Verify that the indicated function y = 9(x) is an explicit solution of the given first-order differential equation.(у — х)у' — у — х+ 18;y = x + 6Vx + 4When y = x + 6Vx + 4,(1+ 3)y' =Vx + 4Thus, in terms of x,(y – x)y' =У — х + 18 %3Since the left and right hand sides of the differential equation are equal when x + 6Vx + 4 is substituted for y, y = x + 6Vx + 4 is a solution.Proceed as in Example 6, by considering p simply as a function and give its domain. (Enter your answer using interval notation.)Then by considering o as a solution of the differential equation, give at least one interval I of definition.(-4, 0)(-0, -4)O [-4, 4]0 (-8, 4)(-8, -4]O O O O O

Answered: Image /qna-images/answer/1f64ff56-1702-4311-b5d7-a49019ec7c4a.jpg

Verify that the indicated function y = p(x) is an explicit solution of the given first-order differential equation. (у — х)у' %3D у — х+ 18; y = x + 6Vx + 4 When y = x + 6Vx + 4, (1+3) y' = Vx+4 Thus, in terms of x, (y – x)y' = у - х + 18 %3 Since the left and right hand sides of the differential equation are equal when x + 6/x + 4 is substituted for y, y = x + 6Vx + 4 is a solution. Proceed as in Example 6, by considering o simply as a function and give its domain. (Enter your answer using interval notation.) Then by considering o as a solution of the differential equation, give at least one interval I of definition. O (-4, ∞) O (-∞, -4) [-4, 4] (-8, 4) O (-8, –4]

Verify that the indicated function y = p(x) is an explicit solution of the given first-order differential equation. (у — х)у' %3D у — х+ 18; y = x + 6Vx + 4 When y = x + 6Vx + 4, (1+3) y' = Vx+4 Thus, in terms of x, (y – x)y' = у - х + 18 %3 Since the left and right hand sides of the differential equation are equal when x + 6/x + 4 is substituted for y, y = x + 6Vx + 4 is a solution. Proceed as in Example 6, by considering o simply as a function and give its domain. (Enter your answer using interval notation.) Then by considering o as a solution of the differential equation, give at least one interval I of definition. O (-4, ∞) O (-∞, -4) [-4, 4] (-8, 4) O (-8, –4]

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

See similar textbooks

Similar questions

The differential equation obtained by eliminating the arbitrary constants A and B from the relation y = Aez + Ber is = O "- 40y + 13y =0 O " - 13y + 40y =0 O "+ 40y - 13y =0 O '+ 13y - 40y = 0 O O D
Match the solution curve with one of the differential equations. y" + 4y = 0 О у" + 2y' + у %3D 0 O y" + y = 0 O y" – 2y' – 3y = 0 О у" - 7y' + 12у %3D 0 y" + 2y' + 2y = 0 Explain your reasoning. (Assume that k, k1, and k2 are all positive.) O The auxiliary equation should have a repeated negative root, so that the solution has the form cje-kx + c2xe¯kx. The auxiliary equation should have a pair of complex roots a ± ßi where a < 0, so that the solution has the form e"x(c1 cos Bx + c2 sin Bx). The differential equation should have the form y" + k?y = 0 where k = 1, so that the period of the solution is 2n. O The auxiliary equation should have one positive and one negative root, so that the solution has the form cqek1× + c2e¬k2×, The differential equation should have the form y" + k2y = 0 where k = 2, so that the period of the solution is n. O The auxiliary equation should have two positive roots, so that the solution has the form c1eki× + c2ek2×.
Suppose tổx1 + 2x2 + sec(t), sin(t)æ1 + tx2 – 3. - This system of linear differential equations can be put in the form æ' = P(t)x + g(t). Determine P(t) and g(t). P(t) = g(t) = %3D
Which one of the following is the Exact differential equation ? ㅇ (교2 +3ary)dy + (1y' + z®)dz = 0 (22 + 3zy)dy + (y?+ 능교2)dz=D0 ㅇ (22 + 3zy)dz + (y' + 교2)dy=0 (교2 +3zg)dz + (y'+ 교)dy= 0 2
The complementary function(C.F.) of the differential equation(D – 6)²(D+5)y= 6x is_ O (c1 + c2x)eôz + c3e5z O (c1 + c2x)e6z + c3e¯ O (c1 + c2x)e-6z + c3e5z O (c1 + c2x)e-6x + c3e¬5z