{"id":22441,"date":"2024-01-17T15:49:24","date_gmt":"2024-01-17T10:19:24","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22441"},"modified":"2024-01-18T15:55:32","modified_gmt":"2024-01-18T10:25:32","slug":"inter-2nd-year-maths-2b-differential-equations-solutions-ex-8d","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2b-differential-equations-solutions-ex-8d\/","title":{"rendered":"Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(d) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(d)<\/h2>\n

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

Question 1.
\n\\(\\frac{dy}{dx}=-\\frac{(12x+5y-9)}{5x+2y-4}\\)
\nSolution:
\nA non-homogenous equation
\n\\(\\frac{dy}{dx}=-\\frac{(ax+by-9)}{a’x+b’y-c’}\\) where b = -a’
\nb = -5, a = 5 \u21d2 b = -a
\n(5x + 2y-4)dy = -(12x + 5y-9) dx
\n(5x + 2y – 4)dy + (12x + 5y – 9) dx = 0
\n5 (x dy + y dx) + 2y dy – 4 dy + 12x dx – 9 dx = 0
\nintegrating 5xy + y\u00b2 – 4y + 6x\u00b2 – 9x = c<\/p>\n

Question 2.
\n\\(\\frac{dy}{dx}=-\\frac{-3x-2y+5}{2x+3y+5}\\)
\nSolution:
\nb = – 2, a = 2 \u21d2 b = -a
\n(2x + 3y + 5) dy = (- 3x – 2y + 5) dx
\n2x dy + 3y dy + 5 dy = -3x dx- 2y dx + 5 dx
\n2(x.dy + y dx) + By dy + 3x dx + 5 dy – 5 dx = 0
\nIntegrating
\n2xy + \\(\\frac{3}{2}\\)y\u00b2 + \\(\\frac{3}{2}\\)x\u00b2 + 5y – 5x = c
\n4xy + 3y\u00b2 + 3x\u00b2 – 10x + 10y = 2c = c’
\nSolution is
\n4xy + 3(x\u00b2 + y\u00b2)- 10(x – y) = c<\/p>\n

Question 3.
\n\\(\\frac{dy}{dx}=\\frac{-3x-2y+5}{2x+3y-5}\\)
\nSolution:
\n\\(\\frac{dy}{dx}=\\frac{-(3x-2y+5)}{2x+3y-5}\\)
\nHere b = – 2, a\u00b9 = 2
\n\u2235 b = -a\u00b9
\n(2x + 3y – 5) dy = (-3x – 2y + 5) dx \u201e
\n\u21d2 2(x dy + y dx) + (3y – 5) dy + (3x – 5) dx – 0
\n\u21d2 2d (xy) + (3y- 5) dy + (3x- 5) dx = 0
\nNow integrating term by term, we get
\n\u21d2 2 \u222bd (xy) + \u222b(3y – 5)dy + \u222b(3x – 5)dx = 0
\n\u21d2 2xy + 3.\\(\\frac{y^2}{2}\\) – 5y + 3\\(\\frac{x^2}{2}\\) – 5x = \\(\\frac{c}{2}\\)
\nor) 3x\u00b2 + 4xy + 3y\u00b2 – 10x – 10y = c
\nWhich is the required solution.<\/p>\n

\"Inter<\/p>\n

Question 4.
\n2(x – 3y + 1) \\(\\frac{dy}{dx}\\) = 4x – 2y + 1
\nSolution:
\n(2x – 6y + 2) dy = (4x – 2y + 1) dx
\n(2x – 6y + 2) dy – (4x – 2y + 1) dx = 0
\n2 (x dy + y dx) – 6y dy + 2 dy – 4x dx – dx = 0
\nIntegrating
\n2xy – 3y\u00b2 – 2x\u00b2 + 2y – x = c<\/p>\n

Question 5.
\n\\(\\frac{dy}{dx}=\\frac{x-y+2}{x+y-1}\\)
\nSolution:
\nb = -1, a’ = 1 \u21d2 b = -a’
\n(x + y – 1) dy = (x – y + 2) dx
\n(x + y – 1) dy = (x – y + 2) dx = 0
\n(x dy + y dx) + y dy – dy – x dx – 2 dx = 0
\nintegrating
\nxy + \\(\\frac{y^2}{2}\\) – \\(\\frac{x^2}{2}\\) – y – 2x = c
\n2xy + y\u00b2 – x\u00b2 – 2y – 4x = 2c = c’<\/p>\n

