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The Cauchy Euler Equation: The second order Cauchy Euler Equation is: ax2 d2y dx2 + bx dy dx + cy = 0 This equation can be transformed into a second order liner differential equa- tion with constant coefficients with use of the substitution: x = et Differentiating with respect to t gives: dx dt = et So dx dt = x In the Cauchy Euler equation we see the term x dydx which we need to substi- tute for, so I will multiply both sides of dxdt = x by dy dx : dx dt dy dx = x dy dx This simplifies to dy dt = x dy dx Differentiating dydt = x dy dx with respect to x gives: d dx dy dt = dy dx + x d2y dx2 In the Cauchy Euler equation we see the term x2 d 2y dx2 which we need to substitute for, so I will multiply the left side of the above equation by dxdt and the right hand side by x (Remember they are equal): d dx dy dt dx dt = ( dy dx + x d2y dx2 ) x This simplifies to d2y dt2 = x dy dx + x2 d2y dx2 Replacing x dydx with dy dt and solving for x 2 d 2y dx2 gives: x2 d2y dx2 = d2y dt2 − dy dt So under this substitution the Cauchy Euler equation becomes: a ( d2y dt2 − dy dt ) + b dy dt + cy = 0 This simplifies to the second order linear equation with constant coefficients: a d2y dt2 + (b− a)dy dt + cy = 0 Which we can solve by finding the roots of the characteristic polynomial: ar2 + (b− a)r + c = 0 Again there are three cases: the roots are real and distinct, the roots are real and repeated or the roots are complex. Case 1: we have two real and distinct roots r1 and r2. Then the solutions to the differential equation are: y1 = e r1t and y2 = e r2t Since x = et, t = ln(x) Making the solution: y1 = e r1 ln(x) = eln(x r1 ) and y2 = e r2 ln(x) = eln(x r2 ) So y1 = x r1 and y2 = x r2 Case 2: The roots are real and repeated r1 = r2. Then the solutions to the differential equation are: y1 = e r1t and y2 = te r1 Since x = et, t = ln(x) Making the solution: y1 = e r1 ln(x) = eln(x r1 ) and y2 = ln(x)e r1 ln(x) = ln(x)eln(x r1 ) So y1 = x r1 and y2 = ln(x)x r1 Case 3: The roots are complex r1 = α + βı and r2 = α − βı. Then the solutions to the differential equation are: y1 = e αt cos(βt) and y2 = e αt sin(βt) Since x = et, t = ln(x) Making the solution: y1 = x α cos(β ln(x)) and y2 = x α sin(β ln(x)) An Example: Solve: x2y′′ − 5xy′ + 13y = 0 Forming the characteristic polynomial r2 + (−5− 1)r + 13 = 0 r2 − 6r + 13 = 0 (r − 3)2 = −4 r = 3± 2ı So we have complex roots so the solution is: y = C1e 3x cos(2x) + C2e 3x sin(2x) 1. Solve: x2 d2y dx2 + 7x dy dx + 8y = 0 2. Solve: x2 d2y dx2 + 9x dy dx + 12y = 0 3. Solve: x2 d2y dx2 − 11xdy dx + 36y = 0 4. Solve: x2 d2y dx2 − 4xdy dx + 6y = x3 ln(x) 5. Solve: x2 d2y dx2 + y = x2
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