{"id":12333,"date":"2024-03-11T10:20:58","date_gmt":"2024-03-11T04:50:58","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=12333"},"modified":"2024-03-13T17:28:53","modified_gmt":"2024-03-13T11:58:53","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-8-ex-8b","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-8-ex-8b\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)"},"content":{"rendered":"

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

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

I.
\nQuestion 1.
\nFind the general solution of \\(\\sqrt{1-x^2} d y+\\sqrt{1-y^2} d x\\) = 0.
\nSolution:
\nGiven equation is \\(\\sqrt{1-x^2} d y+\\sqrt{1-y^2} d x\\) = 0<\/p>\n

\"TS<\/p>\n

\u21d2 sin-1<\/sup> y = – sin-1<\/sup> x + c
\n\u21d2 sin-1<\/sup> x + sin-1<\/sup> y + c is the general solution.<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the general solution of \\(\\frac{d y}{d x}=\\frac{2 y}{x}\\).
\n?Solution:
\nThe given equation can be written in variable separable form as \\(\\frac{d y}{d x}=\\frac{2 y}{x}\\).
\n\u21d2 \\(\\frac{\\mathrm{dy}}{\\mathrm{y}}=2\\left(\\frac{\\mathrm{dx}}{x}\\right)\\)
\n\u21d2 log |y| = 2 log |x| + log c1<\/sub>
\n\u21d2 log y = log x2<\/sup> + log c
\n\u21d2 log \\(\\left(\\frac{y}{x^2}\\right)\\) = log c
\n\u21d2 y = cx2<\/sup>
\n\u21d2 x2<\/sup> = \\(\\frac{1}{c}\\) y
\n\u21d2 x2<\/sup> = c1<\/sub>y where c1<\/sub> is a constant is the general solution.<\/p>\n

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

Question 1.
\n\\(\\frac{d y}{d x}=\\frac{1+y^2}{1+x^2}\\)
\nSolution:
\nThe given equation can be written in variable seperable form as
\n\\(\\frac{d y}{d x}=\\frac{1+y^2}{1+x^2}\\)
\n\u2234 \\(\\int \\frac{d y}{1+y^2}=\\int \\frac{d x}{1+x^2}\\)
\n\u21d2 tan-1<\/sup> y = tan-1<\/sup> x + tan-1<\/sup> c
\n\u21d2 tan-1<\/sup> y = tan-1<\/sup> x + tan-1<\/sup> c is the solution of the given differential equation.<\/p>\n

\"TS<\/p>\n

Question 2.
\n\\(\\frac{d y}{d x}\\) = ey-x<\/sup>
\nSolution:
\nThe given equation can be written in variable separable form as
\n\\(\\frac{d y}{d x}\\) = ey-x<\/sup>
\n\u21d2 \\(\\frac{d y}{e^y}=\\frac{d x}{e^x}\\)
\n\u21d2 \u222b e-y<\/sup> dy = \u222b e-x<\/sup> dx
\n\u21d2 – e-y<\/sup> dy = – e-x<\/sup> + c
\n\u21d2 e-x<\/sup> – e-y<\/sup> = c is the solution 0f the given differential equation.<\/p>\n

Question 3.
\n(ex<\/sup> + 1) y dy + (y + 1) dx = 0.
\nSolution:
\nThe given differential equation can be written as (ex<\/sup> + 1)y dy = – (y + 1) dx
\n\u21d2 \\(\\frac{y d y}{y+1}=-\\frac{d x}{e^x+1}\\)
\n\u21d2 \\(\\left[\\frac{(y+1)-1}{y+1}\\right] d y=\\frac{-e^{-x}}{1+e^{-x}} d x\\)
\n\u2234 \u222b dy – \u222b \\(\\frac{d y}{y+1}\\) dx = \u222b \\(\\frac{\\mathrm{e}^{-\\mathrm{x}}}{1+\\mathrm{e}^{-\\mathrm{x}}}\\) dx
\n\u21d2 y – log (y + 1) = log (1 + e-x<\/sup>)
\n\u21d2 y = log (y + 1) + log (1 + e-x<\/sup>) + log c
\n= log [(y + 1) (e-x<\/sup> + 1) c]
\n\u2234 ey<\/sup> = c(y + 1) (e-x<\/sup> + 1)
\n\u2234 The solution of the given equation is
\ney<\/sup> = c (y + 1) (1 + e-x<\/sup>).<\/p>\n

