{"id":32100,"date":"2024-12-12T12:17:12","date_gmt":"2024-12-12T06:47:12","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=32100"},"modified":"2024-12-13T09:56:59","modified_gmt":"2024-12-13T04:26:59","slug":"samacheer-kalvi-12th-business-maths-guide-chapter-4-ex-4-5","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-12th-business-maths-guide-chapter-4-ex-4-5\/","title":{"rendered":"Samacheer Kalvi 12th Business Maths Guide Chapter 4 Differential Equations Ex 4.5"},"content":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Business Maths Guide<\/a> Pdf Chapter 4 Differential Equations Ex 4.5 Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5<\/h2>\n

Question 1.
\n\\(\\frac { d^2y }{dx^2}\\) – 6\\(\\frac { dy }{dx}\\) + 8y = 0
\nSolution:
\nGiven (D2<\/sup> – 6D + 8) y = 0, D = \\(\\frac{d}{d x}\\)
\nThe auxiliary equations is
\nm2<\/sup> – 6m + 8 = 0
\n(m – 4)(m – 2) = 0
\nm = 4, 2
\nRoots are real and different
\nThe complementary function (C.F) is (Ae4x<\/sup> + Be2x<\/sup>)
\nThe general solution is y = Ae4x<\/sup> + Be2x<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n\\(\\frac { d^2y }{dx^2}\\) – 4\\(\\frac { dy }{dx}\\) + 4y = 0
\nSolution:
\nThe auxiliary equations A.E is m2<\/sup> – 4m + 4 = 0
\n(m – 2)2<\/sup> = 0
\nm = 2, 2
\nRoots are real and equal
\nThe complementary function (C.F) is (Ax + B) e2x<\/sup>
\nThe general solution is y = (Ax + B) e2x<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 3.
\n(D\u00b2 + 2D + 3) y = 0
\nSolution:
\nThe auxiliary equation is m\u00b2 + 2m + 3 = 0
\nHere a = 1, b = 2, c = 3
\n\"Samacheer
\nThe complementary function is
\neax<\/sup> (Acos\u00dfx + Bsin\u00dfx)
\n\u2234 C.F = e-x<\/sup> [Acos\u221a2x + Bsin \u221a2x]
\n\u2234 The general solution is
\ny = e-x<\/sup> (Acos\u221a2x + Bsin\u221a2x)<\/p>\n

Question 4.
\n\\(\\frac { d^2y }{dx^2}\\) – 2k\\(\\frac { dy }{dx}\\) + k\u00b2y = 0
\nSolution:
\nGiven (D2<\/sup> – 2kD + k2<\/sup>)y = 0, D = \\(\\frac{d}{d x}\\)
\nThe auxiliary equations is m2<\/sup> – 2km + k = 0
\n\u21d2 (m – k)2<\/sup> = 0
\n\u21d2 m = k, k
\nRoots are real and equal
\nThe complementary function (C.F) is (Ax + B) ekx<\/sup>
\nThe general solution is y = (Ax + B) ekx<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 5.
\n(D\u00b2 – 2D – 15) y = 0
\nSolution:
\nThe auxiliary equation is
\nm\u00b2 – 2m + 15 = 0
\nm\u00b2 + 3m – 5m – 15 = 0
\nm (m + 3) – 5 (m + 3) = 0
\n(m + 3) (m – 5) = 0
\nm = -3, 5
\nRoots are real and different
\n\u2234 The complementary function is
\nAem1<\/sub>x<\/sup> + Bem2<\/sub>x<\/sup>
\nC.F = Ae-3x<\/sup> + Be5x<\/sup>
\n\u2234 The general solution is
\ny = (Ae-3x<\/sup> + Be5x<\/sup>) ………… (1)
\n\\(\\frac { dy }{dx}\\) = Ae-3x<\/sup> (-3) + Be5x<\/sup> (5)
\n\\(\\frac { dy }{dx}\\) = -3Ae-3x<\/sup> + 5Be5x<\/sup> ………… (2)
\n\\(\\frac { d^2y }{dx^2}\\) = 9Ae-3x<\/sup> + 25Be5x<\/sup> ……….. (3)
\nwhen x = 0; \\(\\frac { dy }{dx}\\) = 0
\n-3 Ae\u00b0 + 5Be\u00b0 = 0
\n-3A + 5B = 0 ………. (4)
\nwhen x = 0; \\(\\frac { d^2y }{dx^2}\\) = 2
\nEqn (3) \u21d2 9Ae\u00b0 + 25Be\u00b0 = 2
\n9A + 25B = 2 ……… (5)
\nSolving equation (4) & (5)
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\n(4D\u00b2 + 4D – 3) y = e2x<\/sup>
\nSolution:
\nThe auxiliary equation is
\n4m\u00b2 + 4m – 3 = 0
\n4m\u00b2 + 6m – 2m – 3 = 0
\n2m (2m + 3) – 1 (2m + 3) = 0
\n(2m + 3) (2m – 1) = 0
\n2m = -3; 2m = 1
\nm = -3\/2, 1\/2
\nRoots are real and different
\nThe complementary function is
\nAem1<\/sub>x<\/sup> + Bem2<\/sub>x<\/sup>
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\n\\(\\frac { d^2y }{dx^2}\\) + 16y = 0
\nSolution:
\nGiven (D2<\/sup> + 16) y =0
\nThe auxiliary equation is m2<\/sup> + 16 = 0
\n\u21d2 m2<\/sup> = -16
\n\u21d2 m = \u00b1 4i
\nIt is of the form \u03b1 \u00b1 i\u03b2, \u03b1 = 0, \u03b2 = 4
\nThe complementary function (C.F) is e0x<\/sup> [A cos 4x + B sin 4x]
\nThe general solution is y = [A cos 4x + B sin 4x]<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\n(D\u00b2 – 3D + 2) y = e3x<\/sup> which shall vanish for x = 0 and for x = log 2
\nSolution:
\n(D\u00b2 – 3D + 2) y = e3x<\/sup>
\nThe auxiliary equation is
\nm\u00b2 – 3m + 2 =0
\n(m – 1) (m – 2) = 0
\nm = 1, 2
\nRoots are real and different
\nThe complementary function is
\nC.F = Aem1<\/sub>x<\/sup> + Bem2<\/sub>x<\/sup>
\nC.F = Ax<\/sup> + Be2x<\/sup>
\n\"Samacheer
\nwhen x = log 2; y = 0
\nAelog 2<\/sup> + Be2log 2<\/sup> + \\(\\frac { e^{3xlog2} }{2}\\) = 0
\nAelog 2<\/sup> + Belog (2)\u00b2<\/sup> + \\(\\frac { e^{log2\u00b3} }{2}\\) = 0
\n2A + 4B + \\(\\frac { 8 }{2}\\) = 0
\n2A + 4B + 4 = 0
\n2A + 4B = -4 ……… (3)
\nSolving equation (2) & (3)
\nEqn (2) \u00d7 2 \u21d2 2A + 2B = -1
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\n(D\u00b2 + D – 2) y = e3x<\/sup> + e-3x<\/sup>
\nSolution:
\nThe auxiliary equation is
\nm\u00b2 + m – 6 = 0
\n(m + 3) (m – 2) = 0
\nRoots are real and different
\nThe complementary function is
\nC.F = Aem1<\/sub>x<\/sup> + Bem2<\/sub>x<\/sup>
\nC.F = Ae-3x<\/sup> + Be2x<\/sup>
\n\"Samacheer<\/p>\n

