Cauchy-Euler Equations Bernd Schr oder logo1 Bernd Schr oder - PowerPoint PPT Presentation

Overview An Example Double Check Discussion Cauchy-Euler Equations Bernd Schr¨ oder logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Definition and Solution Method logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Definition and Solution Method 1. A second order Cauchy-Euler equation is of the form a 2 x 2 d 2 y dx 2 + a 1 xdy dx + a 0 y = g ( x ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Definition and Solution Method 1. A second order Cauchy-Euler equation is of the form a 2 x 2 d 2 y dx 2 + a 1 xdy dx + a 0 y = g ( x ) . If g ( x ) = 0, then the equation is called homogeneous . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Definition and Solution Method 1. A second order Cauchy-Euler equation is of the form a 2 x 2 d 2 y dx 2 + a 1 xdy dx + a 0 y = g ( x ) . If g ( x ) = 0, then the equation is called homogeneous . 2. To solve a homogeneous Cauchy-Euler equation we set y = x r and solve for r . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Definition and Solution Method 1. A second order Cauchy-Euler equation is of the form a 2 x 2 d 2 y dx 2 + a 1 xdy dx + a 0 y = g ( x ) . If g ( x ) = 0, then the equation is called homogeneous . 2. To solve a homogeneous Cauchy-Euler equation we set y = x r and solve for r . 3. The idea is similar to that for homogeneous linear differential equations with constant coefficients. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Definition and Solution Method 1. A second order Cauchy-Euler equation is of the form a 2 x 2 d 2 y dx 2 + a 1 xdy dx + a 0 y = g ( x ) . If g ( x ) = 0, then the equation is called homogeneous . 2. To solve a homogeneous Cauchy-Euler equation we set y = x r and solve for r . 3. The idea is similar to that for homogeneous linear differential equations with constant coefficients. We will use this similarity in the final discussion. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r − x r logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r − x r = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r − x r = 0 2 r ( r − 1 )+ r − 1 = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r − x r = 0 2 r ( r − 1 )+ r − 1 = 0 2 r 2 − r − 1 = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r − x r = 0 2 r ( r − 1 )+ r − 1 = 0 2 r 2 − r − 1 = 0 ( − 1 ) 2 − 4 · 2 · ( − 1 ) � − ( − 1 ) ± = r 1 , 2 2 · 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r − x r = 0 2 r ( r − 1 )+ r − 1 = 0 2 r 2 − r − 1 = 0 ( − 1 ) 2 − 4 · 2 · ( − 1 ) � − ( − 1 ) ± = r 1 , 2 2 · 2 1 ± 3 = 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r − x r = 0 2 r ( r − 1 )+ r − 1 = 0 2 r 2 − r − 1 = 0 ( − 1 ) 2 − 4 · 2 · ( − 1 ) � − ( − 1 ) ± = r 1 , 2 2 · 2 1 ± 3 = 1 , − 1 = 4 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations

Overview An Example Double Check Discussion Solve the Initial Value Problem 2 x 2 y ′′ + xy ′ − y = 0, y ( 1 ) = 1, y ′ ( 1 ) = 2 2 x 2 y ′′ + xy ′ − y = 0 2 x 2 r ( r − 1 ) x r − 2 + xrx r − 1 − x r = 0 2 r ( r − 1 ) x r + rx r − x r = 0 2 r ( r − 1 )+ r − 1 = 0 2 r 2 − r − 1 = 0 ( − 1 ) 2 − 4 · 2 · ( − 1 ) � − ( − 1 ) ± = r 1 , 2 2 · 2 1 ± 3 = 1 , − 1 = 4 2 y ( x ) = c 1 x 1 + c 2 x − 1 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cauchy-Euler Equations