\"TS<\/p>\n

Question 4.
\n\\(\\frac{d y}{d x}\\) = ex-y<\/sup> + x2<\/sup>e-y<\/sup>
\nSolution:
\n\\(\\frac{d y}{d x}\\) = ex-y<\/sup> + x2<\/sup>e-y<\/sup>
\n= e-y<\/sup> (ex<\/sup> + x2<\/sup>)
\nWriting in variable separable form we get
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{1}{y}\\) (ex<\/sup> + x2<\/sup>)
\n\u21d2 \u222b ey<\/sup> dy = \u222b (ex<\/sup> + x2<\/sup>) dx
\n\u21d2 ey<\/sup> = ex<\/sup> + \\(\\frac{x^3}{3}\\) + c
\nThe solution of the given equation is ey<\/sup> = ex<\/sup> + \\(\\frac{x^3}{3}\\) + c.<\/p>\n

Question 5.
\ntan y dx + tan x dy = 0.
\nSolution:
\nThe given equation can be written as
\n\\(\\frac{d x}{\\tan x}+\\frac{d y}{\\tan y}\\) = 0
\n\u21d2 \\(\\int \\frac{\\mathrm{dx}}{\\tan x}+\\int \\frac{\\mathrm{dy}}{\\tan y}\\) = 0
\n\u21d2 \u222b cot x dx + \u222b cot y dy = 0
\n\u21d2 log (sin x) + log (sin y) = log c
\n\u21d2 sin x sin y = c is the solution of the given differential equation.<\/p>\n

Question 6.
\n\\(\\sqrt{1+x^2} d x+\\sqrt{1+y^2} d y\\) = 0
\nSolution:
\nThe given equation can be written in variable seperable form as<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 7.
\ny – x\\(\\frac{d y}{d x}\\) = 5 (y2<\/sup> + \\(\\frac{d y}{d x}\\))
\nSolution:
\nThe given differential equation is<\/p>\n

\"TS<\/p>\n

\u2234 1 = A (1 – 5y) + By
\n\u21d2 A = 1 and B – 5A = 0
\n\u21d2 B = 5
\n\u2234 \\(\\int \\frac{1}{y-5 y^2} d y=\\int \\frac{1}{y} d y+\\int \\frac{5}{1-5 y} d y\\)
\n= log |y| – log (1 – 5y)
\n\u2234 From (1)
\nlog |x + 5| = log |y| – log (1 – 5y) + log c
\n\u21d2 x + 5 = \\(\\frac{c y}{1-5 y}\\)
\n\u2234 Solution of the given dillerential equation is 5 + x = \\(\\frac{c y}{1-5 y}\\)<\/p>\n

Question 8.
\n\\(\\frac{d y}{d x}=\\frac{y(x+1)}{x(y+1)}\\)
\nSolution:
\nThe given equation \\(\\frac{d y}{d x}=\\frac{y(x+1)}{x(y+1)}\\) writing in variable separable form
\n\\(\\frac{(y+1) d y}{y}=\\frac{(x+1) d x}{x}\\)
\n\\(\\int\\left(\\frac{y+1}{y}\\right) d y=\\int \\frac{(x+1) d x}{x}\\)
\n\u21d2 y + log |y| = x + log |x| + log c
\n\u21d2 y – x = |og |x| – log |y| + log c
\n= log \\(\\left(\\frac{c x}{y}\\right)\\)
\n\u2234 y – x = log \\(\\left(\\frac{c x}{y}\\right)\\) is the solution.<\/p>\n

\"TS<\/p>\n

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

Question 1.
\n\\(\\frac{d y}{d x}=\\frac{1+y^2}{\\left(1+x^2\\right) x y}\\)
\nSolution:
\nThe given equation is \\(\\frac{d y}{d x}=\\frac{1+y^2}{\\left(1+x^2\\right) x y}\\) which can be written in variable separable form as<\/p>\n

\"TS<\/p>\n

\u21d2 (1 + x2<\/sup>) (1 + y2<\/sup>) = c2<\/sup>x2<\/sup>
\n= cx2<\/sup> where c2<\/sup> = c is a constant
\n\u2234 The solution of the given differential equation is
\n(1 + x2<\/sup>) (1 + y2<\/sup>) = cx2<\/sup><\/p>\n

Question 2.
\n\\(\\frac{d y}{d x}\\) + x2<\/sup> = x2<\/sup> e3y<\/sup>
\nSolution:
\nThe given equation can be written in variable separable form as
\n\\(\\frac{d y}{d x}\\) = x2<\/sup> e3y<\/sup> – x2<\/sup>
\n= x2<\/sup> (e3y<\/sup> – 1)
\n\\(\\frac{d y}{e^{3 y}-1}\\) = xsup>2 dx
\n\u2234 \u222b \\(\\frac{d y}{e^{3 y}-1}\\) = \u222b x2<\/sup> dx
\n\u21d2 \u222b \\(\\left(\\frac{e^{-3 y}}{1-e^{-3 y}}\\right)\\) dy = \\(\\frac{x^3}{3}\\) + c
\n\u21d2 log (1 – e-3y<\/sup>) = x3<\/sup> + c
\n\u21d2 1 – e-3y<\/sup> = ex<\/sup>3<\/sup> + ec<\/sup>
\n= k ex<\/sup>3<\/sup>
\n\u2234 The solution of the given differential equation is 1 – e-3y<\/sup> = k ex<\/sup>3<\/sup>.<\/p>\n

