{"id":22142,"date":"2024-01-11T16:46:15","date_gmt":"2024-01-11T11:16:15","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22142"},"modified":"2024-01-12T16:18:06","modified_gmt":"2024-01-12T10:48:06","slug":"inter-2nd-year-maths-2b-differential-equations-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2b-differential-equations-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2B Differential Equations Important Questions"},"content":{"rendered":"

Students get through Maths 2B Important Questions<\/a> Inter 2nd Year Maths 2B Differential Equations Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 2nd Year Maths 2B Differential Equations Important Questions<\/h2>\n

Question 1.
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ex – y<\/sup> + x2<\/sup> e-y<\/sup> [Mar. 06; May 05]
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\nx\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) – y = 2x2<\/sup> sec2<\/sup>2x [May 11]
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 3.
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y tan x = sin x. [T.S. Mar. 16]
\nSolution:
\nI.F. = \\(e^{\\int \\tan x d x}\\) = elog sec x<\/sup> = sec x
\ny.sec x = \\(\\int\\) sin x . sec x dx = \\(\\int\\) tan x dx
\n= log sec x + c<\/p>\n

Question 4.
\ncos x . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y sin x = sec2<\/sup>x [Mar. 14]
\nSolution:
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + tan x . y = sec3<\/sup>x
\nI.F. = e\\(\\int\\)tan x dx<\/sup> = elog sec x<\/sup> = sec x
\ny . sec x = \\(\\int\\) sec4<\/sup>x dx = \\(\\int\\) (1 + tan2<\/sup> x) sec2<\/sup> x
\ndx = tan x + \\(\\frac{\\tan ^{3} x}{3}\\) + c<\/p>\n

Question 5.
\n(x + y + 1)\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 1.
\nSolution:
\n\\(\\frac{\\mathrm{dx}}{\\mathrm{dy}}\\) = x + y + 1
\n\\(\\frac{\\mathrm{dx}}{\\mathrm{dy}}\\) = x + y + 1
\nI.F. = e\\(\\int\\) -dy<\/sup> = e-y<\/sup>
\nx . e-y<\/sup> = \\(\\int\\) e-y<\/sup> (y + 1)dy
\n= – (y + 1) . e-y<\/sup> + \\(\\int\\) e-y<\/sup> . dy
\n= – (y + 1) e-y<\/sup> – e-y<\/sup>
\n= – (y + 2) e-y<\/sup> + c
\nx = – (y + 2) + c. e-y<\/sup><\/p>\n

\"Inter<\/p>\n

Question 6.
\nFind the order and degree of r
\n\\(\\frac{\\mathrm{d}^{3} \\mathrm{~y}}{\\mathrm{dx}^{3}}\\) – 3 (\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)) – ex<\/sup> = 4. [Mar. 14]
\nSolution:
\nThe equation is a polynomial in \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) and \\(\\frac{\\mathrm{d}^{3} \\mathrm{~y}}{\\mathrm{dx}^{3}}\\).
\nThe exponent of \\(\\frac{\\mathrm{d}^{3} \\mathrm{~y}}{\\mathrm{dx}^{3}}\\) is 2.
\nHence the degree is 2.
\n\\(\\frac{\\mathrm{d}^{3} \\mathrm{~y}}{\\mathrm{dx}^{3}}\\) is the highest order derivative occuring in the equation.
\nOrder of the equation is 3.<\/p>\n

Question 7.
\nx\\(\\frac{1}{2}\\)<\/sup>(\\(\\frac{\\mathrm{d}^{2} \\mathrm{~y}}{\\mathrm{dx}^{2}}\\))\\(\\frac{1}{3}\\)<\/sup> + x . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y = 0 has order 2 and degree 1. Prove. [T.S. Mar. 15]
\nSolution:
\nThe given equation can be written as
\n\"Inter
\n\u2234 The order of the equation is 2 and its degree is 1.<\/p>\n

Question 8.
\nFind the order and degree of \\(\\left(\\frac{d^{2} y}{d x^{2}}+\\left(\\frac{d y}{d x}\\right)^{3}\\right)^{\\frac{6}{5}}\\) = 6y [Mar. 16; May 11]
\nSolution:
\nGiven equation is \\(\\left(\\frac{d^{2} y}{d x^{2}}+\\left(\\frac{d y}{d x}\\right)^{3}\\right)^{\\frac{6}{5}}\\) = 6y
\ni.e., \\(\\frac{\\mathrm{d}^{2} \\mathrm{~y}}{\\mathrm{dx}^{2}}\\) + (\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\))3<\/sup> = (6y)\\(\\frac{5}{6}\\)<\/sup>
\nOrder = 2, degree = 1<\/p>\n

