solution_2.1 - Solutions to selected problems in Section...

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Solutions to selected problems in Section 2.1 1. The ODE is linear because it can be put in the form y 0 + p ( t ) y = g ( t ). Another way to see why this ODE is linear is because y and its derivatives appear by themselves or as multiples of themselves and functions of t . 2. The ODE is nonlinear because of the y 2 term. 3. The ODE is nonlinear because of the y 0 /y term. 4. The equation is autonomous. The equilibria are the roots of y 2 - y = 0, i.e. 0 and 1. 5. Nonlinear because of the division by y . 6. Linear, homogeneous 7. Nonlinear because of yy 0 . 8. Nonlinear. 9. Linear, inhomogeneous (multiply by y ). 10. Linear, homogeneous. 13. (a). Note that for the given ODE, we have p ( t ) = t t 2 - 4 , g ( t ) = e t t - 3 . Discontinuous points of p ( t ): t = 2 , - 2. Discontinuous points of g ( t ): t = 3. These discontinuous points, together with -∞ and , can form the following intervals: ( -∞ , - 2) , ( - 2 , 2) , (2 , 3) , (3 , ) . From the initial condition, we have t 0 = 5. The only interval found above that contains t 0 = 5 is (3 , ); hence this is the interval on which Theorem 2.1 guarantees the existence of
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This note was uploaded on 04/27/2010 for the course MATH 13960 taught by Professor Prather during the Spring '09 term at Virginia Tech.

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solution_2.1 - Solutions to selected problems in Section...

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