(x)=cxi +cx2, x >0
Real, Distinct Roots If F(r) has real roots r1 r2 , then are solutions to the Euler equation. Note that Thus y1 and y2 are linearly independent, and the general solution to our differential equation is 1 2 ( ) , ( ) 1 2 r r y x = x y x = x ( ) 0 for all 0. 1 2 1 1 1 1 2 1 2 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 = − = − = = + − + − + − − − r r x x r x r x r x r x x x y y y y W r r r r r r r r r r ( ) , 0 1 2 y x = c1 x + c2 x x r r
Example 1* Consider the equation3x y"+xy'-y=0, x >0* Substituting y = x' into this equation, we obtainy=x", y'=rxr-l,y"=r(r-1)x"-2y(x) = x(-1/3) + )and333r(r -1)x +rx- x = 02.5x'[3r(r-1)+r-1]= 0y(91.5x"[3r? - 2r-1]=00.5x' (3r +1)(r -1)= 0* Thus ri = -1/3, rz = 1, and our general solution isy(x)=cjx-/3 +cx, x>0
Example 1 Consider the equation Substituting y = x r into this equation, we obtain and Thus r1 = -1/3, r2 = 1, and our general solution is 1 2 , , ( 1) − − = = = − r r r y x y r x y r r x (3 1)( 1) 0 3 2 1 0 3 ( 1) 1 0 3 ( 1) 0 2 + − = − − = − + − = − + − = x r r x r r x r r r r r x rx x r r r r r r ( ) 2 , 0 1/3 = 1 + − y x c x c x x 3 0, 0 2 x y + xy − y = x