Question 6.
\n\\(\\frac{dy}{dx}=\\frac{2x-y+1}{x+2y-3}\\)
\nSolution:
\nb = -1, a = 1 \u21d2 b = -a’
\n(x + 2y – 3) dy = (2x – y + 1) dx
\n(x + 2y – 3) dy – (2x – y + 1) dx = 0
\n(x dy + y dx) 4- 2y dy – 3 dy – 2x dx – dx = 0
\nIntegrating
\nxy + y\u00b2 – x\u00b2 – 3y – x = c<\/p>\n

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

Question 1.
\n(2x + 2y + 3) \\(\\frac{dy}{dx}\\) = x + y + 1
\nSolution:
\n\\(\\frac{dy}{dx}=\\frac{x+y+1}{2x+2y+3}\\)
\n\"Inter
\nMultiplying with 9
\n6v + log (3v + 4) = 9x + 9c
\n6(x + y) + log [3(x + y) + 4] = 9x + c
\ni.e., log (3x + 3y + 4) = 3x – 6y + c<\/p>\n

Question 2.
\n\\(\\frac{dy}{dx}=\\frac{4x+6y+5}{3y+2x+4}\\)
\nSolution:
\n\"Inter
\nMultiplying with 64
\n8v + 9log (8v + 23) = 64x + 64c
\n8 (2x + 3y) – 64x + 9 log (16x + 24y + 23) = c’
\nDividing with 8
\n2x + 3y – 8x + \\(\\frac{9}{8}\\) log (16x + 24y + 23) = c”
\n3y – 6x + \\(\\frac{9}{8}\\) log (16x + 24y + 23) = c”
\nDividing with 3, solution is 3
\ny – 2x + \\(\\frac{3}{8}\\) log (16x + 24y + 23) = k<\/p>\n

\"Inter<\/p>\n

Question 3.
\n(2x + y + 1) dx + (4x + 2y – 1) dy = 0
\nSolution:
\n\"Inter
\n\u222b(2 + \\(\\frac{1}{v-1}\\))dv = 3\u222bdx
\n2v + log (v – 1) = 3x + c
\n2v – 3x + log (v – 1) = c
\n2(2x + y) – 3x + log (2x + y – 1) = c
\n4x + 2y – 3x + log (2x + y – 1) = c
\nSolution is x + 2y + log (2x + y – 1) = c<\/p>\n

Question 4.
\n\\(\\frac{dy}{dx}=\\frac{2y+x+1}{2x+4y+3}\\)
\nSolution:
\n\"Inter
\nMultiplying with 8
\n4v + log (4v + 5) = 8x + 8c
\n4(x + 2y) – 8x + log [4(x + 2y) + 5] = c’
\nSolution is
\n4x + 8y – 8x + log (4x + 8y + 5) = c’
\n8y – 4x + log (4x + 8y + 5) = c’<\/p>\n

Question 5.
\n(x + y – 1) dy = (x + y + 1)dx
\nSolution:
\n\"Inter
\nv – log v = 2x + c
\nx + y – log (x + y) = 2x – c
\n(x – y) + log (x + y) = c is the required
\nsolution.<\/p>\n

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

Question 1.
\n\\(\\frac{dy}{dx}=\\frac{3y-7x+7}{3x-7y-3}\\)
\nSolution:
\nLet x = x + h, y = y + k so that \\(\\frac{dy}{dx}=\\frac{dy}{dx}\\)
\n\"Inter
\n\"Inter
\n3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)
\n14ln x – ln c = – 10ln (v + 1) – 4 ln (v – 1)
\nln (v + 1)5<\/sup> + ln (v – 1)\u00b2 + ln x7<\/sup> = ln c
\n(v +1)5<\/sup>. (v – 1)\u00b2. x7<\/sup> = c
\n(\\(\\frac{y}{x}\\) + 1)5<\/sup> (\\(\\frac{y}{x}\\) – 1)\u00b2.x7<\/sup> = c
\n(y – x)\u00b2 (y + x)5<\/sup> = c
\n[y – (x – 1 )]\u00b2 (y + x – 1 )5<\/sup> = c
\nSolution is [y-x + 1 ]\u00b2 (y + x – 1)5<\/sup> = c.<\/p>\n

