{"id":13006,"date":"2024-03-13T09:48:14","date_gmt":"2024-03-13T04:18:14","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=13006"},"modified":"2024-03-16T17:46:05","modified_gmt":"2024-03-16T12:16:05","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-8-ex-8e","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-8-ex-8e\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(e)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions<\/a> Chapter 8 Differential Equations Ex 8(e) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(e)<\/h2>\n

I.
\nFind the I.F. of the following differential equations by transforming them into linear form.<\/p>\n

Question 1.
\nx \\(\\frac{d y}{d x}\\) – y = 2x2<\/sup> sec2<\/sup> 2x.
\nSolution:
\nThe given equation can be expressed as
\n\\(\\frac{d y}{d x}-\\frac{y}{x}\\) = 2x sec2<\/sup> 2x
\nThis is of the form \\(\\frac{d y}{d x}\\) + Py = Q where the
\nIntegrating factor I.F = e\u222b P dx<\/sup>, P = – \\(\\frac{1}{x}\\)
\n= e– \u222b \\(\\frac{1}{x}\\) dx<\/sup>
\n= e– log x<\/sup>
\n= elog x-1<\/sup><\/sup> = \\(\\frac{1}{x}\\)<\/p>\n

Question 2.
\ny \\(\\frac{d y}{d x}\\) – x = 2y3<\/sup>
\nSolution:
\nThe given equation can be written as
\n\\(\\frac{d x}{d y}-\\frac{x}{y}\\) = 2y2<\/sup>
\nand the integrating factor I.F = e\u222b P dy<\/sup>
\n= e– \u222b \\(\\frac{1}{y}\\) dy<\/sup>
\n= e– log y<\/sup>
\n= elog y-1<\/sup><\/sup> = \\(\\frac{1}{y}\\).<\/p>\n

\"TS<\/p>\n

II. Solve the following differential equations.<\/p>\n

Question 1.
\n\\(\\frac{d y}{d x}\\) + y tan x = cos3<\/sup> x
\nSolution:
\nGiven \\(\\frac{d y}{d x}\\) + y tan x = cos3<\/sup> x
\nwhich is of the form \\(\\frac{d y}{d x}\\) + Py = Q where
\nP = tan x and Q = cos3<\/sup> x
\n\u2234 Integrating Factor I.F. = e\u222b P dx<\/sup>
\n= e\u222b tan x dx<\/sup>
\n= elog sec x<\/sup> = sec x
\nGeneral solution is y. sec x = \u222b Q (I.F.) dx
\n= \u222b cos3<\/sup> x sec x dx
\n= \u222b cos2<\/sup> x dx
\n= \\(\\int \\frac{1+\\cos 2 x}{2} d x=\\frac{1}{2} x+\\frac{1}{4} \\sin 2 x\\)
\n= \\(\\frac{1}{2}\\) (x + sin x cos x) + c
\n\\(\\frac{y}{\\cos x}\\) = \\(\\frac{1}{2}\\) (x + sin x cos x) + c
\n\u21d2 2y = cos x (x + sin x cos x) + c cos x
\n= x cos x + sin x cos2<\/sup> x + c cos x is the solution.<\/p>\n

Question 2.
\n\\(\\frac{d y}{d x}\\) + y sec x = tan x
\nSolution:
\nThis is of the form \\(\\frac{d y}{d x}\\) + Py = Q
\nwhere P = sec x and Q = tan x
\n\u2234 Integrating Factor I.F. = e\u222b sec x dx<\/sup>
\n= elog (sec x + tan x)<\/sup>
\n= sec x + tan x
\n\u2234 General solution is
\ny . e\u222b P dx<\/sup> = \u222b Q . e\u222b P dx<\/sup> dx + c
\n\u2234 y (sec x + tan x) = \u222b tan x (sec x + tan x) + c
\n= \u222b sec x tan x dx . \u222b tan2<\/sup> x dx
\n= sec x + \u222b (sec2<\/sup> x – 1) dx + c
\n= sec x + tan x – x + c
\n\u2234 y (sec x + tan x) sec x + tan x – x + c is the solution.<\/p>\n

