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Unformatted text preview: Chapter 4 4. The corresponding homogeneous equation, y 00 + y = 0, has the associated auxiliary equation r 2 + r = r ( r + 1) = 0. This gives r = 0, 1, and a general solution to the homogeneous equation is y h ( t ) = c 1 + c 2 e t . Combining this solution with the particular solution, y p ( t ) = t , we find that a general solution is given by y ( t ) = y p ( t ) + y h ( t ) = t + c 1 + c 2 e t . 6. The corresponding auxiliary equation, r 2 + 5 r + 6 = 0, has the roots r = 3, 2. Therefore, a general solution to the corresponding homogeneous equation has the form y h ( x ) = c 1 e 2 x + c 2 e 3 x . By the superposition principle, a general solution to the original nonhomogeneous equation is y ( x ) = y p ( x ) + y h ( x ) = e x + x 2 + c 1 e 2 x + c 2 e 3 x . 8. First, we rewrite the equation in standard form, that is, y 00 2 y = 2 tan 3 x. The corresponding homogeneous equation, y 00 2 y = 0, has the associated auxiliary equation r 2 2 = 0. Thus r = ± √ 2, and a general solution to the homogeneous equation is y h ( x ) = c 1 e √ 2 x + c 2 e √ 2 x . Combining this with the particular solution, y p ( x ) = tan x , we find that a general solution is given by y ( x ) = y p ( x ) + y h ( x ) = tan x + c 1 e √ 2 x + c 2 e √ 2 x . 10. We can write the nonhomogeneous term as a difference ( e t + t ) 2 = e 2 t + 2 te t + t 2 = g 1 ( t ) + g 2 ( t ) + g 3 ( t ) . The functions g 1 ( t ), g 2 ( t ), and g 3 ( t ) have a form suitable for the method of undetermined coefficients. Therefore, we can apply this method to find particular solutions y p, 1 ( t ), y p, 2 ( t ), and y p, 3 ( t ) to y 00 y + y = g k ( t ) , k = 1 , 2 , 3 , respectively. Then, by the superposition principle, y p ( t ) = y p, 1 ( t ) + y p, 2 ( t ) + y p, 3 ( t ) is a particular solution to the given equation.particular solution to the given equation....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

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