Lesson 5a - Linear

Letusseewhy consider i c x e a x dx c i c x e

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Unformatted text preview: constant (insert shocked face here). Let us see why: Consider I c ( x) e a ( x )dx c I c ( x) e a ( x )dx e c I c ( x) Ce a ( x )dx I c ( x) C ( I ( x)) Step 3: Your solution becomes: y 1 I ( x)b( x)dx I ( x) This means that our final solution would be: 1 y I I ( x) c ( x)b( x)dx c I ( x) Note that is called the integrating factor. No matter what c we have, it will cancel in the end every time! 1 y C ( I ( x))b( x)dx C ( I ( x)) y 1 I ( x)b( x)dx I ( x) Example 1 Determine the general solution to xy ' y 3 x 3 I ( x)b( x)dx x(3 x 2 )dx xy ' y 3 x 3 3x 3dx xy ' y 3 x 3 xx x 1 y ' y 3 x 2 x y ' a ( x) y b( x) 1 a( x) x I ( x) e b( x ) 3 x 2 1 dx x I ( x) e ln| x| I ( x) x 34 x c 4 1 y I ( x)b( x)dx I ( x) 13 y ( x 4 c) x4 y 33 c x 4 x Example 2 Determine the general solution to I ( x)b( x)dx e y '2 xy x y ' a ( x) y b( x) a( x) 2 x 2 xdx I ( x) e I ( x) e x2 y ' xy x y b( x ) x x2 xdx Let u x 2 , then du 2 xdx 1 e u du 2 1u e c 2 1 x2 e c 2 1 y I ( x)b( x)dx I ( x) 1 1 x2 y x 2 ( e c) e2 1 x2 y ce 2...
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This note was uploaded on 02/07/2014 for the course MATH 1004 taught by Professor Mark during the Summer '00 term at Carleton CA.

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