\"TS<\/p>\n

Question 3.
\n\\(\\frac{d y}{d x}\\) – y tan x = ex<\/sup> sec x.
\nSolution:
\nThis is of the form \\(\\frac{d y}{d x}\\) + Py = Q where
\nP = – tan x and Q = ex<\/sup> sec x.
\n\u2234 Integrating Factor IF. = e\u222b P dx<\/sup> dx
\n= e\u222b tan x<\/sup> dx
\n= elog cos x<\/sup> = cos x
\n\u2234 General solution is y . e\u222b P dx<\/sup> dx = \u222b Q . e\u222b P dx<\/sup> dx + c
\n\u2234 y . cos x = \u222b ex<\/sup> sec x cos x dx + c
\n= \u222b ex<\/sup> dx + c
\n= ex<\/sup> dx + c
\ny = ex<\/sup> sec x + c sec x. is the solution.<\/p>\n

Question 4.
\nx \\(\\frac{d y}{d x}\\) + 2y = log x.
\nSolution:
\nThe equation can be written as
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}+\\frac{2}{\\mathrm{x}} \\mathrm{y}=\\frac{\\log \\mathrm{x}}{\\mathrm{x}}\\)
\nThis is of the form \\(\\frac{d y}{d x}\\) + Py = Q where
\nI.F. = e\u222b P dx<\/sup> where P = \\(\\frac{2}{x}\\), and Q = \\(\\frac{\\pi}{2}\\)
\n= e\u222b \\(\\frac{2}{x}\\)<\/sup>
\n= elog x2<\/sup><\/sup> = x2<\/sup>.
\ny . e\u222b P dx<\/sup> = \u222b Q . e\u222b P dx<\/sup> dx + c<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\n(1 + x2<\/sup>)\\(\\frac{d y}{d x}\\) + y = etan-1<\/sup><\/sup> x
\nSolution:
\nThe equation can be written as
\n\\(\\frac{d y}{d x}+\\frac{y}{1+x^2}=\\frac{e^{\\tan ^{-1} x}}{1+x^2}\\)<\/p>\n

\"TS<\/p>\n

Question 6.
\n\\(\\frac{d y}{d x}+\\frac{2 y}{x}\\) = 2x2<\/sup>
\nSolution:
\nThe given equation can be written as
\n\\(\\frac{d y}{d x}+\\frac{2 y}{x}\\) = 2x2<\/sup>
\n\u2234 I.F.= e\u222b P dx<\/sup>
\n= e\u222b \\(\\frac{2}{x}\\) dx<\/sup>
\n= e2 log x<\/sup>
\n= elog x2<\/sup><\/sup> = x2<\/sup>
\n\u2234 Solution is y . x2<\/sup> = \u222b 2x2<\/sup> . x2<\/sup> dx
\n= 2 \u222b x4<\/sup> dx
\n= 2 \\(\\frac{x^5}{5}\\) + c.<\/p>\n

\"TS<\/p>\n

Question 7.
\n\\(\\frac{d y}{d x}+\\frac{4 x}{1+x^2} y=\\frac{1}{\\left(1+x^2\\right)^2}\\)
\nSolution:
\nThis is of the form \\(\\frac{d y}{d x}\\) + Py = Q where
\nP = \\(\\frac{4 x}{1+x^2}\\) and Q = \\(\\frac{1}{\\left(1+x^2\\right)^2}\\)
\n\u2234 I.F = e\u222b P dx<\/sup>
\n= \\(e^{\\int \\frac{4 x}{1+x^2} d x}\\)
\n= e2 log (1 + x2<\/sup>)<\/sup>
\n= (1 + x2<\/sup>)2<\/sup> dx + c
\n= x + c is the solution.<\/p>\n

Question 8.
\nx \\(\\frac{d y}{d x}\\) + y = (1 + x) ex<\/sup>
\nSolution:
\nThe given equation can be written as
\n\\(\\frac{d y}{d x}+\\frac{y}{x}=\\frac{1+x}{x} e^x\\)
\n\u2234 I.F = e\u222b P dx<\/sup>
\n= e\u222b \\(\\frac{1}{x}\\) dx<\/sup>
\n= elog x<\/sup> = x
\n\u2234 Solution is
\ny . x = \u222b \\(\\frac{(1+x)}{x}\\) ex<\/sup> . x dx + c
\n= \u222b (1 + x) ex<\/sup> dx + c
\n= \u222b ex<\/sup> dx + \u222b x ex<\/sup> dx + c
\n= ex<\/sup> + x ex<\/sup> – ex<\/sup> + c
\n= x ex<\/sup> + c is the solution.<\/p>\n

