challenge-firstorderlinearode

challenge-firstorderlinearode - MATH 16B - SPRING 2009 -...

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MATH 16B - SPRING 2009 - CHALLENGE PROBLEM 3 - FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS Problem Find a solution y = f ( t ) to the following initial value problem: y 0 = t ln( t ) y + te t 2 ln( t ) / 2 , (1) y (1) = 1 . (2) Solution The diﬀerential equation in (1) is a ﬁrst order linear diﬀerential equation. Written in the standard form, it is (3) y 0 + ( - t ln t ) y = te t 2 ln( t ) / 2 . The solution method for this type of equation now calls for us to compute an antiderivative A ( t ) for - t ln t . We integrate by parts with f ( t ) = ln t and g ( t ) = - t . The function g ( t ) has antiderivative G ( t ) = - t 2 / 2, and f ( t ) has derivative f 0 ( t ) = 1 /t . Thus, A ( t ) = Z - t ln( t ) dt = Z f ( t ) g ( t ) dt = f ( t ) G ( t ) - Z f 0 ( t ) G ( t ) dt = ln( t ) · - t 2 2 - Z 1 t · - t 2 2 dt = - t 2 ln( t ) 2 + Z t 2 dt = - t 2 ln( t ) 2 + t 2 4 . Here, as usual, we have assumed the constant of integration to be 0. We now multiply the diﬀerential equation in standard form (equation (3)) by the integrating factor

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This note was uploaded on 02/08/2011 for the course MATH 16B taught by Professor Sarason during the Fall '06 term at Berkeley.

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challenge-firstorderlinearode - MATH 16B - SPRING 2009 -...

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