# Consequently y 0 x 6 parenleftbig e x 2 1

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Consequently, y 0 ( x ) = 6 parenleftBig e x 2 1 parenrightBig 1 / 2 , and so y 0 (1) = 6 parenleftBig e 1 parenrightBig 1 / 2 . 021 10.0points If y 0 is the particular solution of the differ- ential equation dy dx + 2 xy + 4 x = 0 , such that y (0) = 3, find the value of y 0 ( ln 2). 1. y 0 ( ln 2 ) = 1 2 correct 2. y 0 ( ln 2 ) = 1 3. y 0 ( ln 2 ) = 5 2
atchison (bma862) – Homework 5 – staron – (53940) 12 4. y 0 ( ln 2 ) = 3 5. y 0 ( ln 2 ) = 7 Explanation: The differential equation dy dx + 2 xy + 4 x = 0 becomes integraldisplay dy y + 2 = integraldisplay 2 x dx after separating variables and integrating. Thus the general solution of this differential equation is ln( y + 2) = x 2 + C, which in explicit form is given by y = 2 + e C x 2 = 2 + Ae x 2 with A an arbitrary constant. For the partic- ular solution y 0 the value of A is determined by the condition y (0) = 3 since y (0) = 3 = 3 = 2 + A, so y 0 ( x ) = 2 + 5 e x 2 . At x = ln 2, therefore, y 0 ( ln 2 ) = 1 2 , since e ln2 = 1 2 . 022(part1of3)10.0points For the differential equation 2 y dy dx = 49 e x +1 , (i) First find the general solution. 1. y = 1 7 ( e x +1 + A ) 2. y = 7( e x +1 + A ) 1 / 2 3. y = 7( e x +1 + A ) 4. y = 7( e x +1 + A ) 1 / 2 correct 5. y = 1 7 ( e x +1 + A ) 1 / 2 Explanation: The differential equation 2 y dy dx = 49 e x +1 becomes 2 integraldisplay y dy = 49 e integraldisplay e x dx after separating variables and integrating. Consequently, the general solution of this dif- ferential equation is given by y 2 = 49 e x +1 + C = 49( e x +1 + A ) , which in explicit form is given by y = 7 parenleftBig e 1+ x + A parenrightBig 1 / 2 with A an arbitrary constant. 023(part2of3)10.0points (ii) Then find the particular solution y 1 such that y ( 1) = 0. 1. y 1 ( x ) = 1 7 ( e x +1 1) 1 / 2 2. y 1 ( x ) = 7( e x +1 1) 3. y 1 ( x ) = 1 7 ( e x +1 1) 4. y 1 ( x ) = 7( e x +1 1) 1 / 2 correct 5. y 1 ( x ) = 7( e x +1 1) 1 / 2
atchison (bma862) – Homework 5 – staron – (53940) 13 Explanation: For the particular solution y 1 of ( ) the arbitrary constant A in y = 7( e x +1 + A ) 1 / 2 is determined by the condition y ( 1) = 0, since y ( 1) = 0 −→ A = 1 . Hence y 1 = 7( e x +1 1) 1 / 2 . 024(part3of3)10.0points (iii) For the particular solution y 1 in (ii), deter- mine the value of y 1 (0). 1. y 1 (0) = 1 7 ( e 1) 1 / 2 2. y 1 (0) = 7( e 1) 1 / 2 correct 3. y 1 (0) = 7( e 1) 4. y 1 (0) = 1 7 ( e 1) 5. y 1 (0) = 7( e 1) 1 / 2 Explanation: When x = 0 the value of the particular solution y 1 = 7( e x +1 1) 1 / 2 is given by y 1 (0) = 7 parenleftBig e 1 parenrightBig 1 / 2 . 025(part1of3)10.0points For the differential equation 2 xy dy dx = 4 + 4 ln x, (i) first find its general solution. 1. y = parenleftBig 4 x + 4(ln x ) 2 + C parenrightBig 1 / 2 2. y = parenleftBig 4 ln x 2(ln x ) 2 + C parenrightBig 1 / 2 3. y = parenleftBig 4 ln x + 2 x 2 + C parenrightBig 1 / 2 4. y = parenleftBig 4 ln x + 2(ln x ) 2 + C parenrightBig 1 / 2 correct 5. y = 4 ln x + 4(ln x ) 2 + C Explanation: The differential equation 2 xy dy dx = 4 + 4 ln x becomes integraldisplay 2 y dy = 4 integraldisplay 1 x dx + 4 integraldisplay ln x x dx after separating variables and integrating.