Question 9.
\n\\(\\frac{d y}{d x}+\\frac{3 x^2}{1+x^3} y=\\frac{1+x^2}{1+x^3}\\)
\nSolution:<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 10.
\n\\(\\frac{d y}{d x}\\) – y = – 2 e-x<\/sup>
\nSolution:
\nHere P = – 1 and Q = – 2 e-x<\/sup>
\n\u2234 I.F. = e\u222b P dx<\/sup> dx
\n= e\u222b – 1 dx<\/sup> dx
\n\u2234 Solution is
\ny . e-x<\/sup> = \u222b – 2 e-x<\/sup> e-x<\/sup> dx
\n= – 2 \u222b e-2x<\/sup> dx
\n= \\(\\frac{(-2)}{(-2)}\\) e-2x<\/sup> + c
\n= e-2x<\/sup> + c
\n\u2234 y = e-x<\/sup> + C ex<\/sup> is the solution.<\/p>\n

Question 11.
\n(1 + x2<\/sup>) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y = Tan-1<\/sup> x
\nSolution:
\nThe equation can be written as<\/p>\n

\"TS<\/p>\n

= \u222b t et<\/sup> dt where t = tan-1<\/sup> x
\n= t et<\/sup> – et<\/sup>
\n= etan-1<\/sup> x (tan-1<\/sup> – 1)<\/sup> + c
\ny = (tan-1<\/sup> x – 1) + c e– tan-1<\/sup> x<\/sup> is the solution.<\/p>\n

Question 12.
\n\\(\\frac{d y}{d x}\\) + y tan x = sin x
\nSolution:
\nWe have P = tan x and Q = sin x
\n\u2234 I.F = e\u222b P dx<\/sup> dx
\n= e\u222b tan x dx<\/sup> dx
\n= elog sec x<\/sup> dx = sec x
\n\u2234 Solution is
\ny sec x = \u222b sin x . sec x dx + c
\n= \u222b tan x dx
\n= log |sec x| + c is the solution.<\/p>\n

\"TS<\/p>\n

III. Solve the following differential equations.<\/p>\n

Question 1.
\ncos x + y sin x = sec2<\/sup> x
\nSolution:
\nThe given equation can be written as
\n\\(\\frac{d y}{d x}\\) + sin x sec x =sec x
\nHere P = sin x sec x – tan x and Q = sec3<\/sup> x
\n\u2234 I.F. = e\u222b P dx<\/sup> dx
\n= e\u222b tan x dx<\/sup> dx
\n= elog (sec x)<\/sup> = sec x
\n\u2234 Solution is
\ny sec x = \u222b sec4<\/sup> x dx + c
\n= \u222b (1 + tan2<\/sup> x) sec2<\/sup> x dx + c
\n= tan x + \\(\\frac{\\tan ^3 x}{3}\\) + c.<\/p>\n

Question 2.
\nsec x dy = (y + sin x) dx
\nSolution:
\nThe given equation can be written as
\nsec x \\(\\frac{d y}{d x}\\) = y + sin x
\n\u21d2 \\(\\frac{d y}{d x}=\\frac{y}{\\sec x}+\\frac{\\sin x}{\\sec x}\\)
\n= y cos x + sin x cos x
\n\\(\\frac{d y}{d x}\\) – y cos x = sin x cos x
\nI.F. = e\u222b P dx<\/sup>
\n= e– \u222b cos x dx<\/sup>
\n= e– sin x<\/sup>
\n\u2234 Solution is y e– sin x<\/sup>
\n= \u222b sin x cos x e– sin x<\/sup> dx
\n= \u222b t e– t<\/sup> dt where t = sin x
\n= t (- e– t<\/sup>) + \u222b e– t<\/sup> dt
\n= – e– t<\/sup> (t + 1) + c
\n= – e– sin x<\/sup> (sin x + 1) + c
\ny = – (sin x + 1) + c esin x<\/sup> is the solution.<\/p>\n

