{"id":28743,"date":"2021-01-05T12:25:57","date_gmt":"2021-01-05T12:25:57","guid":{"rendered":"https:\/\/samacheer-kalvi.com\/?p=28743"},"modified":"2021-12-06T15:53:14","modified_gmt":"2021-12-06T10:23:14","slug":"samacheer-kalvi-12th-maths-guide-chapter-10-ex-10-7","status":"publish","type":"post","link":"https:\/\/samacheer-kalvi.com\/samacheer-kalvi-12th-maths-guide-chapter-10-ex-10-7\/","title":{"rendered":"Samacheer Kalvi 12th Maths Guide Chapter 10 Ordinary Differential Equations Ex 10.7"},"content":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Maths Guide<\/a> Pdf Chapter 10 Ordinary Differential Equations Ex 10.7 Textbook Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.7<\/h2>\n

Question 1.
\nSolve the following Linear differential equations.
\ncos x \\(\\frac { dy }{ dx }\\) + y sin x = 1
\nSolution:
\nThe given differential equation can be written as
\n\"Samacheer
\nThis is the form \\(\\frac { dy }{ dx }\\) + py = Q
\nwhere P = tan x
\nQ = sec x
\nThus, the given differential equation is linear.
\nI.F = e\u222b pdx<\/sup> = e\u222b tan x dx<\/sup> = elog (sec x)<\/sup> = sec x
\nSo, the required solution is given by
\n[y \u00d7 I.F] = \u222b [Q \u00d7 IF] dx + c
\ny \u00d7 sec x = \u222b sec x \u00d7 sec x dx + c
\ny sec x = \u222b sec\u00b2 x dx + c
\ny sec x = tan x + c
\n\"Samacheer
\n= sin x + c cos x
\ny = sin x + c cos x is the required solution.<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n(1 – x\u00b2)\\(\\frac { dy }{ dx }\\) – xy = 1
\nSolution:
\nThe given differential equation can be written as
\n\"Samacheer
\nThus, the given differential equation is linear.
\n\"Samacheer
\nWhich is a required solution.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\n\\(\\frac { dy }{ dx }\\) + \\(\\frac { y }{ x }\\) = sin x
\nSolution:
\nThe given differential equation can be written as
\n\\(\\frac { dy }{ dx }\\) + (\\(\\frac { 1 }{ x }\\))y = sin x
\nThis is of the form \\(\\frac { dy }{ dx }\\) + Py = Q
\nwhere P = \\(\\frac { 1 }{ x }\\)
\nQ = sin x
\nThus, the given differential equation is linear.
\nI.F= e\u222b pdx<\/sup> = e\u222b \\(\\frac { 1 }{ x }\\) dx<\/sup> = elog x<\/sup> = x
\nSo, the required solution is given by
\nyx I.F = \u222b (Q \u00d7 I.F) dx + c
\nyx = \u222b sin x \u00d7 x dx + c
\n= x (-cos x) – (1) (-sin x) + c
\nyx = -x cos x + sin x + c
\nyx + x cos x = sin x + c
\n(y + cos x) x = sin x + c is a required solution.<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\n(x\u00b2 + 1)\\(\\frac { dy }{ dx }\\) + 2xy = \\(\\sqrt { x^2+4 }\\)
\nSolution:
\nThe given differential equation may be written as
\n\"Samacheer
\nThus, the given differential equation is linear.
\nI.F= e\u222b pdx<\/sup> = e\u222b \\(\\frac { 2x }{ x^+1 }\\) dx<\/sup> = elog (x\u00b2 + 2)<\/sup> = x\u00b2 + 1
\nSo, the required solution is given by
\n\"Samacheer
\nIs the required solution.<\/p>\n

Question 5.
\n(2x – 10y\u00b3) dy + y dx = 0
\nSolution:
\nThe given differential equation may be written as
\ny dx = -(2x – 10y\u00b3) dy
\n\"Samacheer
\nSo, the required solution is
\nx \u00d7 I.F = \u222b (Q \u00d7 I.F)dy + c
\nxy\u00b2 = \u222b 10 y\u00b2 \u00d7 y\u00b2 dy + c
\n= \u222b 10 y4<\/sup> dy + c
\n= \\(\\frac { 10y^5 }{ 5 }\\) + c = 2y5<\/sup> + c
\nxy\u00b2 = 2y5<\/sup> + c is a required solution.<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nx sin x \\(\\frac { dy }{ dx }\\) + (x cos x + sin x) y = sin x
\nSolution:
\nThe given differential equation can be written as
\n\"Samacheer
\nThus, the given differential equation is linear.
\nI.F = e\u222b pdx<\/sup> = e\u222b (cot x + 1\/x)dx<\/sup> = elog sin x + log x<\/sup>
\n= elog (x sin x)<\/sup> = x sin x
\nSo, the solution of the given differential equation is given by
\ny \u00d7 I.F = \u222b(Q \u00d7 I.F) dx + c
\ny (x sin x) = \u222b \\(\\frac { 1 }{ x }\\) x sin x dx + c
\n= \u222b sin x dx + c
\ny (x sin x) = -cos x + c
\nxy sin x + cos x = c is the required solution.<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nSolve (y – esin-1<\/sup>x<\/sup>) \\(\\frac { dx }{ dy }\\) + \\(\\sqrt { 1-x^2 }\\) = 0
\nSolution:
\n\"Samacheer
\nThus, the given differential equation is Linear
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\n\\(\\frac { dy }{ dx }\\) + \\(\\frac { y }{ (1-x)\u221ax }\\) = 1 – \u221ax
\nSolution:
\nThe given linear differential equation is of the form
\n\\(\\frac { dy }{ dx }\\) + py = Q
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\n(1 + x + xy\u00b2) \\(\\frac { dy }{ dx }\\) + (y + y\u00b3) = 0
\nSolution:
\n\"Samacheer
\nSo, the solution of the equation is given by
\n\"Samacheer
\nxy + tan-1<\/sup> y = c
\nWhich is the required solution.<\/p>\n

