{"id":31927,"date":"2024-12-11T05:23:11","date_gmt":"2024-12-10T23:53:11","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=31927"},"modified":"2024-12-12T10:21:06","modified_gmt":"2024-12-12T04:51:06","slug":"samacheer-kalvi-12th-business-maths-guide-chapter-4-ex-4-4","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-12th-business-maths-guide-chapter-4-ex-4-4\/","title":{"rendered":"Samacheer Kalvi 12th Business Maths Guide Chapter 4 Differential Equations Ex 4.4"},"content":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Business Maths Guide<\/a> Pdf Chapter 4 Differential Equations Ex 4.4 Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.4<\/h2>\n

Question 1.
\n\\(\\frac { dy }{dx}\\) – \\(\\frac { dy }{dx}\\) = x
\nSolution:
\n\"Samacheer
\nThe required solution is
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n\\(\\frac { dy }{dx}\\) + y cos x = sin x cos x
\nSolution:
\nIt is of the form \\(\\frac { dy }{dx}\\) + Py = Q
\nHere P = cos x; Q = sin x cos x
\n\u222bPdx = \u222bcos x dx = sin x
\nI.F = e\u222bpdx<\/sup> = esinx<\/sup>
\nThe required solution is
\nY(I.F) = \u222bQ (IF) dx + c
\nY(esinx<\/sup>) = \u222bQ (I-F) dx + c
\ny (esinx<\/sup>) = \u222bsin x cos x esinx<\/sup> dx + c
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nx\\(\\frac { dy }{dx}\\) + 2y = x4<\/sup>
\nSolution:
\nThe given equation can be reduced to
\n\\(\\frac { dy }{dx}\\) + \\(\\frac { 2y }{x}\\) = x\u00b3
\nIt is of the form \\(\\frac { dy }{dx}\\) + Py = Q
\nHere P = \\(\\frac { 2 }{x}\\); Q = x\u00b3
\n\u222bpdx = \u222b\\(\\frac { 2 }{x}\\)dx = 2\u222b\\(\\frac { 1 }{x}\\)dx = 2log x – log x\u00b2
\nI.F = e\u222bPdx<\/sup> = elogx\u00b2<\/sup> = x\u00b2
\nThe required solution is
\ny(I.F) = \u222bQ (IF) dx + c
\ny(x\u00b2) = \u222bx\u00b3 (x\u00b2) dx + c
\nx\u00b2y = \u222bx5<\/sup> dx + c
\nx\u00b2y = \\(\\frac { x^6 }{6}\\) + c<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\n\\(\\frac { dy }{dx}\\) + \\(\\frac { 3x^2 }{1+x^3}\\) = \\(\\frac { 1+x^2 }{1+x^3}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 5.
\n\\(\\frac { dy }{dx}\\) + \\(\\frac { y }{x}\\) = xex<\/sup>
\nSolution:
\n\\(\\frac { dy }{dx}\\) + py = Q
\nHere P = \\(\\frac { 1 }{x}\\); Q = xex<\/sup>
\n\u222bPdx = \u222b\\(\\frac { 1 }{x}\\) dx = log x
\nI.F = e\u222bPdx = elog<\/sup> = x
\nThe required solution is
\ny (I.F) = \u222bQ (I.F) dx + c
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\n\\(\\frac { dy }{dx}\\) + y tan x = cos\u00b3 x
\nSolution:
\nIt is of the form \\(\\frac { dy }{dx}\\) + Py = Q
\nHere P = tan x; Q = cos\u00b3 x
\n\u222bPdx = \u222btan x dx = \u222b\\(\\frac { sin x }{cos x}\\) dx = -\u222b\\(\\frac { -sin x }{cos x}\\) dx
\n= -log cos x = log sec x
\nI.F = e\u222bPdx<\/sup> = elog sec x<\/sup> = sec x
\nThe required solution is
\ny(I.F) = \u222bQ(I.F) dx + c
\ny (sec x) = \u222bcos\u00b3x (sec x) dx + c
\ny(sec x) = \u222bcos\u00b3x \\(\\frac { 1 }{cos x}\\) dx + c
\ny (sec x) = \u222bcos\u00b2x dx + c
\ny (sec x)= \u222b(\\(\\frac { 1+cos 2x }{2}\\)) dx + c
\ny (sec x) = \\(\\frac { 1 }{2}\\) \u222b(1 + cos2x) dx + c