\"Inter<\/p>\n

Question 9.
\nSolve \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{x(2 \\log x+1)}{\\sin y+y \\cos y}\\) [Mar. 08]
\nSolution:
\nGiven equation can be written as
\n(sin y + y cos y) dy = x(2 log x + 1) dx
\n\\(\\int\\) sin y dy + \\(\\int\\) y cos y dy = \\(\\int\\) 2x log x dx + \\(\\int\\) x dx
\n\\(\\int\\) sin y dy + y sin y – \\(\\int\\) sin y dy = x2<\/sup> log x – \\(\\int\\) x2<\/sup> . \\(\\frac{1}{x}\\) dx + \\(\\int\\) x dx + c
\ny sin y = x2<\/sup> log x + c<\/p>\n

Question 10.
\n(xy2<\/sup> + x) dx + (yx2<\/sup> + y) dy = 0. [A.P. Mar. 15, 07]
\nSolution:
\n(xy2<\/sup> + x) dx + (yx2<\/sup> + y) dy = 0
\nx(y2<\/sup> + 1) dx + y (x2<\/sup> + 1) dy = 0
\nDividing with (1 + x2<\/sup>) (1 + y2<\/sup>)
\n\\(\\frac{x d x}{1+x^{2}}\\) + \\(\\frac{y d x}{1+y^{2}}\\) = 0
\nIntegrating
\n\\(\\int \\frac{x d x}{1+x^{2}}+\\int \\frac{y d y}{1+y^{2}}=0\\)
\n\\(\\frac{1}{2}\\)[(log (1 + x2<\/sup>) + log (1 + y2<\/sup>)] = log c
\nlog (1 + x2<\/sup>) (1 + y2<\/sup>) = 2 log c = log c2<\/sup>
\nSolution is(1 + x2<\/sup>) (1 + y2<\/sup>) = k when k = c2<\/sup>.<\/p>\n

Question 11.
\nsin-1<\/sup> (\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)) = x + y [Mar. 07]
\nSolution:
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = sin (x + y)
\nx + y = t
\n1 + \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\)
\n\\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\) – 1 = sin t
\n\\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\) = 1 + sin t
\n\\(\\frac{d t}{1+\\sin t}\\) = dx
\nIntegrating both sides we get
\n\\(\\int \\frac{d t}{1+\\sin t}=\\int d x\\)
\n\\(\\int \\frac{1-\\sin t}{\\cos ^{2} t} d t=x+c\\)
\n\\(\\int\\) sec2<\/sup> t dt = \\(\\int\\) tan t . sec t dt = x + c
\ntan t – sec t = x + c
\n\u21d2 tan (x + y) – sec (x + y) = x + c<\/p>\n

\"Inter<\/p>\n

Question 12.
\n(x2<\/sup> – y2<\/sup>) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = xy [May 11]
\nSolution:
\n\"Inter
\n= log y + c
\n\\(\\frac{-x^{2}}{2 y^{2}}\\) = (log y + c)
\n-x2<\/sup> = 2y2<\/sup> (c + log y)
\n\u21d2 Solution is x2<\/sup> + 2y2<\/sup> (c + log y) = 0.<\/p>\n

Question 13.
\nSolve : x dy = (y + x cos2<\/sup> \\(\\frac{y}{x}\\)) dx.
\nSolution:
\n\"Inter<\/p>\n

Question 14.
\n(2x + y + 1) dx + (4x + 2y – 1) dy = 0 [T.S. Mar. 15]
\nSolution:
\n\"Inter
\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

\"Inter<\/p>\n

Question 15.
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y tan x = cos3<\/sup>x [May 11]
\nSolution:
\nI.F. = e\\(\\int\\) tan x dx<\/sup> = elog sec x<\/sup> = sec x
\ny . sec x = \\(\\int\\) sec x. cos3<\/sup> x dx
\n= \\(\\int\\) cos2<\/sup>x dx
\n= \\(\\frac{1}{2}\\) \\(\\int\\) (1 + cos 2x) dx
\n= \\(\\frac{1}{2}\\) (x + \\(\\frac{sin2x}{2}\\)) + c
\n\\(\\frac{2 y}{\\cos x}\\) = x + sin x . cos x + c
\nSolution is 2y = x cos x + sin x . cos2<\/sup> x + c . cos x^<\/p>\n

Question 16.
\n(1 + x2<\/sup>) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y = etan-1 x<\/sup> [May 07] [A.P. Mar. 16] [T.S. Mar. 15]
\nSolution:
\n\"Inter<\/p>\n

Question 17.
\nSolve (1 + y2<\/sup>)dx = (tan-1<\/sup>y – x)dy. [A.P. Mar. 15]
\nSolution:
\nGiven \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{\\tan ^{-1} y-x}{1+y^{2}}\\)
\n\"Inter
\n\"Inter<\/p>\n

