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Overview An Example Double Check Series Solutions Bernd Schr¨ oder logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check What are Series Solutions? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check What are Series Solutions? 1. A series solution to a differential equation is a solution of ∞ ∑ c n ( x − x 0 ) n . the form y = n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check What are Series Solutions? 1. A series solution to a differential equation is a solution of ∞ ∑ c n ( x − x 0 ) n . That is, any solution that can the form y = n = 0 be expanded into a Taylor series is a series solution. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check What are Series Solutions? 1. A series solution to a differential equation is a solution of ∞ ∑ c n ( x − x 0 ) n . That is, any solution that can the form y = n = 0 be expanded into a Taylor series is a series solution. 2. For differential equations of the form y ′′ + P ( x ) y ′ + Q ( x ) y = 0, there are mild technical conditions ( P and Q must be analytic at x 0 ) that guarantee that there are series solutions. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check What are Series Solutions? 1. A series solution to a differential equation is a solution of ∞ ∑ c n ( x − x 0 ) n . That is, any solution that can the form y = n = 0 be expanded into a Taylor series is a series solution. 2. For differential equations of the form y ′′ + P ( x ) y ′ + Q ( x ) y = 0, there are mild technical conditions ( P and Q must be analytic at x 0 ) that guarantee that there are series solutions. 3. If there is a series solution, then we can substitute the series and its derivatives into the equation to obtain a set of equations for the c n . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check What are Series Solutions? 1. A series solution to a differential equation is a solution of ∞ ∑ c n ( x − x 0 ) n . That is, any solution that can the form y = n = 0 be expanded into a Taylor series is a series solution. 2. For differential equations of the form y ′′ + P ( x ) y ′ + Q ( x ) y = 0, there are mild technical conditions ( P and Q must be analytic at x 0 ) that guarantee that there are series solutions. 3. If there is a series solution, then we can substitute the series and its derivatives into the equation to obtain a set of equations for the c n . 4. These equations will allow us to compute the c n . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check What are Series Solutions? 1. A series solution to a differential equation is a solution of ∞ ∑ c n ( x − x 0 ) n . That is, any solution that can the form y = n = 0 be expanded into a Taylor series is a series solution. 2. For differential equations of the form y ′′ + P ( x ) y ′ + Q ( x ) y = 0, there are mild technical conditions ( P and Q must be analytic at x 0 ) that guarantee that there are series solutions. 3. If there is a series solution, then we can substitute the series and its derivatives into the equation to obtain a set of equations for the c n . 4. These equations will allow us to compute the c n . That’s it. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check What are Series Solutions? 1. A series solution to a differential equation is a solution of ∞ ∑ c n ( x − x 0 ) n . That is, any solution that can the form y = n = 0 be expanded into a Taylor series is a series solution. 2. For differential equations of the form y ′′ + P ( x ) y ′ + Q ( x ) y = 0, there are mild technical conditions ( P and Q must be analytic at x 0 ) that guarantee that there are series solutions. 3. If there is a series solution, then we can substitute the series and its derivatives into the equation to obtain a set of equations for the c n . 4. These equations will allow us to compute the c n . That’s it. (Except convergence analysis. Separate topic.) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 � ′′ � ′ � � � � ∞ ∞ ∞ ∑ c n x n ∑ c n x n ∑ c n x n − x + x = 0 n = 0 n = 0 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 � ′′ � ′ � � ∞ ∞ ∞ ∑ c n x n ∑ c n x n ∑ c n x n − x + x = 0 n = 0 n = 0 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 � ′′ � ∞ ∞ ∞ c n nx n − 1 + x ∑ c n x n ∑ ∑ c n x n − x = 0 n = 0 n = 1 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 ∞ ∞ ∞ c n nx n − 1 + x c n n ( n − 1 ) x n − 2 − x ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 ∞ ∞ ∞ c n nx n − 1 + x c n n ( n − 1 ) x n − 2 − x ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ nc n x n + n ( n − 1 ) c n x n − 2 − c n x n + 1 ∑ ∑ ∑ = 0 n = 2 n = 1 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 ∞ ∞ ∞ c n nx n − 1 + x c n n ( n − 1 ) x n − 2 − x ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ nc n x n + n ( n − 1 ) c n x n − 2 − c n x n + 1 ∑ ∑ ∑ = 0 n = 2 n = 1 n = 0 k : = n − 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 ∞ ∞ ∞ c n nx n − 1 + x c n n ( n − 1 ) x n − 2 − x ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ nc n x n + n ( n − 1 ) c n x n − 2 − c n x n + 1 ∑ ∑ ∑ = 0 n = 2 n = 1 n = 0 k : = n − 2 k : = n logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 ∞ ∞ ∞ c n nx n − 1 + x c n n ( n − 1 ) x n − 2 − x ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ nc n x n + n ( n − 1 ) c n x n − 2 − c n x n + 1 ∑ ∑ ∑ = 0 n = 2 n = 1 n = 0 k : = n − 2 k : = n k : = n + 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions

Overview An Example Double Check Solve the Initial Value Problem y ′′ − xy ′ + xy = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1 y ′′ − xy ′ + xy = 0 ∞ ∞ ∞ c n nx n − 1 + x c n n ( n − 1 ) x n − 2 − x ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ nc n x n + n ( n − 1 ) c n x n − 2 − c n x n + 1 ∑ ∑ ∑ = 0 n = 2 n = 1 n = 0 k : = n − 2 k : = n k : = n + 1 n = k + 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Series Solutions