Math Euler Cauchy Equations

Transcription

1 1 Math Euler Cauchy Equations Erik Kjær Pedersen November 29, 2005 We shall now consider the socalled Euler Cauchy equation x 2 y + axy + by = 0 where a and b are constants. To solve this we put y = x m to get x 2 m(m 1)x m 2 +axmx m 1 +bx m = m 2 x m +(a 1)mx m +bx m = 0

2 So we see that y = x m is a solution if 2 m 2 + (a 1)m + b = 0 There are now three cases corresponding to two real roots, one real double root, and only complex roots. Case I: Two distinct real roots If we have two distinct real roots m 1 and m 2 they are linearly independent (not proportional), so the complete solution is given by Example Solve the initial value problem y = c 1 x m 1 + c 2 x m 2 x 2 y + 2xy 30y = 0 y(1) = 3 y (1) = 4

3 We first get the equation 3 m 2 + (2 1)m 30 = m 2 + m 30 = 0 This gives m = 1± so m = 5 and m = 6. The complete solution is thus We calculate that y = c 1 x 5 + c 2 x 6 y = 5c 1 x 4 6c 2 x 7 The initial conditions give c 1 + c 2 = 3 and 20c 1 6c 2 = 4 We calculate that c 1 = 2 and c 2 = 1 so the final answer is y = 2x 5 + x 6

4 4 Notice that the solution is not defined for x = 0 To explain this recall that the theory is only claiming there are two linearly independent solutions for equations of type y + p(x)y + q(x)y = 0 This equation is only of that type after dividing by x 2. But after dividing by x 2 the equation itself is not defined for x = 0. Case I: One real double root We now only get one solution and that solution is 1 2 (1 a), so the first solution is y 1 = x 1 2 (1 a) Attempting to find one more solution we write y = uy 1 and put that into the equation.

5 We get 5 y = uy 1 + u y 1 and y = uy 1 + 2u y 1 + u y 1 Putting this into the equation we get x 2 y = ux 2 y 1 +2x 2 u y 1 +x 2 u y 1 axy = uaxy 1 axu y 1 by = uby 1 The left column adds up to 0 because y 1 is a solution so we are left with 2x 2 u y 1 + x 2 u y 1 + axu y 1 = 0

6 Recalling that y 1 = x 1 2 (1 a) we get 6 2x 2 1 a 2 x 1 2 ( 1 a) u + x 2 u x 1 2 (1 a) + axx 1 2 (1 a) u = 0 Calculate and switch the last two terms around to get and reduce to (1 a)x 1 2 (3 a) u + ax 1 2 (3 a) u + xu x 1 2 (3 a) = 0 u + xu = 0 We now put U = u We get du U = dx x This means ln(u) = ln(x) = ln(x 1 ) So u = U = x 1 Therefore u = ln(x).

7 The complete solution is thus 7 where m is the root. Example Solve the initial value problem y = c 1 x m + c 2 x m ln(x) x 2 y xy + y = 0, y(1) = 1 y (1) = 2 We put y = x m and get the equation m 2 2m + 1 = 0 which has 1 as double root. We thus know the general solution is y = c 1 x + c 2 x ln(x) We calculate that y = c 1 + c 2 (ln(x) + 1) The initial conditions give c 1 = 1 and c 1 + c 2 = 2 so c 1 = c 2 = 1

8 The final solution is thus 8 y = x + x ln(x) Notice at this stage we can always just check the solution against the original equation to see if it is right. Case III: Complex roots α + iβ For completeness we carry this through. If m = α + iβ we get Continuing we get y = x α+iβ = x α x iβ = x α e ln(x)iβ y = x α (cos(ln(x)β) + i sin(ln(x)β)) The equation is linear so since this is a solution so is the real part and the imaginary part, so we get two real solutions

9 9 y 1 = x α cos(ln(x)β) and y 2 = x α sin(ln(x)β) so the general solution is y = c 1 x α cos(ln(x)β) + c 2 x α sin(ln(x)β) If we have a nonzero term on the right hand side of an Euler Cauchy equation there is only one method to use, namely the formula. Let us consider an example like that Example Solve the boundary value problem x 2 y 2y = 4x 3, y(1) = 2, y(2) = 12

10 10 First we solve the corresponding homogeneous equation. Put y = x m to get m 2 m 2 = 0 which has solutions y 1 = x 1 and y 2 = x 2. To find a particular solution we want to use the formula y 2 r y = y 1 W (x) + y y 1 r 2 W (x) Here W (x) = y 1 y 2 y 2y 1 = x 1 2x + x 2 x 2 = 3. We have to be very careful when we use this formula, because the formula relates to an equation of the form y + p(x)y + q(x)y = r(x) We have to divide by x 2 to get our equation on this form. This means in our case r(x) = 4x.

11 We now get the particular solution by the formula 11 y = x 1 x 2 4x 3 + x 2 x 1 4x 3 = 1 3 x x 3 = x 3 To get the complete solution we have to add the solution of the homogeneous equation to this particular solution The boundary values give us y = x 3 + c 1 x 1 + c 2 x c 1 + c 2 = 2, c 1 + 4c 2 = 12 This gives c 1 = 0 and c 2 = 1 so the final answer is y = x 3 + x 2

12 Let us consider one more example Example Solve the initial value problem 12 x 2 y 3xy + 4y = x, y(1) = 2 y (1) = 4 We start out finding solutions to the corresponding homogeneous equation by putting y = x m. We get m 2 4m + 4 = 0 and we see that m = 2 is a double root. This means the solutions to the homogeneous equation are y 1 = x 2 y 2 = x 2 ln(x) To find a particular solution we want to use the formula again y 2 r y p = y 1 W (x) + y y 1 r 2 W (x)

13 13 Here W (x) = y 1 y 2 y 2y 1 = x 2 (2x ln(x) + x) 2x 3 ln(x) = x 3. Once again we have to remember that we have to divide by x 2 to see that r(x) = x 1. The formula gives We reduce this to y p = x 2 x 2 ln(x)x 1 x 3 y p = x 2 ln(x) x 2 x + x 2 2 x 1 ln(x) x 3 + x 2 ln(x) x 2 We calculate ln(x) by partial integration x 2 ln(x) x 2 = ln(x)x 1 + x 1 1 x = x 1 ln(x) x 1

14 We see that 14 y p = x ln(x) + x + x 2 ln(x)( x 1 ) = x The complete solution is obtained by adding the particular solution to the solution to the homogeneous equation. We get This gives The initial conditions now give y = x + c 1 x 2 + c 2 x 2 ln(x) y = 1 + c 1 2x + c 2 (2x ln(x) + x) 1 + c 1 = 2, 1 + 2c 1 + c 2 = 4

15 so c 1 = c 2 = 1 and the final answer is 15 y = x + x 2 + x 2 ln(x) Once again we can of course check this by entering this in the original equation.