Consequently y x parenleftBig 4 4 x 2 parenrightBig 1 2 in which case y 2 20 1

# Consequently y x parenleftbig 4 4 x 2 parenrightbig 1

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Consequently, y 0 ( x ) = parenleftBig 4 + 4 x 2 parenrightBig 1 / 2 , in which case y 0 (2) = 20 1 / 2 is the required value of y 0 . 013(part1of3)10.0points For the differential equation 4 dy dx + 4 x y 3 = 0 , ( x, y > 0) , find the general solution. 1. y = ( C + 4 ln x ) 1 / 3 2. y = ( C 4 ln x ) 1 / 4 correct 3. y = ( C + 4 ln x ) 1 / 4 4. y = ( C + 5 ln x ) 1 / 4 5. y = ( C 4 ln x ) 1 / 3 Explanation: The differential equation 4 dy dx + 4 x y 3 = 0 becomes 4 integraldisplay y 3 dy = 4 integraldisplay 1 x dx
feuge (ejf557) – Homework 5 – staron – (53940) 9 after separating variables and integrating. Thus the general solution of this differential equatoin is y 4 = 4 ln x + C which can be written in explicit form as y = parenleftBig C 4 ln x parenrightBig 1 / 4 with C an arbitrary constant. 014(part2of3)10.0points Then find the particular solution y 1 such that y 1 (1) = 2. 1. y = (2 4 4 ln x ) 1 / 4 correct 2. y = (2 2 + 4 ln x ) 1 / 4 3. y = (2 3 4 ln x ) 1 / 3 4. y = (2 4 + 4 ln x ) 1 / 4 5. y = (2 3 + 4 ln x ) 1 / 3 Explanation: The value of C specifying the particular solution y 1 is determined by the condition y (1) = 2 since y (1) = 2 = 2 4 = C. Thus y 1 ( x ) = (2 4 4 ln x ) 1 / 4 . 015(part3of3)10.0points For the particular solution y 1 in part 2, determine the value of y 1 ( e ). 1. y 1 ( e ) = 12 1 / 4 correct 2. y 1 ( e ) = 13 1 / 3 3. y 1 ( e ) = 15 1 / 3 4. y 1 ( e ) = 14 1 / 4 5. y 1 ( e ) = 16 1 / 4 Explanation: At x = e the value of y 1 is given by y 1 ( e ) = (2 4 4) 1 / 4 = 12 1 / 4 . 016 10.0points If y = y 0 ( x ) satisfies the equations 2 y dy dx = 4 e x +4 , y ( 4) = 0 , and y > 0, find the value of y 0 (0). 1. y 0 (0) = 3 ( e 4 1) 1 / 2 2. y 0 (0) = 2 ( e 5 1) 1 / 2 3. y 0 (0) = 2 ( e 4 1) 1 / 2 correct 4. y 0 (0) = 3 ( e 4 + 1) 1 / 2 5. y 0 (0) = 2 ( e 4 + 1) 1 / 2 Explanation: The differential equation becomes integraldisplay 2 y dy = 4 integraldisplay e x +4 dx after separating variables and integrating. Thus its general solution is y 2 = 4 e x +4 + C which in explicit form is given by y ( x ) = (4 e x +4 + C ) 1 / 2 with C an arbitrary constant. For the partic- ular solution y 0 the value of C is determined by the condition y ( 4) = 0 since y ( 4) = 0 = 0 = 4 + C.
feuge (ejf557) – Homework 5 – staron – (53940) 10 Consequently, y 0 ( x ) = (4 e x +4 4) 1 / 2 , in which case y 0 (0) = 2( e 4 1) 1 / 2 . 017(part1of3)10.0points Consider the differential equation x + 2 dy dx = x radicalbig y 4 , Find its general solution. 1. y = 4+ parenleftBig 2 3 ( x +2) 3 / 2 2( x +2) 1 / 2 + A parenrightBig 2 2. y = 4 parenleftBig 1 3 ( x +2) 3 / 2 2( x +2) 1 / 2 + A parenrightBig 2 3. y = 4+ parenleftBig 1 3 ( x +2) 3 / 2 2( x +2) 1 / 2 + A parenrightBig 2 correct 4. y = parenleftBig 1 3 ( x +2) 3 / 2 2( x +2) 1 / 2 + A parenrightBig 2 4 5. y = 4 parenleftBig 2 3 ( x +2) 3 / 2 2( x +2) 1 / 2 + A parenrightBig 2 Explanation: The differential equation x + 2 dy dx = x radicalbig y 4 becomes integraldisplay 1 y 4 dy = integraldisplay x x + 2 dx, after separating variables and integrating. Consequently, 2 radicalbig y 4 = integraldisplay x x + 2 dx. To integrate the right hand side we use the substitution u = x + 2. Then du = dx , so integraldisplay x x + 2 dx = integraldisplay u 2 u 1 / 2 du = integraldisplay parenleftBig u 1 / 2 2 u 1 / 2 parenrightBig du = 2 3 ( x + 2) 3 / 2 4( x + 2) 1 / 2 + C.