{"id":35551,"date":"2022-09-19T17:33:21","date_gmt":"2022-09-19T12:03:21","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35551"},"modified":"2022-09-19T17:33:21","modified_gmt":"2022-09-19T12:03:21","slug":"inter-2nd-year-maths-2b-differential-equations-solutions-ex-8c","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-differential-equations-solutions-ex-8c\/","title":{"rendered":"Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)"},"content":{"rendered":"

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

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

I.<\/span><\/p>\n

Question 1.
\nExpress x dy – ydx = \\(\\sqrt{x^2+y^2}\\) dx in the form F (\\(\\frac{y}{x}\\)) = (\\(\\frac{dy}{dx}\\)).
\nSolution:
\nx. dy – y dx = \\(\\sqrt{x^2+y^2}\\) dx
\n\\(\\frac{dy}{dx}\\) – y = \\(\\sqrt{x^2+y^2}\\)
\n\"Inter<\/p>\n

Question 2.
\nExpress (x – y Tan-1<\/sup> \\(\\frac{y}{x}\\))dx + x Tan-1<\/sup>\\(\\frac{y}{x}\\) dy = 0 in the form F(\\(\\frac{y}{x}\\)) = \\(\\frac{dy}{dx}\\).
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\nExpress x\\(\\frac{dy}{dx}\\) = y(log y – log x + 1) in the from F (\\(\\frac{y}{x}\\)) = \\(\\frac{dy}{dx}\\)
\nSolution:
\nx . \\(\\frac{dy}{dx}\\) = y(log y – log x + 1)
\n\\(\\frac{dy}{dx}\\) = \\(\\frac{y}{x}\\)(log\\(\\frac{y}{x}\\) + 1)<\/p>\n

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

Question 1.
\n\\(\\frac{dy}{dx}=\\frac{x-y}{x+y}\\)
\nSolution:
\n\\(\\frac{dy}{dx}=\\frac{x-y}{x+y}\\)
\nPut y = vx
\n\\(\\frac{dy}{dx}\\) = v + x\\(\\frac{dv}{dx}\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 2.
\n(x\u00b2 + y\u00b2) dy = 2xy dx
\nSolution:
\n(x\u00b2 + y\u00b2) dy = 2xy dx
\n\"Inter
\n1 + v\u00b2 = A(1 – v\u00b2) + Bv(1 – v) + Cv(1 + v)
\nv = 0 \u21d2 = A
\nv = 1 \u21d2 1 + 1 = C(2) \u21d2 c = 1
\nv = -1 \u21d2 1 + 1 = B(-1) (2) \u21d2 2 = -2B
\nB = -1
\n\"Inter<\/p>\n

Question 3.
\n\\(\\frac{dy}{dx}=\\frac{-(x^2+3y^2)}{3x^2+y^2}\\)
\nSolution:
\n\\(\\frac{dy}{dx}=\\frac{-(x^2+3y^2)}{3x^2+y^2}\\)
\nPut y = vx
\n\"Inter
\nMultiplying with (v + 1)\u00b3
\n3 + v\u00b2 = A(v + 1)\u00b2 + B(v + 1) + C
\nv = – 1 \u21d2 3 + 1 = C \u21d2 C = 4
\nEquating the co-efficients of v\u00b2
\nA = 1
\nEquating the co-efficients of v
\n0 = 2A + B
\nB = -2A = -2
\n\"Inter<\/p>\n

Question 4.
\ny\u00b2dx + (x\u00b2 – xy)dy = 0
\nSolution:
\n\"Inter
\nv – log v = log x + log k
\nv = log v + log x + log k
\n= log k (vx)
\n\\(\\frac{y}{x}\\) = log ky
\nSolution is ky = ey\/x<\/sup><\/p>\n

Question 5.
\n\\(\\frac{dy}{dx}=\\frac{(x+y)^2}{2x^2}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 6.
\n(x\u00b2 – y\u00b2)dx – xy dy = 0
\nSolution:
\n(x\u00b2 – y\u00b2)dx – xy dy = 0
\n(x\u00b2 – y\u00b2)dx = xy . dy
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 7.
\n(x\u00b2y – 2xy\u00b2)dx = (x\u00b3 – 3x\u00b2y)dy
\nSolution:
\n(x\u00b2y – 2xy\u00b2)dx = (x\u00b3 – 3x\u00b2y)dy
\n\"Inter
\n\"Inter<\/p>\n

