{"id":32214,"date":"2021-08-16T17:49:42","date_gmt":"2021-08-16T12:19:42","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=32214"},"modified":"2021-08-16T17:49:42","modified_gmt":"2021-08-16T12:19:42","slug":"ncert-solutions-for-class-12-maths-chapter-9-ex-9-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-2\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 9 Differential Equations Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\ny = ex<\/sup> + 1 : y”- y’ = 0
\nSolution:
\ny = ex<\/sup> + 1
\nDifferentiating w.r.t. x, we get
\ny’ = ex<\/sup>
\ni.e., y” = y’
\n\u21d2 y” – y’ = 0
\nHence y = ex<\/sup> is a solution of y” – y’ = 0.<\/p>\n

Question 2.
\ny = x\u00b2 + 2x + C : y’ – 2x – 2 = 0
\nSolution:
\ny = x\u00b2 + 2x + C
\nDifferentiating w.r.t. x, we get y’ = 2x + 2
\ni.e., y\u2019 – 2x – 2 = 0
\nHence y = x\u00b2 + 2x + C is a solution of y’ – 2x – 2 = 0<\/p>\n

\"NCERT<\/p>\n

Question 3.
\ny = cos x + C : y’ + sin x = 0
\nSolution:
\ny = cos x + C
\nDifferentiating w.r.t., x, we get y’ = – sin x
\ni.e., y’ + sin x = 0
\nHence y = cos x + C is a solution of y’ + sin x = 0<\/p>\n

Question 4.
\ny = \\(\\sqrt{1+x^{2}} \\quad: \\quad y^{\\prime}=\\frac{x y}{1+x^{2}}\\)
\nSolution:
\n\\(\\sqrt{1+x^{2}}\\)
\nDifferentiating w.r.t. x, we get
\n\"NCERT<\/p>\n

Question 5.
\ny – Ax : xy’ = y (x \u2260 0)
\nSolution:
\ny = Ax … (1)
\nDifferentiating w.r.t., x, we get y’ = A
\ni.e., y’ = \\(\\frac { y }{ x }\\) [From(1), A = \\(\\frac { y }{ x }\\)]
\ni.e., xy’ = y (x \u2260 0)
\n\u2234 y = Ax is a solution of xy’ = y.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\ny = x sin x : xy’= y + x \\(\\sqrt{x^{2}-y^{2}}\\)
\n(x \u2260 0 and x > y or x < – y)
\nSolution:
\ny = x sinx … (1)
\nDifferentiating (1) w.r.t. x, we get
\ny’ = x cos x + sin x … (2)
\nFrom (1) sin x = \\(\\frac { y }{ x }\\)
\n\"NCERT
\n\u2234 y = x sinx is a solution of xy’ = x\\(\\sqrt{x^{2}-y^{2}}\\) + y<\/p>\n

Question 7.
\nAy = logy + C : y’ = \\(\\frac{y^{2}}{1-x y}(x y \\neq 1)\\)
\nSolution:
\nxy = logy + C
\nDifferentiating w.r.t. x, we get
\nxy’ + y = \\(\\frac { 1 }{ y }\\) y’
\ni.e., xyy’ + y\u00b2 = y’
\ny’ – xyy = y\u00b2
\ny'(1 – xy) = y\u00b2
\ni.e., y’ = \\(\\frac{y^{2}}{1-x y}\\)
\nHence xy = logy + C is a solution of y’ = \\(\\frac{y^{2}}{1-x y}\\)<\/p>\n

Question 8.
\ny – cos y = x : (ysiny + cosy + x)y’ = y.
\nSolution:
\ny – cosy = x … (1)
\nDifferentiating (1) w.r.t. x, we get
\ny’ + sin y . y’ = 1
\ni. e., yy’ + yy’sin y = y
\n(x + cos y)y’ + yy’sin y = y
\n[since from (1) y = x + cosy]
\nxy’ + y’cosy+ yy’siny = y
\n(y sin y + cos y+ x)y’ = y
\nHence y – cosy = x is a solution of
\n(y sin y + cos y + x)y’ = y<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nx + y = tan-1<\/sup> y : y\u00b2y’ + y\u00b2 + 1 = 0
\nSolution:
\nx + y = tan-1<\/sup>
\nDifferentiating w.r.t. x, we get,
\n1 + y’ = \\(\\frac{1}{1+y^{2}}\\)
\ni.e., (1 + y\u00b2)(1 + y’) = y ‘
\n1 + y’ + y\u00b2 + y\u00b2y’ = y’
\ni.e., y\u00b2y’ + y\u00b2 + 1 = 0
\nHence x + y = tan-1<\/sup> is a solution of y\u00b2y’ + y\u00b2 + 1 = 0<\/p>\n

Question 10.
\n\\(y=\\sqrt { { a }^{ 2 }-{ x }^{ 2 } } x\\in (-a,a);x+y\\frac { dy }{ dx } =0,(y\\neq 0)\\)
\nSolution:
\ny = \\(\\sqrt{a^{2}-x^{2}}\\) , x \u2208 (- a, a)
\nDifferentiating w.r.t. x, we get
\n\"NCERT
\nHence y = \\(\\sqrt{a^{2}-x^{2}}\\) is a solution of x + y\\(\\frac { dy }{ dx }\\) = 0<\/p>\n

Question 11.
\nThe number of arbitrary constants in the general solution of a differential equation of fourth order are
\na. 0
\nb. 2
\nc. 3
\nd. 4
\nSolution:
\nd. 4
\nThe general solution of a differential equation contains as many arbitrary constants as the order of the differential equation. Since the differential equation is of order 4, its solution contains 4 arbitrary constants.<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nThe number of arbitrary constants in the particular solution of a differential equation of third order are
\na. 3
\nb. 2
\nc. 1
\nd. 0
\nSolution:
\nd. 0
\nThe particular solution is free from the arbitrary constants.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.2 Question 1. y = ex + 1 : y”- y’ = 0 Solution: y = ex + 1 Differentiating …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts. 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