Question 18.
\n(x2<\/sup> – y2<\/sup>)dx – xy dy = 0 [May 06]
\nSolution:
\n(x2<\/sup> – y2<\/sup>)dx – xy dy = 0
\n(x2<\/sup> – y2<\/sup>)dx = xy . dy
\n\"Inter
\n–\\(\\frac{1}{4}\\) [log (x2<\/sup> – 2y2<\/sup>) – log x2<\/sup>] = log x + log c
\n–\\(\\frac{1}{4}\\) log (x2<\/sup> – 2y2<\/sup>) + \\(\\frac{1}{4}\\) . 2 log x = log x + log c
\n–\\(\\frac{1}{4}\\) log (x2<\/sup> – 2y2<\/sup>) = \\(\\frac{1}{2}\\) log x + log c
\n– log (x2<\/sup> – 2y2<\/sup>) = – 2 log x – 4 log c
\nlog (x2<\/sup> – 2y2<\/sup>) = – 2 log x + log k where
\nk = \\(\\frac{1}{c^{4}}\\) = log \\(\\frac{\\mathrm{k}}{\\mathrm{x}^{2}}\\)
\nx2<\/sup> – 2y2<\/sup> = \\(\\frac{\\mathrm{k}}{\\mathrm{x}^{2}}\\)
\nSolution is x2<\/sup> (x2<\/sup> – 2y2<\/sup>) = k<\/p>\n

\"Inter<\/p>\n

Question 19.
\n\\(\\frac{d y}{d x}=\\frac{3 y-7 x+7}{3 x-7 y-3}\\) [T.S. Mar. 16]
\nSolution:
\n\"Inter
\n\"Inter
\n= 3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)
\n14ln x – ln c = – 10 ln (v + 1) – 4 ln (v – 1)
\nln (v + 1)5<\/sup> + ln (v – 1)2<\/sup> + ln x7<\/sup> = ln c
\n(v + 1)5<\/sup> . (v – 1)2<\/sup> . x7<\/sup> = c
\n(\\(\\frac{y}{x}\\) + 1)5<\/sup> (\\(\\frac{y}{x}\\) – 1)2<\/sup> . x7<\/sup> = c
\n(y – x)2<\/sup> (y + x)5<\/sup> = c
\n[y – (x – 1)]2<\/sup> (y + x – 1)5<\/sup> = c
\nSolution is [y – x + 1]2<\/sup> (y + x – 1)5<\/sup> = c.<\/p>\n

\"Inter<\/p>\n

Question 20.
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) (x2<\/sup>y3<\/sup> + xy) = 1 [Mar. 11]
\nSolution:
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = xy + x2<\/sup>y3<\/sup>
\nThis is Bernoulli’s equation
\nx-2<\/sup> . \\(\\frac{\\mathrm{dx}}{\\mathrm{dy}}\\) – \\(\\frac{1}{x}\\) . y = y3<\/sup>
\n\"Inter
\n\"Inter<\/p>\n

Question 21.
\nForm the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [AP Mar. 15]
\nSolution:
\nWe have y = A cos 3x + B sin 3x
\nDifferentiating w.r.to x
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = -3A sin 3x + 3B cos 3x
\nDifferentiating again w.r.to. x
\n\\(\\frac{\\mathrm{d}^{2} \\mathrm{~y}}{\\mathrm{dx}^{2}}\\) = -9A cos 3x – 9B sin 3x
\n= – 9(A cos 3x + B sin 3x)
\n= -9y .
\nis \\(\\frac{\\mathrm{d}^{2} \\mathrm{~y}}{\\mathrm{dx}^{2}}\\) + 9y = 0.
\nAlternate method:
\nEliminating A, B from the equation
\ny = A cos 3x + B sin 3x
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = – 3A sin 3x + 3B sin cos 3x
\n\"Inter
\nThis is the required differential equation.<\/p>\n

Question 22.
\nSolve (x2<\/sup> + y2<\/sup>) dx = 2xy dy [A.P. Mar. 16]
\nSolution:
\nGiven equation can be written as
\n\"Inter
\nlog cx(1 – v2<\/sup> = log 1
\ncx (1 – v2<\/sup>) = 1
\ncx (1 – \\(\\frac{\\mathrm{y}^{2}}{\\mathrm{x}^{2}}\\)) = 1
\nc(x2<\/sup> – y2<\/sup>) = x is the required solution.<\/p>\n

\"Inter<\/p>\n

Question 23.
\nGive the solution of x sin2<\/sup> \\(\\frac{y}{x}\\) dx = y dx – x dy which passes through the point (1, \\(\\frac{\\pi}{4}\\)). [Mar. 14]
\nSolution:
\nThe given equation can be written as
\n\"Inter
\nThe given curve passes through (1, \\(\\frac{\\pi}{4}\\))
\ncot \\(\\frac{\\pi}{4}\\) = log 1 + c
\n1 = 0 + c \u21d2 c = 1
\nSolution is cot \\(\\frac{y}{x}\\) = log x + 1<\/p>\n

Question 24.
\nFind the order and degree of the differential equation \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) = – p2<\/sup>y.
\nSolution:
\nThe given equation is a polynomial equation in \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\)
\nHence the degree is 1
\n\\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) is the highest order derivative occuring in the equation.
\nIts order is 2.<\/p>\n