Question 8.
\ny\u00b2dx + (x\u00b2 – xy + y\u00b2) dy = 0
\nSolution:
\ny\u00b2dx = – (x\u00b2 – xy + y\u00b2)dy
\n\"Inter
\n1 – v + v\u00b2 = A(1 + v\u00b2) + (Bv + C)v
\nv\u00a0 =\u00a0 0 \u21d2 1 = A
\nEquating the co-efficients of v\u00b2
\n1 = A + .B \u21d2 B = 0
\nEquating the co-efficients of v
\n-1 = C
\n\"Inter<\/p>\n

Question 9.
\n(y\u00b2 – 2xy)dx + (2xy – x\u00b2)dy = 0
\nSolution:
\n(y\u00b2 – 2xy)dx + (2xy – x\u00b2)dy = 0
\n(2xy – x\u00b2)dy = – (y\u00b2 – 2xy)dx
\n\"Inter
\n2v – 1 = A(1 – v) + Bv
\nv = 0 \u21d2 -1 = A \u21d2 A = -1
\nv = 1 \u21d2 1 = B \u21d2 B = 1
\n\"Inter<\/p>\n

Question 10.
\n\\(\\frac{dy}{dx}+\\frac{y}{x}=\\frac{y^2}{x^2}\\)
\nSolution:
\n\\(\\frac{dy}{dx}+\\frac{y}{x}=\\frac{y^2}{x^2}\\)
\n\"Inter
\n\"Inter<\/p>\n

Question 11.
\nx dy – y dx = \\(\\sqrt{x^2+y^2}\\)
\nSolution:
\nx dy – y dx = \\(\\sqrt{x^2+y^2}\\)
\n\"Inter<\/p>\n

Question 12.
\n(2x – y)dy = (2y – x)dx
\nSolution:
\n\"Inter
\n\"Inter
\n\u21d2 (y – x)\u00b2 = c\u00b2(y + x)\u00b2(y\u00b2 – x\u00b2)
\n\u21d2 y – x = c\u00b2(y + x)\u00b3
\n\u21d2 (x + y)\u00b3 = c(x – y) where c = \\(\\frac{-1}{c^2}\\)
\n\u2234 (x + y)\u00b3 = c(x – y)<\/p>\n

\"Inter<\/p>\n

Question 13.
\n(x\u00b2 – y\u00b2)\\(\\frac{dy}{dx}\\) = xy
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 14.
\n2\\(\\frac{dy}{dx}=\\frac{y}{x}+\\frac{y^2}{x^2}\\)
\nSolution:
\nPut y = vx
\n\"Inter
\n(y – x)\u00b2 = y\u00b2c\u00b2x
\nSolution is y\u00b2x = c(x – y)\u00b2<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nSolve : (1 + ex\/y<\/sup>)dx + ex\/y<\/sup>(1 – \\(\\frac{x}{y}\\))dy = 0.
\nSolution:
\nPut x = vy
\n\"Inter<\/p>\n

Question 2.
\nSolve : x sin \\(\\frac{y}{x}.\\frac{dy}{dx}\\) = y sin \\(\\frac{y}{x}\\) – x
\nSolution:
\nDividing with x, we have
\n\"Inter<\/p>\n

Question 3.
\nSolve : x dy = (y + x cos\u00b2 \\(\\frac{y}{x}\\))dx
\nSolution:
\nx.\\(\\frac{dy}{dx}\\) = y + xcos\u00b2 \\(\\frac{y}{x}\\)
\n\"Inter<\/p>\n

Question 4.
\nSolve : (x – y log y + y log x)dx + x(log y – log x)dy = 0
\nSolution:
\nDividing with x. dx we get
\n\"Inter
\n\"Inter<\/p>\n

Question 5.
\nSolve : (y dx + x dy) x cos \\(\\frac{y}{x}\\) = (x dy – y dx) y sin \\(\\frac{y}{x}\\).
\nSolution:
\nThe given equation can be written as
\n\"Inter
\n\u2234 This is a homogeneous equation
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 6.
\nFind the equation of a curve whose gradient is \\(\\frac{dy}{dx}=\\frac{y}{x}-\\cos^2\\frac{y}{x}\\), where x > 0, y > 0 and which passes through the point (1, \\(\\frac{\\pi}{4}\\)).
\nSolution:
\n\"Inter
\ntan v = – log |x| + c
\nThis curve passes through (1, \\(\\frac{\\pi}{4}\\))
\ntan \\(\\frac{\\pi}{4}\\) = c – log 1
\nc = 1
\n\u2234 Equation of the curve is
\ntan v = 1 – log |x|
\ntan \\(\\frac{y}{x}\\) = 1 – log |x|<\/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(c) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c) I. Question 1. Express x dy – ydx = dx in the form F () = (). Solution: …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35551"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35551"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35551\/revisions"}],"predecessor-version":[{"id":35552,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35551\/revisions\/35552"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35551"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35551"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35551"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}