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Unformatted text preview: Please report typos to LM Page 1 (20) 1. Solve the following initial value problem: z tan( x ) z = sec( x ) ; z (0) = The integrating factor is ( x ) = e R tan( x ) dx = e ln(cos( x )) = cos( x ) Thus the equation is reduced to d dx (cos( x ) z ) = 1 which yield z = x + c cos( x ) By applying the initial conditions one finds c = thus the solution of the problem is: z = x + cos( x ) Note: The integral of tan( x ) is one the simplest example of integrals that are easily solved by substitution. Since 80% of the class could not solve this integral... here is how substitution goes: Let u = cos( x ), then Z tan( x ) dx = Z du u = ln(cos( x )) Please report typos to LM Page 2 (20) 2. Solve the following initial value problem: y 00 + 2 4 y = f ( ) ; y (0) = y (0) = 0 . where f ( ) is a continuous function. Let the solution be decomposed into the sum of the homogeneous and a particular solution....
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 Spring '06
 LuigiMartinelli

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