\ny (sec x) = \\(\\frac { 1 }{2}\\) [x + \\(\\frac { sin2x }{2}\\)] + c<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nIf \\(\\frac { dy }{dx}\\) + 2y tan x = sinx and if y = 0 when x = \u03c0\/3 express y in terms of x
\nSolution:
\n\\(\\frac { dy }{dx}\\) + 2y tan x = sinx
\nIt is of the form \\(\\frac { dy }{dx}\\) + Py = Q
\nHere P = 2tan x ; Q = sin x
\n\u222bPdx = \u222b2 tan x dx = 2\u222btan xdx = 2 log sec x
\nlog sec\u00b2 x
\nI.F = e\u222bPdx<\/sup> = elog(sec\u00b2x)<\/sup> = sec\u00b2 x
\nThe required solution is
\ny(I.F) = \u222bQ(I.F) dx + c
\ny (sec\u00b2 x) = \u222bsin x (sec\u00b2x) dx + c
\ny(sec\u00b2x) = \u222bsin x(\\(\\frac { 1 }{cos x}\\)) sec x dx + c
\ny sec\u00b2x = \u222b(\\(\\frac { sin x }{cos x}\\)) sec x dx + c
\ny(sec\u00b2x) = \u222btan x sec x dx + c
\n\u21d2 y(sec\u00b2x) = sec x + c ………. (1)
\nIf y = 0 when x = \/3, then (1) \u21d2
\n0(sec\u00b2(\u03c0\/3)) = sec(\u03c0\/3) + c
\n0 = 2 + c
\n\u21d2 c = -2
\n\u2234 Eqn (1) \u21d2 y sec\u00b2x = sec x – 2<\/p>\n

Question 8.
\n\\(\\frac { dy }{dx}\\) + \\(\\frac { y }{x}\\) = xex<\/sup>
\nSolution:
\nIt is of the form \\(\\frac { dy }{dx}\\) + Py = Q
\nHere P = \\(\\frac { 1 }{x}\\); Q = xex<\/sup>
\n\u222bPdx = \u222b\\(\\frac { 1 }{x}\\) dx = log x
\nI.F = e\u222bPdx<\/sup> = elog x<\/sup> = x
\nThe required solution is
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nA bank pays interest by contionous compounding, that is by treating the interest rate as the instantaneous rate of change of principal. A man invests Rs 1,00,000 in the bank deposit which accures interest, 8% over year compounded continuously. How much will he get after 10 years?
\nSolution :
\nLet P(t) denotes the amount of money in the account at time t. Then the differential equation govemning the growth of money is
\n\\(\\frac { dp }{dt}\\) = \\(\\frac { 8 }{100}\\)p = 0.08 p
\n\u21d2 \\(\\frac { dp }{p}\\) = 0.08 dt
\nIntegrating on both sides
\n\u222b\\(\\frac { dp }{p}\\) = \u222b0.08 dt
\nloge<\/sub> P = 0.08 t + c
\nP = e0.08 t<\/sup> + c
\nP = e0.08 t<\/sup>. ec<\/sup>
\nP = C1<\/sub> e0.08 t<\/sup> ………. (1)
\nwhen t = 0, P = Rs 1,00,000
\nEqn (1) \u21d2 1,00,000 = C1<\/sub> e\u00b0
\nC1<\/sub> = 1,00,000
\n\u2234 P = 100000 e0.08 t<\/sup>
\nAt t = 10
\nP= 1,00,000 . e0.08(10)<\/sup>
\n= 1,00,000 e0.8<\/sup> {\u2235 e0.8<\/sup> = 2.2255}
\n= 100000 (2.2255)
\np = Rs 2,25,550<\/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 4.4 Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 4 Differential Equations Ex 4.4 Question 1. – = x Solution: The required solution is Question 2. + y cos x = sin …<\/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-31927","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\/31927"}],"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=31927"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/31927\/revisions"}],"predecessor-version":[{"id":41566,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/31927\/revisions\/41566"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=31927"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=31927"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=31927"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}