\"TS<\/p>\n

Question 3.
\nx log x \\(\\frac{d y}{d x}\\) + y = 2 log x
\nSolution:
\nThe equation can be written as<\/p>\n

\"TS<\/p>\n

Question 4.
\n(x + y + 1) \\(\\frac{d y}{d x}\\) = 1
\nSolution:
\nFrom the given equation
\n\\(\\frac{d y}{d x}=\\frac{1}{x+y+1}\\)
\n\\(\\frac{d y}{d x}\\) = x + y + 1
\n\u2234 \\(\\frac{d y}{d x}\\) – x = (y + 1)
\nThis is of the form \\(\\frac{d x}{d y}\\) + x = Q(y)
\nP = – 1 and Q = (y + 1)
\n\u2234 I.F. = e\u222b P dy<\/sup> = e– y<\/sup>
\n\u2234 Solution is
\nx e– y<\/sup> = \u222b (y + 1) e– y<\/sup> dy
\n= \u222b e– y<\/sup> y dy + \u222b e– y<\/sup> dy
\n= – y e– y<\/sup> + \u222b e– y<\/sup> dy – e– y<\/sup>
\n= – y e– y<\/sup> – e– y<\/sup> – e– y<\/sup> + c
\n= – y e– y<\/sup> – 2 e– y<\/sup> + c
\nx = – y – 2 + cey<\/sup>
\n= – (y + 2) cey<\/sup> is the solution.<\/p>\n

\"TS<\/p>\n

Question 5.
\nx (x – 1) \\(\\frac{d y}{d x}\\) – y = x3<\/sup> (x – 1)3<\/sup>
\nSolution:
\nThe equation can be written as<\/p>\n

\"TS<\/p>\n

Question 6.
\n(x + 2y3<\/sup>) \\(\\frac{d y}{d x}\\) = y.
\nSolution:
\nThe given differential equation is
\n(x + 2y3<\/sup>) \\(\\frac{d y}{d x}\\) = y<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 7.
\n(1 – x2<\/sup>) \\(\\frac{d y}{d x}\\) + 2xy = x \\(\\sqrt{1-\\mathbf{x}^2}\\)
\nSolution:
\nDividing by (1 – x2<\/sup>) both sides<\/p>\n

\"TS<\/p>\n

Question 8.
\nx (x – 1) \\(\\frac{d y}{d x}\\) – (x – 2) y = x3<\/sup> (2x – 1)
\nSolution:
\nDividing the given equation by x(x – 1) we get
\n\\(\\frac{d y}{d x}+\\frac{(-x+2)}{x(x-1)} y=\\frac{x^2(2 x-1)}{x-1}\\)
\nNow \\(\\frac{2-x}{x(x-1)}=\\frac{A}{x}+\\frac{B}{x-1}\\)
\n\u2234 2 – x = A (x – 1) + Bx
\nPut x = 1 both sides, 1 = B
\nand A + B = – 1
\n\u21d2 A = – 2
\n\u2234 \\(\\frac{2-x}{x(x-1)}=\\frac{-2}{x}+\\frac{1}{x-1}\\)
\n\u2234 I.F = \\(e^{\\int \\frac{2-x}{x(x-1)} d x}=e^{\\int\\left(\\frac{-2}{x}+\\frac{1}{x-1}\\right) d x}\\)
\n= elog (x – 1) – 2 log x<\/sup>
\n= elog (x – 1) – log x2<\/sup><\/sup>
\n= \\(\\frac{x-1}{x^2}\\)
\n\u2234 General Solution is \\(y\\left(\\frac{x-1}{x^2}\\right)=\\int \\frac{x^2(2 x-1)}{x-1}\\left(\\frac{x-1}{x^2}\\right) d x\\)
\n= \u222b (2x – 1) dx + c
\n= x2<\/sup> – x + c
\n\u2234 y (x – 1) = x2<\/sup> (x2<\/sup> – x + c) is the solution of the differential equation.<\/p>\n