Question 10.
\n\\(\\frac { dy }{ dx }\\) + \\(\\frac { y }{ x log x }\\) = \\(\\frac { sin 2x }{ log x }\\)
\nSolution:
\nThe given differential equation may be written as
\n\\(\\frac { dy }{ dx }\\) + \\(\\frac { 1 }{ x log x }\\)y = \\(\\frac { sin 2x }{ log x }\\)
\nThis is of the form \\(\\frac { dy }{ dx }\\) + Py = Q
\nWhere P = \\(\\frac { 1 }{ x log x }\\); Q = \\(\\frac { sin 2x }{ log x }\\)
\nThus, the differential equation is linear.
\nI.F = e\u222bpdx<\/sup>
\n= e\u222b \\(\\frac { 1 }{ x log x }\\) dx<\/sup>
\n= e\u222b \\(\\frac { 1 }{ t }\\) dt<\/sup>
\n= elog t<\/sup>
\n= log x
\nSo, the solution of the given differential equation is
\ny \u00d7 I.F = \u222b (Q \u00d7 I.F) dx + c
\ny \u00d7 log x = \u222b \\(\\frac { sin 2x }{ log x }\\) \u00d7 log x dx + c
\n= \u222b sin 2x dx + c
\ny log x = \\(\\frac { -cos 2x }{ 2 }\\) + c
\ny log x + \\(\\frac { cos 2x }{ 2 }\\) = c is a required solution.<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\n(x + a) \\(\\frac { dy }{ dx }\\) – 2y = (x + a)4<\/sup>
\nSolution:
\n\"Samacheer
\n= -2 log (x + a)
\n= log (x + a)-2<\/sup>
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 12.
\n\\(\\frac { dy }{ dx }\\) = \\(\\frac { sin^x }{ 1+x^3 }\\) – \\(\\frac { 3x^2 }{ 1+x^3 }\\)y
\nSolution:
\nThe equation can be written as
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 13.
\nx \\(\\frac { dy }{ dx }\\) + y = x log x
\nSolution:
\nThe given differential equation may be written as
\n\"Samacheer
\nThus, the given differential equation is linear.
\nI.F = e\u222bpdx<\/sup> = e\u222b\\(\\frac { 1 }{ x }\\) dx<\/sup> = elog x<\/sup> = x
\nSo, the solution of the given differential equation is given by
\ny \u00d7 I.F = \u222b(Q \u00d7 I.F)dx + c
\nyx = \u222b log x x dx + c
\nyx = \u222b x log x dx + c
\nu = log x \u222bdv = \u222bx dx
\n\"Samacheer
\nMultiply by 4
\n4yx = 2x\u00b2 log x – x\u00b2 + 4c
\n4xy = 2x\u00b2 log x – x\u00b2 + 4c is a required solution.<\/p>\n

\"Samacheer<\/p>\n

Question 14.
\nx \\(\\frac { dy }{ dx }\\) + 2y – x\u00b2 log x = 0
\nSolution:
\nThe given differential equation may be written as
\n\\(\\frac { x }{ x }\\) \\(\\frac { dy }{ dx }\\) + \\(\\frac { 2y }{ x }\\) = \\(\\frac { x log x }{ x }\\)
\nThis is of the form \\(\\frac { dy }{ dx }\\) + Py = Q
\nwhere P = \\(\\frac { 2 }{ x }\\); Q = x log x
\nThus, the given equation is linear.
\nI.F = e\u222bpdx<\/sup> = e\u222b\\(\\frac { 2 }{ x }\\) dx<\/sup>
\nelog x<\/sup> = elog x\u00b2<\/sup> = x\u00b2
\nSo the required solution is
\ny \u00d7 I.F = \u222b(Q \u00d7 I.F) dx + c
\nyx\u00b2 = \u222bx log x x\u00b2 dx + c
\nyx\u00b2 = \u222bx\u00b3 logx dx + c
\n\"Samacheer
\nMultiply by 16
\n16x\u00b2y = 4x4<\/sup> log x – x4<\/sup> + 16c is a required solution<\/p>\n

\"Samacheer<\/p>\n

Question 15.
\n\\(\\frac { dy }{ dx }\\) + \\(\\frac { 3y }{ x }\\) = \\(\\frac { 1 }{ x^2 }\\), given that y = 2 when x = 1
\nSolution:
\nThe given differential equation can be written as
\n\"Samacheer
\nSo, its solution is given by
\ny \u00d7 I.F = \u222b(Q \u00d7 I.F) dx + c
\nyx\u00b3 = \u222b \\(\\frac { 1 }{ x^2 }\\) + x\u00b3 dx + c
\n= \u222bx dx + c
\ny x\u00b3 = \\(\\frac { x^2 }{ 2 }\\) + c
\n2yx\u00b3 = x\u00b2 + c
\nGiven that y = 2 when x = 1
\n2 (2) (1)\u00b3 = 1 + c
\n4 – 1 = c
\nc = 3
\n\u2234 2yx\u00b3 = x\u00b2 + 3 is a required solution.<\/p>\n

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

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.7 Textbook Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.7 Question 1. Solve the following Linear differential equations. cos x + y sin x = 1 Solution: The given differential …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/28743"}],"collection":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/comments?post=28743"}],"version-history":[{"count":1,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/28743\/revisions"}],"predecessor-version":[{"id":47825,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/28743\/revisions\/47825"}],"wp:attachment":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/media?parent=28743"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/categories?post=28743"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/tags?post=28743"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}