soln_hw1 - Solutions to Homework 1 Section 1.3 1 Second...

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Solutions to Homework 1 Section 1.3 1. Second order linear. 2. Second order nonlinear. The term (1 + y 2 ) d 2 y dt 2 is nonlinear. 5. Second order nonlinear. The term sin ( t + y ) is nonlinear. 6. Third order linear. 8. To show that y 1 and y 2 are solutions we take two derivatives, y 1 = e 3 t , y 1 = - 3 e 3 t , y ′′ 1 = 9 e 3 t , substitute into the equation y ′′ + 2 y - 3 y = 0 and verify that it does in fact equal 0, y ′′ 1 + 2 y 1 - 3 y 1 = 9 e 3 t - 6 e 3 t - 3 e 3 t = 0 . For y 2 , y 2 = y 2 = y ′′ 2 = e t , and y ′′ 2 + 2 y 2 - 3 y 2 = e 3 t + 2 e 3 t - 3 e 3 t = 0 . 9. To show that y = 3 t + t 2 is a solution take the first derivative, y = 3+2 t , and substitute in ty - y = t (3 + 2 t ) - (3 t + t 2 ) = 3 t + 2 t 2 - 3 t - t 2 = t 2 . 16. To find the values for r that are solutions we simply substitute y = e rt into the equation and solve for r , 1
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y ′′ - y = r 2 e rt - e rt = 0 . Note that e rt never equals zero so it can safely be divided out, r 2 - 1 = 0 . The two values of r that give solutions are r = 1 and r = - 1. 18. As above substitute in y = e rt , y ′′′ - 3 y ′′ + 2 y = r 3 e rt - 3 r 2 e rt + 2 re rt .
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