\"TS<\/p>\n

Question 9.
\n\\(\\frac{d y}{d x}\\) (x2<\/sup> y3<\/sup> + xy) = 1
\nSolution:<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 10.
\n\\(\\frac{d y}{d x}\\) + x sin 2y = x3<\/sup> cos2<\/sup> y
\nSolution:
\nDividing by cos2<\/sup> y we get
\nsec2<\/sup> y \\(\\frac{d y}{d x}\\) + 2x tan y = x3<\/sup> …………….(1)
\n[\u2234 \\(\\frac{\\sin 2 y}{\\cos ^2 y}=\\frac{2 \\sin y \\cos y}{\\cos ^2 y}\\) = 2 tan y]
\nLet tan y = t then sec2<\/sup> y \\(\\frac{d y}{d x}\\) = \\(\\frac{d t}{d x}\\)
\n\u2234 \\(\\frac{d t}{d x}\\) + 2xt = x3<\/sup>
\nHere, P = 2x and Q = x3<\/sup>
\n\u2234 I.F = e\u222b P dx<\/sup> = ex2<\/sup><\/sup>
\n\u2234 t . ex2<\/sup><\/sup> = \u222b x3<\/sup> ex2<\/sup><\/sup> dx
\n= \u222b x2<\/sup> . x . ex2<\/sup><\/sup> dx
\n= \\(\\frac{1}{2}\\) \u222b z . ez<\/sup> dz
\nwhere z = x2<\/sup>
\n= \\(\\frac{1}{2}\\) ez<\/sup> (z – 1) + c
\n= \\(\\frac{1}{2}\\) ex2<\/sup><\/sup> (x2<\/sup> – 1) + c
\n\u2234 tan y ex2<\/sup><\/sup> = \\(\\frac{1}{2}\\) ex2<\/sup><\/sup> (x2<\/sup> – 1) + c
\n\u2234 Solution of the given differential equation is tan y = \\(\\frac{1}{2}\\) (x2<\/sup> – 1) + c e-x2<\/sup><\/sup><\/p>\n

\"TS<\/p>\n

Question 11.
\ny2<\/sup> + (x – \\(\\frac{1}{y}\\)) \\(\\frac{d y}{d x}\\) = 0.
\nSolution:<\/p>\n

\"TS<\/p>\n

Where P = y-2<\/sup> and Q = y-3<\/sup>
\n\u2234 I.F. = e\u222b P dx<\/sup>
\n= e\u222by-2<\/sup> dy<\/sup>
\n= \\(e^{-\\frac{1}{y}}\\)
\n\u2234 Solution is
\nx \\(e^{-\\frac{1}{y}}\\) = \u222b \\(e^{-\\frac{1}{y}}\\) y-3<\/sup> dy + c
\n= \u222b e-y-1<\/sup><\/sup> y-3<\/sup> dy + c
\n= \u222b e-y-1<\/sup><\/sup> y-2<\/sup> y-1<\/sup> dy + c
\nLet y-1<\/sup> = t then – y-2<\/sup> dy = dt
\n\u2234 x \\(e^{-\\frac{1}{y}}\\) = – \u222b t e-t<\/sup> dt + c
\n= – [- t e-t<\/sup> + \u222b e-t<\/sup> dt] + c
\n= t e-t<\/sup> + e-t<\/sup> + c
\n= e-t<\/sup> (t + 1) + c
\n= \\(e^{-\\frac{1}{y}}\\) (\\(\\frac{1}{y}\\) + 1) + c
\n\u2234 x = (\\(\\frac{1}{y}\\) + 1) + c \\(e^{\\frac{1}{y}}\\)
\n\u21d2 xy = 1 + y + y . c\\(e^{\\frac{1}{y}}\\)
\n\u2234 Solution of the given differential equation is xy = 1 + y + cy \\(e^{\\frac{1}{y}}\\).<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(e) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(e) I. Find the I.F. of the following differential equations by transforming them into linear form. Question 1. x … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13006"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=13006"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13006\/revisions"}],"predecessor-version":[{"id":13019,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13006\/revisions\/13019"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=13006"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=13006"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=13006"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}