Question 10.
\n(D\u00b2 – 10D + 25) y = 4e5x<\/sup> + 5
\nSolution:
\nThe auxiliary equation is
\nm\u00b2 – 10m + 25 = 0
\n(m – 5) (m – 5) = 0
\nm = 5, 5
\nRoots are real and equal
\nC.F = (Ax + B) emx<\/sup>
\nC.F = (Ax + B) e5x<\/sup>
\nP.I(1)<\/sub> = x. \\(\\frac { 4 }{2D-10}\\) e5x<\/sup>
\nReplace D by 5, 2D – 10 = 0 when D = 5
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\n(4D\u00b2 + 16D +15) y = 4e\\(\\frac { -3 }{2}\\)<\/sup>x
\nSolution:
\nThe auxiliary equation is 4m\u00b2 + 16m + 15 = 0
\n4m\u00b2 + 16m + 10m + 15 = 0
\n2m (2m + 3) + 5 (2m + 3) = 0
\n(2m + 3) (2m + 5) = 0
\n2m = -3, -5
\n\u2234 m = -3\/2, -5\/2
\nRoots are real and different
\nC.F = (Ax + B) em1<\/sub>x<\/sup> + Bem2<\/sub>x<\/sup>
\nC.F = Ae-3\/2 x<\/sup> + Be-5\/2 x<\/sup>
\n\"Samacheer<\/p>\n

Question 12.
\n(3D\u00b2 + D – 14) y – 13 e2x<\/sup>
\nSolution:
\nThe auxiliary equation is 3m\u00b2 + m – 14 = 0
\n3m\u00b2 – 6m + 7m – 14 = 0
\n3m (m – 2) + 7 (m – 2) = 0
\n(m – 2) (3m + 7) = 0
\nm = 2; 3m = -7
\nm = 2, -7\/3
\nRoots are real and different
\nC.F = (Ax + B) em1<\/sub>x<\/sup> + Bem2<\/sub>x<\/sup>
\nC.F = Ae2x<\/sup> + Be-7\/3 x<\/sup>
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 13.
\nSuppose that the quantity demanded Qd = 13 – 6p + 2\\(\\frac { dp }{dt}\\) + \\(\\frac { d^2p }{dt^2}\\) = and quantity supplied Qd = -3 + 2p where is the price. Find the equilibrium price for market clearence.
\nSolution:
\nFor market clearance, the required condition is Qd = Qs
\n\"Samacheer
\nThe auxiliary equation is
\nm\u00b2 + 2m – 8 = 0
\n(m + 4) (m – 2) = 0
\nm = -4, 2
\nRoots are real and different
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 4 Differential Equations Ex 4.5 Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.5 Question 1. – 6 + 8y = 0 Solution: Given (D2 – 6D + 8) y = 0, …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[5],"tags":[],"class_list":["post-32100","post","type-post","status-publish","format-standard","hentry","category-class-12"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/32100"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=32100"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/32100\/revisions"}],"predecessor-version":[{"id":41579,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/32100\/revisions\/41579"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=32100"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=32100"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=32100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}