As before we set c y 0 multiplying by y 1 and

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As before, we set c = y (0). Multiplying by y + 1 and integrating, we find tan( T ) = R T 0 sec 2 ( t ) dt = R T 0 ( y ( t ) + 1) y 0 ( t ) dt = R u = y ( T ) u = C ( u + 1) du = 1 2 y ( T ) 2 + y ( T ) - 1 2 c 2 - c 19
10: Differential equations 10.1:Solutions of elementary and separable differential equations Solution, continued So, y satisfies the equation y 2 + 2 y - c 2 - 2 c - 2 tan( t ) = 0 20
10: Differential equations 10.1:Solutions of elementary and separable differential equations Solution, continued So, y satisfies the equation y 2 + 2 y - c 2 - 2 c - 2 tan( t ) = 0 From the quadratic formula, we compute that y = - 2 ± 2 2 - 4( - c 2 - 2 c - 2 tan( t )) 2 = - 1 ± p 1 + c 2 + 2 c + 2 tan( t ) = - 1 ± p ( c + 1) 2 + 2 tan( t ) 20
10: Differential equations 10.1:Solutions of elementary and separable differential equations Undefined solutions, multiple solutions I For each choice of c 6 = - 1, we found two solutions to the initial value problem y 0 = sec 2 ( t ) y +1 , namely y = - 1 + p ( c + 1) 2 + 2 tan( t ) and y = - 1 - p ( c + 1) 2 + 2 tan( t ). However, only y = - 1 + p ( c + 1) 2 + 2 tan( t ) satisfies y (0) = c . 21
10: Differential equations 10.1:Solutions of elementary and separable differential equations Undefined solutions, multiple solutions I For each choice of c 6 = - 1, we found two solutions to the initial value problem y 0 = sec 2 ( t ) y +1 , namely y = - 1 + p ( c + 1) 2 + 2 tan( t ) and y = - 1 - p ( c + 1) 2 + 2 tan( t ). However, only y = - 1 + p ( c + 1) 2 + 2 tan( t ) satisfies y (0) = c . I Strictly speaking, there is no solution with y (0) = - 1, but there is a solution having lim t 0 + y ( t ) = - 1. 21
10: Differential equations 10.1:Solutions of elementary and separable differential equations Undefined solutions, multiple solutions I For each choice of c 6 = - 1, we found two solutions to the initial value problem y 0 = sec 2 ( t ) y +1 , namely y = - 1 + p ( c + 1) 2 + 2 tan( t ) and y = - 1 - p ( c + 1) 2 + 2 tan( t ). However, only y = - 1 + p ( c + 1) 2 + 2 tan( t ) satisfies y (0) = c . I Strictly speaking, there is no solution with y (0) = - 1, but there is a solution having lim t 0 + y ( t ) = - 1. I No solution to the differential equation is defined for all values of t . If t 0 so that tan( t ) < - ( c +1) 2 2 , then p ( c + 1) 2 + 2 tan( t ) is not a real number. 21
10: Differential equations 10.1:Solutions of elementary and separable differential equations Another Example Find a function y satisfying y (0) = 5 and y 0 = t e y 22
10: Differential equations 10.1:Solutions of elementary and separable differential equations Another Example Find a function y satisfying y (0) = 5 and y 0 = t e y As the exponential function never attains the value zero, there are no constant solutions to this differential equation. 22
10: Differential equations 10.1:Solutions of elementary and separable differential equations Another Example Find a function y satisfying y (0) = 5 and y 0 = t e y As the exponential function never attains the value zero, there are no constant solutions to this differential equation. Multiplying both sides of the equation by e y and integrating, we obtain: 1 2 T 2 = Z T 0 tdt = Z T 0 e y ( t ) y 0 ( t ) dy = Z y ( T ) 5 e y dy = e y ( T ) - e 5 22
10: Differential equations 10.1:Solutions of elementary and separable differential equations Solution, continued So we have e y ( T ) - e 5 = 1 2 T 2 .

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