Question 25.
\nFind the order and degree of
\n(\\(\\frac{\\mathrm{d}^{3} \\mathrm{y}}{\\mathrm{dx}^{3}}\\))2<\/sup> – 3 (\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\))2<\/sup> – ex<\/sup> = 4 [Mar. 14]
\nSolution:
\nThe equation is a polynomial in \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) and \\(\\frac{\\mathrm{d}^{3} \\mathrm{y}}{\\mathrm{dx}^{3}}\\).
\nThe exponent of \\(\\frac{\\mathrm{d}^{3} \\mathrm{y}}{\\mathrm{dx}^{3}}\\) is 2.
\nHence the degree is 2.
\n\\(\\frac{\\mathrm{d}^{3} \\mathrm{y}}{\\mathrm{dx}^{3}}\\) is the highest \u00f5rder derivative occuring in the equation.
\nOrder of the equation is 3.<\/p>\n

\"Inter<\/p>\n

Question 26.
\nx\\(\\frac{1}{2}\\)<\/sup>(\\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\))\\(\\frac{1}{3}\\)<\/sup> + x . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y = 0 has order 2 and degree 1. Prove. [T.S. Mar. 15]
\nSolution:
\nThe given equation can be written as
\nx\\(\\frac{1}{2}\\)<\/sup>(\\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\))\\(\\frac{1}{3}\\)<\/sup> = -[x . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y]
\nCubing both sides
\nx\\(\\frac{3}{2}\\)<\/sup>(\\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\)) = -[x . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y]3<\/sup>
\n\u2234 The order of the equation is 2 and its degree is 1.<\/p>\n

Question 27.
\nFind the order and degree of \\(\\left(\\frac{d^{2} y}{d x^{2}}+\\left(\\frac{d y}{d x}\\right)^{3}\\right)^{\\frac{6}{5}}\\) = 6y [A.P. Mar. 16; May 11]
\nSolution:
\nGiven equation is \\(\\left(\\frac{d^{2} y}{d x^{2}}+\\left(\\frac{d y}{d x}\\right)^{3}\\right)^{\\frac{6}{5}}\\) = 6y
\ni.e., \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) + (\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\))3<\/sup> = (6y)\\(\\frac{5}{6}\\)<\/sup>
\nOrder = 2; degree = 1<\/p>\n

Question 28.
\nFind the order of the differential equation corresponding to y = c(x – c)2<\/sup>, where c is an arbitrary constant.
\nSolution:
\nThe given differential equation is
\ny = c(x – c)2<\/sup>
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 2c(x – c)
\n\u2234 Order of the differential equation is 1.<\/p>\n

\"Inter<\/p>\n

Question 29.
\nFind the order of the differential equation corresponding to y = Aex<\/sup> + Be3x<\/sup> + Ce5x<\/sup>; (A, B, C being parameters) is a solution.
\nSolution:
\nRequired differential equation is obtained by eliminating A, B, C from y,
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\), \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\), \\(\\frac{\\mathrm{d}^{3} \\mathrm{y}}{\\mathrm{dx}^{3}}\\)
\nHighest order deviation = \\(\\frac{\\mathrm{d}^{3} \\mathrm{y}}{\\mathrm{dx}^{3}}\\)
\nOrder of the differential equation = 3.<\/p>\n

Question 30.
\nForm the differential equation corresponding to y = cx – 2c2<\/sup>, where c is a parameter.
\nSolution:
\nGiven y = cx – 2c2<\/sup> ………………. (1)
\nDifferentiating (1) w.r.to
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = c
\nSubstituting in (1), required differential equation is
\ny = x . (\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)) – 2(\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\))2<\/sup><\/p>\n

Question 31.
\nForm the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [A.P. Mar. 15]
\nSolution:
\nWe have y = A cos 3x + B sin 3x
\nDifferentiating w.r.to x
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = -3A sin 3x + 3B cos 3x
\nDifferentiating again w.r.to. x
\n\\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) = -9A cos 3x – 9B sin 3x
\n= -9(A cos 3x + B sin 3x)
\n= -9y
\nis \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) + 9y = 0<\/p>\n

\"Inter<\/p>\n

Alternate Method:
\nEliminating A, B from the equation
\ny = A cos 3x + B sin 3x
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = -3A sin 3x = 3B sin cos 3x
\n\\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) = -9A cos 3x – 9B sin 3x
\nWe get \\(\\left|\\begin{array}{ccc}
\ny & -\\cos 3 x & -\\sin 3 x \\\\
\n\\left(\\frac{d y}{d x}\\right) & 3 \\sin 3 x & -3 \\cos 3 x \\\\
\n\\left(\\frac{d^{2} y}{d x^{2}}\\right) & 9 \\cos 3 x & 9 \\sin 3 x
\n\\end{array}\\right|\\) = 0
\ny(27 sin2<\/sup> 3x + 27 cos2<\/sup> 3x) – (-9 sin 3x. cos 3x + 9 cos 3x. sin 3x) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + (3 cos2<\/sup> 3x + 3 sin2<\/sup> 3x) \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) = 0
\n= 27y + 3 .\\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) = 0 or \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) + 9y = 0
\nThis is the required differential equation.<\/p>\n

