Sec_2.2 - 2.2 Separable Variables Ex1. Solve dy = 1 + e2 x...

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Unformatted text preview: 2.2 Separable Variables Ex1. Solve dy = 1 + e2 x a. dx Solution: dy = 1 + e 2 x dy = ( 1 + e 2 x ) dx dx 1 = + e 2 x ) dx y = x + e 2 x + c. dy (1 2 b. dy = sin ( x ) dx Solution: dy = sin ( x ) dy = sin ( x ) dx dx = ( x ) dx y = - cos ( x ) + c. dy sin Definition 2.1 Separable Variables A differential equation of the form dy g ( x ) = dx h ( y ) is said to be separable. Ex2. Solve ( 1 + x ) dy - ydx = 0 Solution: ( 1 + x ) dy - ydx = 0 ( 1 + x ) dy = ydx dy dx dy dx = = y 1+ x y 1+ x ln y = ln 1 + x + c eln y = eln 1+ x +c y = eln 1+ x c y = c1 ln 1+ x e e y = c1 + x ) . (1 Ex3. An Initial-Value Problem Solve the initial-value problem Solution: dy x = - ydy = - xdx = - ydy xdx dx y 1 1 y 2 = - x 2 + c, y ( 4 ) = 3 2 2 1 1 25 2 = - 2 + c c = 3 4 2 2 2 2 2 x + y = 25. e- y Ex4. Solve x sin ( x ) dx - ydy = 0 . Solution: x - y sin ( x ) dx - ydy = 0 x sin ( x ) dx = e y ydy e ( x ) dx = ydy, ( Integrate by parts) x sin ey - x cos ( x ) + sin ( x ) = ye y - e y + c. 4 2 -3 x Ex5. Solve xy dx + ( y + 2 ) e dy = 0 . Solution: 4 2 -3 x dy x = - , y ( 4 ) = 3. dx y y2 + 2 xy dx + ( y + 2 ) e dy = 0 e xdx = - dy y4 1 2 e3 x xdx = - 2 + 4 dy y y 9 6 e3 x ( 3 x - 1) = + 3 + c. y y 3x Ex5. Solve Solution: dy = y 2 - 4, y ( 0 ) = -2 . dx dy dy dy = y2 - 4 2 = dx 2 = dx dx y -4 y -4 1 1 1 - dy = x + c 4 -2 y+2 y 1 y - 2 - ln y + 2 x + c ln = 4 y-2 y-2 ln = 4 x + c1 = c2e 4 x ( *) y+2 y+2 1 + ce 4 x 1+ c y=2 y ( 0 ) = -2 -2 = 2 1 - ce 4 x 1- c -1 + c = 1 + c ???? How are we going to make the conclusion? It can be seen that dy = y2 - 4 = ( y + 2) ( y - 2) = 0 dx y = -2 or y = 2. Those constant functions y = -2 or y = 2 will satisfy the differential equation and only y = -2 will also satisfy initial condition. Homework: 1~47 odd. ...
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