MATH150T5

MATH150T5 - MATH 150 Second-order linear homogeneous ODE...

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Unformatted text preview: MATH 150 Second-order linear homogeneous ODE with constant coefficients Cheung Man Wai Tutorial 5 Please refer to http://ihome.ust.hk/ mandywai for updates. The general second-order linear differential equation is given by ¨ x + p ( t ) ˙ x + q ( t ) x = g ( t ) , where ˙ x = dx/dt and ¨ x = d 2 x/dt 2 . Unique solution requires initial values x ( t ) = x , ˙ x ( t ) = u . The equation with constant coefficients assumes that p ( t ) and q ( t ) are constants. The second-order linear ODE is said to be homoge- neous if g ( t ) = 0, in which the equation becomes ¨ x + p ( t ) ˙ x + q ( t ) x = 0 . Suppose that X 1 ( t ) and X 2 ( t ) are solutions, then the principle of superposition states that any linear combination of X 1 and X 2 , say X ( t ) := c 1 X 1 ( t ) + c 2 X 2 ( t ) , is also a solution, where c 1 and c 2 are constants. Now consider the equation a ¨ x + b ˙ x + cx = 0 (*) with a , b and c constants. Let x = e rt , we obtain ar 2 + br + c = 0 (**) which is called the characteristic equation . By quadratic formula, the two roots are r ± =- b ± √ b 2- 4 ac 2 a . 1 1 Real, distinct roots When r + 6 = r- are real roots, then the general solution to (*) can be written as a linear combination of the two solutions e r + t and e r- t . In other words, x ( t ) = c 1 e r + t + c 2 e r- t . The unknown constants c 1 and c 2 can be determined by the initial conditions x ( t ) = x and ˙ x ( t ) = u . Let us give some examples of this method. Example 1 Find the solution of the given initial value problem ¨ x + ˙ x- 2 x = 0 , x (0) = 2 , ˙ x (0) =- 1 and describe its behavior as t increases. We assume that x = e rt and it follows that r must be a root of the charac- teristic equation r 2 + r- 2 = ( r- 1)( r + 2) = 0 ....
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MATH150T5 - MATH 150 Second-order linear homogeneous ODE...

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