Question 32.
\nForm the differential equation corresponding to the family of circles of radius r given by (x – a)2<\/sup> + (y – b)2<\/sup> = r2<\/sup>, where a and b are parameters.
\nSolution:
\nWe have (x – a)2<\/sup> + (y – b)2<\/sup> = r2<\/sup> ………………… (1)
\nDifferentiating (1) w.r.to x
\n2(x – a) + 2(y – b) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 0 ……………….. (2)
\nDifferentiating (2) w.r.to. x
\n1 + (y – b) \\(\\frac{\\mathrm{d}^{2} \\mathrm{y}}{\\mathrm{dx}^{2}}\\) + (\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\))2<\/sup> = 0 ……………… (3)
\nFrom (2) (x – a) = -(y – b) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)
\nSubstituting in (1), we get
\n\"Inter
\ni.e., \\(r^{2}\\left(\\frac{d^{2} y}{d x^{2}}\\right)^{2}=\\left(1+\\left(\\frac{d y}{d x}\\right)^{2}\\right)^{3}\\)
\nWhich is the required differential equation.<\/p>\n

Question 33.
\nForm the differential equation corresponding to the family of circles passing through the origin and having centres on Y-axis.
\nSolution:
\nThe equation of the family of all circles passing through the origin and having centres on Y\u2014axis is
\nx2<\/sup> + y2<\/sup> + 2hy = 0 …………………. (1)
\nWhere h is a parameter
\nDifferentiating (1) w.r. to x
\n2x + 2y . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + 2h . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 0
\nor x + y . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + h . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 0
\n-(x + y . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)) = h. \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)
\nh = \\(\\frac{-\\left(x+y \\cdot \\frac{d y}{d x}\\right)}{\\frac{d y}{d x}}\\)
\nSubstituting in (1)
\nWe get x2<\/sup> + y2<\/sup> – 2y \\(\\frac{\\left(x+y \\cdot \\frac{d y}{d x}\\right)}{\\frac{d y}{d x}}\\) = 0
\nx2<\/sup> . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y2<\/sup> . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) – 2xy – 2y2<\/sup> . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 0
\nor (x2<\/sup> – y2<\/sup>) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) – 2xy = 0
\nThis is the required differential equation.<\/p>\n

\"Inter<\/p>\n

Question 34.
\nExpress the following differential equations in the form f(x) dx + g(y) dy = 0.
\ni) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}=\\frac{1+\\mathrm{y}^{2}}{1+\\mathrm{x}^{2}}\\)
\nSolution:
\n\u21d2 \\(\\frac{d y}{1+y^{2}}=\\frac{d x}{1+x^{2}}\\)
\n\\(\\frac{d x}{1+x^{2}}-\\frac{d y}{1+y^{2}}\\) = 0<\/p>\n

ii) y – x \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = a (y2<\/sup> + \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\))
\nSolution:
\ny – x . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ay2<\/sup> + a. \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)
\ny – ay2<\/sup> = (x + a) . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)
\n\\(\\frac{d x}{x+a}=\\frac{d y}{y-a y^{2}}\\)<\/p>\n

iii) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ex-y<\/sup> + x2<\/sup> e-y<\/sup>
\nSolution:
\nMultiplying in the ey<\/sup>
\ney<\/sup> . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ex<\/sup> + x2<\/sup>
\ney<\/sup> . dy = (ex<\/sup> + x2<\/sup>) dx
\n(ex<\/sup> + x2<\/sup>) dx – ey<\/sup> . dy = 0<\/p>\n

iv) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + x2<\/sup> = x2<\/sup> e3y<\/sup>
\nSolution:
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = x2<\/sup> . e3y<\/sup> – x2<\/sup> = x2<\/sup>(e3y<\/sup> – 1)
\n\\(\\frac{d y}{e^{3 y}-1}\\) = x2<\/sup> dx \u21d2 x2<\/sup> dx – \\(\\frac{d y}{e^{3 y}-1}\\) = 0
\nor x2<\/sup>dx + \\(\\frac{1}{\\left(1-e^{3 y}\\right)}\\) . dy = 0<\/p>\n

Question 35.
\nFind the general solution of
\nx + y\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 0.
\nSolution:
\nGiven equation is x + y . \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 0
\nx . dx + y . dy = 0
\nIntegrating \\(\\frac{x^{2}}{2}\\) + \\(\\frac{y^{2}}{2}\\) = c
\nor x2<\/sup> + y2<\/sup> = 2 c = c’<\/p>\n

Question 36.
\nFind the general solution of \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ex+y<\/sup>.
\nSolution:
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ex+y<\/sup> = ex<\/sup> . ey<\/sup>
\n\\(\\frac{d y}{e^{y}}\\) = ex<\/sup> dx
\n\\(\\int\\) e-y<\/sup> dy = \\(\\int\\) ex<\/sup> dx
\ne – e-y<\/sup> = ex<\/sup>
\nor ex<\/sup> + e-y<\/sup> = c is the required solution.<\/p>\n

