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4. [-/1 Points] DETAILS SCALCCC4 7.1.006. Consider the following differential equation. ry + y = 5 (a) Show that every member of the family of functions y = (51n(x) + C)/x is a solution of the differential equation. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Illustrate part (a) by graphing several members of the family of solutions on a common screen. (Do this on paper. Your instructor may ask you to turn in this work.) (c) Find a solution of the differential equation that satisfies the initial condition y(1) = 4. (Enter the argument of the logarithmic function in parentheses.) y(x) = (d) Find a solution of the differential equation that satisfies the initial condition y(4) = 1. (Enter the argument of the logarithmic function in parentheses.) Need Help? Read It Talk to a Tutor Show My Work (Optional) 5. [-/1 Points) DETAILS SCALCCC4 7.1.007. (a) What can you say about a solution of the equation y' = -(1/4)y? just by looking at the differential equation? The function y must be increasing (or equal to 0) on any interval on which it is defined. The function y must be decreasing (or equal to 0) on any interval on which it is defined. (6) Verify that all members of the family y = 4/(x + C) are solutions of the equation in part (a). (c) Find a solution of the initial-value problem. y = - - (a) x2 y . y (0) - 0.1 y(x) =

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Answer 1

astayl + nya ar y + y = 5 It is linear egin 2 here P(n) at and lins I.F es pensan leyn es tan e - Solh Y. If Qin Iof da te or y.n- 5.x da t + c DY yn = ay yn - 5 ft da slayn y = (hegn + c) ta or when yel) - 4 then Y 4 = Slylte (4=ch :lol y = 5leyn + 4 n when y(4) = 1 then 1- Shey are 4. 3) 4-5 lyt and hence gol is y= sley 4 + 4 AS

5. y! - ty? non positive no, Since dy an 7 1 2 4 2 is a neg no # for any internal on which it is aletined an : jun y must be decreasing (or equal to 0) for any interval. yl= y 2 or dy an y2 ( In & 4 44 TO yoo or A+ C 4 y 4 ntc EN and when yo) zool then 0.1 = 4 > c= 40 Soln y = 4 n+40