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# L42 - Fall 2003 Math 308/501502 4 Linear Second-Order...

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Fall 2003 Math 308/501–502 4 Linear Second-Order Equations 4.2 Homogeneous Linear Equations Wed, 24/Sep c 2003, Art Belmonte Summary In the Section 6.1 lecture handout, there is an overview of the theory of n th order linear differential equations for n 1, q.v. Accordingly, the homogeneous linear second-order constant coefficient equations dealt with here are just a special case. Rather than recapitulate what is covered in §6 . 1, we simply highlight how to actually find solutions to these special types of equations. Terminology and Solutions Consider the ODE ay 00 + by 0 + cy = 0, where a , b , c are constants (with a 6= 0) and t is the independent variable (for specificity). The ODE has an associated auxiliary or characteristic equation ar 2 + br + c = 0. This quadratic equation has roots r = r 1 , r 2 = - b ± p b 2 - 4 ac 2 a , the discriminant of which is b 2 - 4 ac . In the following , c 1 and c 2 are arbitrary constants. When b 2 - 4 ac > 0, the roots are real and distinct. In this case, y 1 ( t ) = e r 1 t and y 2 ( t ) = e r 2 t form a fundamental set of solutions to the ODE. A general solution is y ( t ) = c 1 e r 1 t + c 2 e r 2 t . When b 2 - 4 ac = 0, there is a repeated or double root. In this case y 1 ( t ) = e rt and y 2 ( t ) = te rt form a fundamental set of solutions to the ODE. A general solution is y ( t ) = c 1 e rt + c 2 te rt . When b 2 - 4 ac < 0, the roots are complex conjugate numbers. We shall deal with this last case in Section 4.3. Hand Examples Example A Find a general solution of 2 y 00 - y 0 - y = 0. Solution The characteristic equation is 0 = 2 r 2 - r - 1 = ( 2 r + 1 )( r - 1 ) , whence r = - 1 2 or r = 1. A general solution of the DE is y ( t ) = c 1 e - t / 2 + c 2 e t . Example B Find a general solution of 4 y 00 + 12 y 0 + 9 y = 0. Solution The characteristic equation is 4 r 2 + 12 r + 9 = 0. The quadratic formula gives r = - 12 ± 144 - 144 8 = - 3 2 , a double root. Accordingly, a general solution is y ( t ) = c 1 e - 3 2 t + c 2 te - 3 2 t . Example C Explain why y 1 ( t ) = e - 3 t and y 2 ( t ) = te - 3 t are linearly independent solutions of the DE y 00 + 6 y 0 + 9 y = 0 by using the definition of linear independence. Also show why the Wronskian of y 1 and y 2 implies this linear independence. Solution By computing derivatives and substituting into the differential equation, we see that both y 1 and y 2 are solutions of the DE.

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L42 - Fall 2003 Math 308/501502 4 Linear Second-Order...

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