\"Inter<\/p>\n

Question 37.
\nSolve y2<\/sup> – x \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = a(y + \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\))
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 38.
\nSolve \\(\\frac{d y}{d x}=\\frac{y^{2}+2 y}{x-1}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 39.
\nSolve \\(\\frac{d y}{d x}=\\frac{x(2 \\log x+1)}{\\sin y+y \\cos y}\\) [Mar. 08]
\nSolution:
\nGiven equation can be written as
\n(sin y + y cos y) dy = x(2 log x + 1) dx
\n\\(\\int\\) sin y dy + \\(\\int\\) y cos y dy = \\(\\int\\) 2x log x dx + \\(\\int\\) x dx
\n\\(\\int\\) sin y dy + y sin y – \\(\\int\\) sin y dy = x2<\/sup> log x – \\(\\int\\) x2<\/sup> . \\(\\frac{1}{x}\\) dx + \\(\\int\\) x dx + c
\ny sin y = x2<\/sup> log x + c<\/p>\n

\"Inter<\/p>\n

Question 40.
\nFind the equation of the curve whose slope, at any point, (x, y) is \\(\\frac{y}{x^{2}}\\) and which satisfies the condition y = 1 when x = 3.
\nSolution:
\n\"Inter<\/p>\n

Question 41.
\nSolve y(1 + x) dx + x(1 + y) dy = 0
\nSolution:
\nThe given equation can be written as
\n\\(\\frac{(1+x)}{x}\\) dx + \\(\\frac{(1+y)}{y}\\) . dy = 0
\n\\(\\int\\) (1 + \\(\\frac{1}{x}\\))dx + \\(\\int\\) (1 + \\(\\frac{1}{y}\\)) dy = 0
\nx + log x + y + log y = c
\nx + y + log (xy) = c is the required solution.<\/p>\n

Question 42.
\nSolve \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = sin (x + y) + cos (x + y)
\nSolution:
\nPut x + y = t
\n\"Inter
\nx = log (1 + tan \\(\\frac{t}{2}\\)) + c
\nBut t = x + y
\nSolution is x = log (1 + tan \\(\\frac{x+y}{2}\\)) + c<\/p>\n

Question 43.
\nSolve (x – y)2<\/sup> \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = a2<\/sup>
\nSolution:
\nPut x – y = t
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 44.
\nSolve \\(\\sqrt{1+x^{2}} \\sqrt{1+y^{2}}\\) dx + xy dy = 0
\nSolution:
\nGiven equation can be written as
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

Question 45.
\nSolve \\(\\frac{d y}{d x}=\\frac{x-2 y+1}{2 x-4 y}\\)
\nSolution:
\nPut x – 2y = t
\n\"Inter<\/p>\n

Question 46.
\nSolve \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\sqrt{y-x}\\)
\nSolution:
\nPut y – x = t2<\/sup>
\n\"Inter<\/p>\n

Question 47.
\nSolve \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + 1 = ex+y<\/sup>
\nSolution:
\nPut t = x + y
\n\\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\) = 1 + \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = et<\/sup>
\n\\(\\int \\frac{d t}{e^{t}}=\\int d x\\)
\n\\(\\int\\) e-t<\/sup> dt = \\(\\int\\) dx
\n-e-t<\/sup> = x + c
\nx + e-t<\/sup> + c = 0
\nSolution is x + e-(x+y)<\/sup> + c = 0<\/p>\n

\"Inter<\/p>\n

Question 48.
\nSolve \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = (3x + y + 4)2<\/sup>
\nSolution:
\nPut t = 3x + y + 4
\n\\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\) = 3 + \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 3 + t2<\/sup>
\n\\(\\frac{\\mathrm{dt}}{\\mathrm{t}^{3}+3}\\) = dx
\n\\(\\int \\frac{d t}{t^{2}+3}=\\int d x\\)
\n\\(\\frac{1}{\\sqrt{3}}\\) tan-1<\/sup> (\\(\\frac{t}{\\sqrt{3}}\\)) = x + c
\nSolution is \\(\\frac{1}{\\sqrt{3}}\\) tan-1<\/sup> (\\(\\frac{3 x+y+4}{\\sqrt{3}}\\)) = x + c<\/p>\n

Question 49.
\nSolve \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) – x tan (y – x) = 1
\nSolution:
\nPut y – x = t
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) – 1 = \\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\)
\n\\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\) = x tan t + 1 – 1 = x tan t
\n\\(\\frac{\\mathrm{dt}}{\\tan \\mathrm{t}}\\) = x dx
\n\\(\\int\\) cot dt = \\(\\int\\) x dx
\nlog |sin | = \\(\\frac{x^{2}}{2}\\) + c
\nSolution is log |sin (y – x)| = \\(\\frac{x^{2}}{2}\\) + c<\/p>\n

