Initial conditionsy(t) = -5t2+10t+2026y(0)=024V22· y'(0)=10200.51201.5t100y(t) = -10t+10-10510.51.520t-9-10y"(t) = -10-1110.51.520tkshum6
Initial conditions • y(0)=0 • y’(0)=10 kshum 6 0 0.5 1 1.5 2 20 22 24 26 t y 0 0.5 1 1.5 2 -10 0 10 t v 0 0.5 1 1.5 2 -11 -10 -9 t a y(t) = –5t2+10t+20 y(t) = –10t+10 y’’(t) = –10
Initial conditionsy(t) = -5(t+1)20y(0)=-5-20y'(0)=-10-40-605100.51.52t-10> -20y(t) = -10t -10-3010.51.520t-9y"(t) = -10-10-115100.51.52tkshum7
Initial conditions • y(0)= –5 • y’(0)= –10 kshum 7 y(t) = –5(t+1)2 y(t) = –10t –10 y’’(t) = –10 0 0.5 1 1.5 2 -60 -40 -20 0 t y 0 0.5 1 1.5 2 -30 -20 -10 t v 0 0.5 1 1.5 2 -11 -10 -9 t a
Variables and parameters The dependent variable is called the system state, orthe phase of the system. The independent variable isusually time.A constant which does not change with time is calleda parameter. In the example Newton's law of motion y"(t) = g- Phase = system state = height of the mass-Independent variable =time g is parameter.kshum8
Variables and parameters kshum 8 • The dependent variable is called the system state, or the phase of the system. The independent variable is usually time. • A constant which does not change with time is called a parameter. • In the example Newton’s law of motion y’’(t) = g – Phase = system state = height of the mass – Independent variable = time – g is parameter
Solutions to a differential eguation: A solution is a function which satisfies the givendifferential equation.In solving differential equation, the solutions arefunction of time. In general, there are many solutions to a givendifferential equation. We have different solutionsfor different initial condition Deriving a solution is difficult, but checkingwhether a given function is a solution is easy.y(t) =-5t2+15 is a solution toy"(t) = -10because after differentiating-5t2+15 twice,we get -10y(t) = 4t2 is not a solution to y"(t) = -10because afterdifferentiating 4t2twice,weget 8,not -109kshum
Solutions to a differential equation • A solution is a function which satisfies the given differential equation. • In solving differential equation, the solutions are function of time. • In general, there are many solutions to a given differential equation. We have different solutions for different initial condition. • Deriving a solution is difficult, but checking whether a given function is a solution is easy. kshum 9 y(t) = –5t2+15 is a solution to y’’(t) = –10 because after differentiating –5t2+15 twice, we get –10. y(t) = 4t2 is not a solution to y’’(t) = –10 because after differentiating 4t2 twice, we get 8, not –10
General solution If every solution to a differential equation canbe obtained from a family of solutionsf(t,C1,C2,C3...,Cn)by choosing the constants C1, C2, C3,.., Cnappropriately, then we say thatf(t,C1,C2,C3..,Cn) is a general solution10kshum
General solution • If every solution to a differential equation can be obtained from a family of solutions f(t,c1 ,c2 ,c3 ,.,cn ) by choosing the constants c1 , c2 , c3 ,., cn appropriately, then we say that f(t,c1 ,c2 ,c3 ,.,cn ) is a general solution. kshum 10