Question 2.
\n\\(\\frac{dy}{dx}=\\frac{6x+5y-7}{2x+18y-14}\\)
\nSolution:
\n\"Inter
\n\"Inter
\n\"Inter
\nMultiplying with (3V – 2)(2V + 1)
\n2 + 18V = A(2V + 1) + B(3V – 2)
\n\"Inter
\n2 log (3V- 2)+ log (2V+ 1) = – 3 log X + log c
\nlog (3V – 2)\u00b2.(2V + 1) + log X\u00b3 = log c
\nlog X\u00b3(3V – 2)\u00b2 (2V + 1) = log c
\nx\u00b3(3V – 2)\u00b2 (2V + 1) = c
\n\"Inter
\nSolution is (3y – 2x – 1)\u00b2 (x + 2y – 2) = 343c = c”.<\/p>\n

\"Inter<\/p>\n

Question 3.
\n\\(\\frac{dy}{dx}=\\frac{10x+8y-12}{7x+5y-9}\\) = 0
\nSolution:
\n\\(\\frac{dy}{dx}=\\frac{10x+8y-12}{7x+5y-9}\\) = 0
\nx = X + h, y = Y + k
\n\"Inter
\n\"Inter
\n5V + 7 = A(V + 2) + B (V + 1)
\nV = -1 \u21d2 2 = A(-1 + 2) = A \u21d2 A = 2
\nV = -2 \u21d2 -3 = B(-2 + 1) = -B, B = 3
\n\u222b(\\(\\frac{2}{(V+1)}+\\frac{3}{(V+2)}\\))dv = \u222b\\(\\frac{dx}{X}\\)
\n2 log (V + 1) + 3 log (V + 2) = – 5 log X + c
\nc = 2 log (V + 1) + 3 log (V + 2) + 5 log X
\n= log (V + 1)\u00b2. (V + 2)\u00b3. X5<\/sup>
\n= log(\\(\\frac{2}{(V+1)})\\))\u00b2.(\\(\\frac{3}{(V+2)}\\))\u00b3. X5<\/sup>
\n= log\\(\\frac{(Y+X)^2}{X^2}\\) \\(\\frac{(Y+2X)^3}{X^3}\\) . X5<\/sup>
\n\u21d2 (Y + X)\u00b2 . (Y + 2X)\u00b3 = ec<\/sup> = c’
\n(Y + 1 – X – 2)\u00b2 (Y + 1 – 2x – 4)\u00b3 = c
\nSolution is (x + y – 1)\u00b2 (2x + y – 3)\u00b3 = c.<\/p>\n

Question 4.
\n(x – y – 2) dx + (x – 2y – 3) dy = 0
\nSolution:
\nGiven equation is \\(\\frac{dy}{dx}=\\frac{-x+y+2}{x-2y-3}\\)
\nLet x = X + h, y = Y + k
\n\"Inter
\n\"Inter
\n\"Inter
\n\"Inter
\nis the required solution.<\/p>\n

Question 5.
\n(x – y) dy = (x + y + 1) dx
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 6.
\n(2x + 3y – 8) dx = (x + y – 3) dy
\nSolution:
\n\"Inter
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 7.
\n\\(\\frac{dy}{dx}=\\frac{x+2y+3}{2x+3y+4}\\)
\nSolution:
\nLet x = X + h, y = Y + k so that \\(\\frac{dY}{dX}=\\frac{dy}{dx}\\)
\n\"Inter
\nChoose h and k so that
\nh + 2k + 3 = 0
\n2h + 3k + 4 = 0
\n\"Inter
\nThis is a homogeneous equation
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 8.
\n\\(\\frac{dy}{dx}=\\frac{2x+9y-20}{6x+2y-10}\\)
\nSolution:
\nGiven equation is \\(\\frac{dy}{dx}=\\frac{2x+9y-20}{6x+2y-10}\\)
\nLet x = X + h, y = Y + k so that \\(\\frac{dY}{dX}=\\frac{dy}{dx}\\)
\n\"Inter
\n\u2234 \\(\\frac{dY}{dX}=\\frac{2X+9Y}{6X+2Y}\\)
\nThis is a homogeneous equation
\n\"Inter
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(d) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(d) I. Solve the following differential equations. Question 1. Solution: A non-homogenous equation where b = -a’ b = … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22441"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22441"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22441\/revisions"}],"predecessor-version":[{"id":22442,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22441\/revisions\/22442"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22441"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22441"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22441"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}