Question 50.
\nShow that f(x, y) = 1 + ex\/y<\/sup> is a homogeneous function of x and y.
\nSolution:
\nf(kx, xy) = 1 + ekx\/ky<\/sup> = 1 + ex\/y<\/sup> = f(x, y)
\nf(x, y) is a homogeneous function degree 0.<\/p>\n

Question 51.
\nShow that f(x, y) = x\\(\\sqrt{x^{2}+y^{2}}\\) – y2<\/sup> is a homogeneous function of x and y.
\nSolution:
\nf(kx, ky) = kx\\(\\sqrt{k^{2} x^{2}+k^{2} y^{2}}\\) – k2<\/sup>y2<\/sup>.
\n= k2<\/sup> (x\\(\\sqrt{x^{2}+y^{2}}\\) – y2<\/sup>) = k2<\/sup> f(x, y)
\nf(x, y) is a homogeneous function of degree 2.<\/p>\n

\"Inter<\/p>\n

Question 52.
\nShow that f(x, y) = x – y log y + y log x is a homogeneous function of x and y.
\nSolution:
\nf(kx, ky) = kx – ky. log ky + ky log (kx)
\n= k(x – y log (ky) + y log kx)
\n= k(x – y log k – y log y + y log k + y log x)
\n= k(x – y log y + y log x)
\n= k. f(x, y)
\nf(x, y) is a homogeneous function of degree 1.<\/p>\n

Question 53.
\nExpress (1 + ex\/y<\/sup>)dx + ex\/y<\/sup> (1 – \\(\\frac{x}{y}\\)) dy = 0 in the form \\(\\frac{\\mathrm{dx}}{\\mathrm{dy}}\\) = F (\\(\\frac{x}{y}\\))
\nSolution:
\n\"Inter<\/p>\n

Question 54.
\nExpress (x\\(\\sqrt{x^{2}+y^{2}}\\) – y2<\/sup>) dx + xy dy = 0 in the form \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = F (\\(\\frac{x}{y}\\))
\nSolution:
\nGiven equation is
\n\"Inter<\/p>\n

Question 55.
\nExpress \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{y}{x+y e^{\\frac{-2 x}{y}}}\\) in the form \\(\\frac{\\mathrm{dx}}{\\mathrm{dy}}\\) = F (\\(\\frac{x}{y}\\))
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 56.
\nSolve \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{y^{2}-2 x y}{x^{2}-x y}\\)
\nSolution:
\nThe given equation is a homogeneous equation.
\nPut y = vx
\n\"Inter
\n\\(\\log v \\sqrt{2 v-3}=-3 \\log \\frac{x}{c}=\\log \\frac{c^{3}}{x^{3}}\\)
\n\"Inter<\/p>\n

Question 57.
\nSolve (x2<\/sup> + y2<\/sup>) dx = 2xy dy [A.P. Mar. 16]
\nSolution:
\nGiven equation can be written as
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{x^{2}+y^{2}}{2 x y}\\)
\nThis is a homogeneous function
\nPut y = vx
\n\"Inter
\n-log (1 – v2<\/sup>) = log x + log c
\n= log cx
\nlog cx + log (1 – v2<\/sup>) = 0
\nlog cx(1 – v2<\/sup>) = log 1
\ncx (1 – v2<\/sup>) = 1
\ncx (1 – \\(\\frac{\\mathrm{y}^{2}}{\\mathrm{x}^{2}}\\)) = 1
\nc(x2<\/sup> – y2<\/sup>) = x is the required solution.<\/p>\n

\"Inter<\/p>\n

Question 58.
\nSolve xy2<\/sup> dy – (x3<\/sup> + y3<\/sup>) dx = 0.
\nSolution:
\nGiven equation is xy2<\/sup> dy = (x3<\/sup> + y2<\/sup>) dx
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{x^{3}+y^{3}}{x y^{2}}\\)
\nThis is a homogeneous equation.
\nPut y = vx
\n\"Inter<\/p>\n

Question 59.
\nSolve \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{x^{2}+y^{2}}{2 x^{2}}\\)
\nSolution:
\nThis is a homogeneous equation.
\n\"Inter<\/p>\n

Question 60.
\nSolve x sec (\\(\\frac{y}{x}\\)) . (y dx + x dy) = y cosec (\\(\\frac{y}{x}\\)) . (x dy – y dx)
\nSolution:
\nThis given equation can be written as
\n\"Inter
\n\"Inter
\nlog(\\(\\frac{\\sin v}{v}\\)) = log cx2<\/sup>
\n\\(\\frac{\\sin v}{v}\\) = cx2<\/sup>
\n\\(\\frac{x}{y}\\) sin (\\(\\frac{y}{x}\\)) = cx2<\/sup>
\nSolution is sin(\\(\\frac{y}{x}\\)) = cxy.<\/p>\n

