Sec_2.5 - 2.5 Linear Equations Definition 2.5 Linear...

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Unformatted text preview: 2.5 Linear Equations Definition 2.5 Linear Equation A differential equation of the form dy a1 ( x ) + a0 ( x ) y = g ( x ) dx is said to be a linear equation. Integrating Factor: a1 ( x ) g ( x) dy dy a ( x ) + a0 ( x ) y = g ( x ) + 0 y= dx dx a1 ( x ) a1 ( x ) dy dy + p ( x) y = f ( x) + p( x) y - f ( x) = 0 dx dx dy + ( x ) y - f ( x ) = 0 p dx ( x ) dy + ( x ) ( x ) y - f ( x ) = 0 p dx It is an exact differential equation. y p { ( x ) ( x ) y - f ( x ) = } x x ( x) ( x) p( x) = pdx = = ce Ex. Solve x Solution: ( x) pdx d ln = pdx + c1 Choose c = 1 = e pdx . dy - 4 y = x 6e x . dx dy dy 4 - 4 y = x 6 e x - y = x 5e x dx dx x 4 p ( x ) = - . So, the integrating factor x x ( x) = e 4 - dx x ( x ) = e -4ln x = x -4 . x -4 dy d - 4 x -5 y = xe x ( x -4 y ) = xe x dx dx d ( x -4 y ) = xe x dx x -4 y = xe x - e x + c y = x 5e x - x 4e x + cx 4 . Ex. Solve dy - 3y = 0 . dx Solution: p ( x ) = -3 . The integrating factor is = e -3 x . dy And e -3 x - 3e-3 x y = 0 dx d -3 x y 0 e -3 x y = c y = ce3 x . e = dx =e -3 dx Ex. Find the general solution of Solution: 2 Rewrite ( x + 9 ) (x 2 + 9) dy + xy = 0 . dx factor is dy dy x + xy = 0 + 2 y = 0 . Then the integrating dx dx x + 9 1 ln ( x 2 + 9 ) x / ( x 2 + 9 ) dx x =e = e2 . x2 + 9 dy x dy And x 2 + 9 + y = 0 x 2 + 9 0 y = dx dx x2 + 9 c x2 + 9 = c y = y . x2 + 9 p( x) = Ex. Solve the IVP dy + 2 xy = x, y ( 0 ) = -3 . dx Solution: p ( x ) = 2 x . The integrating factor is = e 2 xdx = e x2 . 2 dy 2 2 2 2 d ex + 2 xe x y = xe x x y xe x e = dx dx 2 2 1 2 e x y = xe x dx = e x + c 2 2 1 y = + ce - x . 2 Since y ( 0 ) = -3 -3 = Ex. Solve the IVP x 1 7 1 7 2 + c c = - y = - e- x . 2 2 2 2 dy + y = 2 x, y ( 1) = 0 . dx Solution: dy dy 1 1 x + y = 2x + y = 2 p ( x ) = . The integrating factor is dx dx x x 1/ xdx e y = eln x = x . And x dy d + y = 2 x [ xy ] = 2 x dx dx xy = 2 xdx xy = x 2 + c. And y ( 1) = 0 0 = 1 + c c = -1 1 y=x- . x Ex. Solve the IVP Solution: dy 1 = , y ( -2 ) = 0 . dx x + y 2 The original equation is not separable, homogeneous, exact, or linear. If dx dx = x + y 2 - x = y 2 p ( y ) = 1. So, the integrating factor we look at dy dy is = e -1dy = e - y . Then, we can obtain dx d e- y - xe - y = y 2e - y - y y 2e - y xe = dy dy xe - y = y 2e - y dy Let u = y 2 , du = 2 ydy dv = e - y dy, v = -e - y xe- y = - y 2 e- y + 2 y e - y dy xe - y = - y 2 e - y + 2 ye - y + e - y dy - xe - y = - y 2 e - y - 2 ye - y - 2e - y + c. Since, y ( -2 ) = 0 -2 = -2 + c c = 0 x = - y 2 - 2 y - 2. Ex. Find a continuous solution satisfying dy + y = f ( x ) , where dx 1, 0 y x 1 f ( x) = and the initial condition y ( 0 ) = 0 . 0, x >1 Solution: p ( x ) = 1 . The integrating factor is = e 1dx = e x . Then, 1. f ( x ) = 1 : dy dy + y = 1 ex + ex y = ex dx dx d x y e x e x y = e x + c, y ( 0 ) = 0 e = dx 0 = 1 + c c = -1 y = 1 - e- x . 2. f ( x ) = 0 : dy dy + y = 0 ex + ex y = 0 dx dx d x y 0 e x y = c1 y = c1e - x . e = dx Then, the solution is 1 1 - e - x ,0 x 1 - e -1 y = -x When x = 1, y = -1 c1e x >1 c1e , c1e-1 = 1 - e -1 c1 = e - 1. 1 - e - x ,0 x 1 y= is constinuous at x = 1 but not differentiable -x x > 1 ( e - 1) e , at x = 1. . Homework: 1~57odd. ...
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