\"TS<\/p>\n

Question 3.
\n(xy2<\/sup> + x) dx + (yx2<\/sup> + y) dy = 0
\nSolution:
\nThe given dilferential equation can be written as
\nx (y2<\/sup> + 1) dx + y(x2<\/sup> + 1) dy = 0 which can be expressed in variable separable form as
\n\\(\\frac{1}{2} \\int \\frac{2 x d x}{x^2+1}+\\frac{1}{2} \\int \\frac{2 y d y}{y^2+1}\\) = 0
\n\u21d2 \\(\\frac{1}{2}\\) log(x2<\/sup> + 1) + log (y2<\/sup> + 1) = log c
\n\u21d2 log \\(\\sqrt{\\mathrm{x}^2+1}\\) + log \\(\\sqrt{\\mathrm{y}^2+1}\\) = log c
\n\u21d2 (1 + x2<\/sup>) (1 + y2<\/sup>) = c2<\/sup>.<\/p>\n

Question 4.
\n\\(\\frac{d y}{d x}\\) = 2y tanh x
\nSolution:
\nThe given equation is \\(\\frac{d y}{d x}\\) = 2y tanh x.
\n\\(\\frac{d y}{y}\\) = 2 tanh x (variable separable form)
\n\u222b \\(\\frac{d y}{y}\\) = 2 \u222b tanh x dx
\n\u21d2 log y = 2 log |cosh x| + log c
\n= log |cosh2<\/sup> x| + log c
\nlog y = log (c cosh2<\/sup> x)
\ny = c .cosh2<\/sup> x which is the solution of the given differential equation.<\/p>\n

Question 5.
\nSin-1<\/sup> (\\(\\frac{d y}{d x}\\)) = x + y
\nSolution:
\nThe given equation is Sin-1<\/sup> (\\(\\frac{d y}{d x}\\)) = x + y
\n\u21d2 sin (x + y) = \\(\\frac{d y}{d x}\\) …………..(1)<\/p>\n

\"TS<\/p>\n

\u21d2 \u222b sec2<\/sup> dz – \u222b sec z tan x dx = \u222b dx + c
\n\u21d2 tan z – sec z = x + c
\n\u21d2 tan (x + y) – sec (x + y) = x + c is the solution of the given differential equation.<\/p>\n

\"TS<\/p>\n

Question 6.
\n\\(\\frac{d y}{d x}+\\frac{y^2+y+1}{x^2+x+1}\\) = 0
\nSolution:
\nGiven equation in variable separable form is
\n\\(\\frac{d y}{y^2+y+1}=-\\frac{d x}{x^2+x+1}\\)<\/p>\n

\"TS<\/p>\n

Question 7.
\n\\(\\frac{d y}{d x}\\) = tan2<\/sup> (x + y)
\nSolution:
\nLet x + y = z then 1 + \\(\\frac{d y}{d x}\\) = \\(\\frac{d z}{d x}\\)
\nfrom the given equation
\n\u2234 1 + \\(\\frac{d y}{d x}\\) = 1 + tan2<\/sup> (x + y)
\n\u21d2 \\(\\frac{d z}{d x}\\) = sec2<\/sup> z
\n\u21d2 \u222b \\(\\frac{\\mathrm{d} z}{\\sec ^2 z}\\) = \u222b dx + c
\n\u21d2 \u222b cos2<\/sup> z dz = x + c
\n\u21d2 \u222b \\(\\left(\\frac{1+\\cos 2 z}{2}\\right)\\) dz = x + c
\n\u21d2 \\(\\frac{1}{2}\\) z + \\(\\frac{1}{4}\\) sin 2z = x + c
\n\u21d2 2z + sin 2z = 4x + 4c
\n\u21d2 2 (x + y) + sin 2 (x + y) = 4x + 4c
\n\u21d2 sin 2 (x + y) = 2x – 2y + 4c
\n= 2x – 2y + c1<\/sub>
\nwhere c1<\/sub> = 4c.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) I. Question 1. Find the general solution of = 0. Solution: Given equation is = 0 \u21d2 sin-1 … 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\/12333"}],"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=12333"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/12333\/revisions"}],"predecessor-version":[{"id":12353,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/12333\/revisions\/12353"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=12333"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=12333"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=12333"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}