\"Inter<\/p>\n

Question 61.
\nGive the solution of x sin2<\/sup> \\(\\frac{y}{x}\\) dx = y dx – x dy which passes through the point (1, \\(\\frac{\\pi}{4}\\)). [Mar. 14]
\nSolution:
\nThe given equation can be written as
\n\"Inter
\nThe given curve passes through (1, \\(\\frac{\\pi}{4}\\))
\ncot \\(\\frac{\\pi}{4}\\) = log 1 + c
\n1 = 0 + c \u21d2 c = 1
\nSolution is cot \\(\\frac{y}{x}\\) = log x + 1<\/p>\n

Question 62.
\nSolve (x3<\/sup> – 3xy2<\/sup>) dx + (3x2<\/sup>y – y3<\/sup>) dy = 0
\nSolution:
\n(x3<\/sup> – 3xy2<\/sup>) dx = -(3x2<\/sup>y – y3<\/sup>) dy
\n\"Inter
\n\"Inter<\/p>\n

Question 63.
\nTransform the following two differential equations Into linear form.
\nx log x \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y = 2 log x
\nSolution:
\nGiven equation can be written as
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + \\(\\frac{1}{x \\log x}\\) . y = \\(\\frac{2}{x}\\)
\nThis is of the form \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + Py = Q<\/p>\n

\"Inter<\/p>\n

Question 64.
\n(x + 2y3<\/sup>) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = y
\nSolution:
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{x+2 y^{3}}{y}\\) = \\(\\frac{x}{y}\\) + 2y2<\/sup>
\n\\(\\frac{\\mathrm{dx}}{\\mathrm{dy}}\\) – \\(\\frac{1}{y}\\) . x = 2y2<\/sup>
\nThis is of the form \\(\\frac{\\mathrm{dx}}{\\mathrm{dy}}\\) + Px = Q.<\/p>\n

Question 65.
\n(cos x) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y sin x = tan x
\nSolution:
\nGiven equation can be written as
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y(tan x) = (tan x) (sec x)
\nP = tan x \u21d2 \\(\\int\\) P dx = \\(\\int\\) tan x dx = log sec x
\nI.F. = e\\(\\int\\) log sec x<\/sup> = sec x<\/p>\n

Question 66.
\nSolve (2x – 10y3<\/sup>) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y = o
\nSolution:
\n\"Inter<\/p>\n

Question 67.
\nSolve (1 + x2<\/sup>) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + 2xy – 4x2<\/sup> = 0
\nSolution:
\nGiven equation can be written as
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 68.
\nSolve \\(\\frac{1}{x}\\) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y . ex<\/sup> = e(1 – x)ex<\/sup>
\nSolution:
\nGiven equation can be written as
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + (x.ex<\/sup>) . y = x . e(1 – x)ex<\/sup>
\nI.F. = e\\(\\int\\).ex dx<\/sup> = e(x – 1) ex<\/sup>
\ny . e(x – 1) ex<\/sup> = \\(\\int\\) x dx
\n= \\(\\frac{\\mathrm{x}^{2}}{2}\\) + c
\n2y. e(x – 1) ex<\/sup> = x2<\/sup> + 2c is the required solution.<\/p>\n

Question 69.
\nSolve sin2<\/sup> x. \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y = cot x
\nSolution:
\nGiven equation can be written as
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) + y cosec2<\/sup> x = cot x . cosec2<\/sup> x
\nI.F. = e\\(\\int\\) cosec2 x dx<\/sup> = e-cot x<\/sup>
\ny . e-cot x<\/sup> = \\(\\int\\) e-cot x<\/sup> . cosec2<\/sup> x. cot x dx ……………… (1)
\nConsider \\(\\int\\) e-cot x<\/sup> . cosec2<\/sup> x . cot x dx
\nPut -cot x = t \u21d2 cosec2<\/sup> x dx = dt
\n(1) becomes y . et<\/sup> = \\(\\int\\) -t. et<\/sup> dt
\n= -(t – 1) et<\/sup> + c
\ny . e-cot x<\/sup> = -(-cot x – 1) e-cot x<\/sup> + 1
\n= (cot x + 1) e-cot x<\/sup> + c is the required solution.<\/p>\n

\"Inter<\/p>\n

Question 70.
\nFind the solution of the equation
\nx(x – 2) \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) – 2(x – 1)y = x3<\/sup>(x – 2)
\nwhich satisfies the condition that y = 9 when x = 3.
\nSolution:
\nThe given equation can be written as
\n\"Inter
\n\"Inter<\/p>\n

Question 71.
\nSolve (1 + y2<\/sup>)dx = (tan-1<\/sup>y – x)dy. [A.P. Mar. 15]
\nSolution:
\n\"Inter
\nis the solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Differential Equations Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2B Differential Equations Important Questions Question 1. = ex – y + x2 e-y [Mar. 06; May 05] Solution: Question 2. x – y = … 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\/22142"}],"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=22142"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22142\/revisions"}],"predecessor-version":[{"id":22143,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22142\/revisions\/22143"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22142"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22142"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22142"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}