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Unformatted text preview: ANSWERS LINEAR EQUATIONS EXAMPLES 1 . dy dx =- y + 5 , y (0) = 1 . The homogeneous problem, y =- y , has general solution y h ( x ) = K e- x . Since the forcing term is constant, we look for constant solutions of the non- homogeneous equation. This equation has the constant solution y nh 5. Hence the general solution y ( x ) = K e- x + 5. In order to satisfy the initial condition: 1 = y (0) = K e + 5 so we must take K =- 4. So the solution of the initial value problem is y ( x ) =- 4 e- x + 5 . 2 . dy dx = 2 y- 10 . The homogeneous problem is y = 2 y which has general solution y h ( x ) = K e 2 x . As in the previous problem, the non-homogeneous has a constant solution as particular solution, namely y nh ( x ) 5. So the general solution for the full problem is y ( x ) = K e 2 x + 5 . 3 . dv dt = 9 . 8- v 5 , v (0) = 0 . In this example, the homogeneous problem has the general solution v h ( t ) = K e- t 5 , while, again, the constant solution to the non-homogeneous problem, derived by setting...
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