{"id":32233,"date":"2021-08-18T14:46:46","date_gmt":"2021-08-18T09:16:46","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=32233"},"modified":"2021-08-18T14:49:23","modified_gmt":"2021-08-18T09:19:23","slug":"ncert-solutions-for-class-12-maths-chapter-9-ex-9-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 9 Differential Equations Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\n\\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
\nSolution:
\n\\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
\ni.e., bx + ay = ab … (1)
\nDifferentiating (1) w.r.t. x, we get
\nb + ay’ = 0
\ni.e., ay\u2019 = – b
\ny = \\(\\frac { – b }{ a }\\)
\nDifferentiating again w.r.t. x, we get y\u2019 = 0
\nwhich is the required differential equation.<\/p>\n

Question 2.
\ny\u00b2 = a(b\u00b2 – x\u00b2)
\nSolution:
\ny\u00b2 = a(b\u00b2 – x\u00b2)
\ni.e., y\u00b2 = ab\u00b2 – x\u00b2a
\nDifferentiating w.r.t. x, we get 2yy’ = – 2ax
\ni.e., yy’ = – ax
\n\\(\\frac { yy’ }{ x }\\) = – a
\nDifferentiating again w.r.t. x, we get
\n\\(\\frac{x\\left[y y^{\\prime \\prime}+\\left(y^{\\prime}\\right)^{2}\\right]-y y^{\\prime}}{x^{2}}\\) = 0
\ni.e., x yy\u201d + x(y’)\u00b2 – yy’ = 0
\nwhich is the required differential equation.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\ny = ae3x<\/sup> + be-2x<\/sup>
\nSolution:
\nGiven that
\ny = ae3x<\/sup> + be-2x<\/sup> … (i)
\nDifferentiating w.r.t. x two times, we get
\ny\u2019 = 3ae3x<\/sup> – 2be-2x<\/sup>
\ny” = 9ae3x<\/sup> + 4be-2x<\/sup>
\ni.e., y” = 6ae3x<\/sup> + 3ae3x<\/sup> + 6be-2x<\/sup> – 2be-2x<\/sup>
\n= 3ae3x<\/sup> – 2be-2x<\/sup> + 6(ae3x<\/sup> + be-2x<\/sup>)
\n= y\u2019 + 6y
\ni.e., y” – y\u2019 – 6y = 0 is the required differential equation.<\/p>\n

Question 4.
\ny = e2x<\/sup> (a + bx)
\nSolution:
\ny = e2x<\/sup> (a + bx)
\ny’ = be2x<\/sup> + (a + bx)e2x<\/sup> x 2
\ny’ = be2x<\/sup> + 2y
\nDifferentiating again w.r.t. x, we get
\ny” = 2be2x<\/sup> + 2y’
\ni.e., y” = 2(y’ – 2y) + 2y’ [\u2235 y’ – 2y = be2x<\/sup>]
\ny” = 4y’ – 4y
\ny” – 4y’ + 4y = 0 is the required differential equation.<\/p>\n

Question 5.
\ny = ex<\/sup>(a cosx + b sinx)
\nSolution:
\nThe curve y = ex<\/sup>(a cosx+b sinx) …(i)
\nDifferentiating (1) w.r.t. x, we get
\ny’ = ex<\/sup>(- a sinx + b cosx) + ex<\/sup>(a cosx + b sinx)
\ny’ = ex<\/sup>(b cos x – a sin y) + y
\ni.e., y’ – y = ex<\/sup>(b cosx – a sinx) … (2)
\nDifferentiating (2) w.r.t. x, we get
\ny”- y’ = – ex<\/sup>(b sinx + a cosx) + ex<\/sup>(b cosx – a sinx)
\ny” – y’ = – y + y’- y [from (1) and (2)]
\ny” – 2y’ + 2y = 0 is the required differential equation.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nForm the differential equation of the family of circles touching the y axis at origin
\nSolution:
\n\"NCERT
\nCircles touching y axis at the orgin will have its centre on x-axis and will pass through the origin.
\nSo the centre of circle will be (a, 0) and radius a.
\n\u2234 Equation of the circle is (x – a)\u00b2 + (y – 0)\u00b2 = a\u00b2
\ni.e., x\u00b2 + y\u00b2 – 2ax = 0
\ni.e., \\(\\frac{x^{2}+y^{2}}{x}\\) = 2a
\n\\(\\frac{x\\left(2 x+2 y y_{1}\\right)-\\left(x^{2}+y^{2}\\right)}{x^{2}}\\) = 0
\nDifferentiating w.r.t. x, we get
\n2x\u00b2 + 2xyy1<\/sub> – x\u00b2 – y\u00b2 = 0
\nx\u00b2 – y\u00b2 + 2xyy1<\/sub> = 0 or y\u00b2 – x\u00b2 – 2xyy1<\/sub> = 0<\/p>\n

Question 7.
\nForm the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
\nSolution:
\n\"NCERT
\nConsider the family of parabolas having focus (0, a) at the positive y-axis, where a is an arbitrary constant.
\n\u2234 The equation of family of parabolas is x\u00b2 = 4ay … (1)
\nDifferentiating both sides w.r.t. x, we get
\n2x = 4 ay’
\ni.e., 4a = \\(\\frac { 2x }{ y’ }\\) … (2)
\nSubstituting (2) in (1) we get x\u00b2 = (\\(\\frac { 2x }{ y’ }\\))y
\nx\u00b2y’ – 2xy = 0
\ni.e., xy’ – 2y = 0 is the required differential equation.<\/p>\n

Question 8.
\nForm the differential equation of family of ellipses having foci on y-axis and centre at origin.
\nSolution:
\nThe equation of family ellipses having foci at y- axis is
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nForm the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
\nSolution:
\nThe equation of family of hyperbolas having foci on x axis is \\(\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}\\) = 1, a and b are the parameters.
\n\"NCERT
\ni.e., xyy” + x(y’)\u00b2 – yy’ = 0 is the required differential equation.<\/p>\n

Question 10.
\nForm the differential equation of the family of circles having centre on y-axis and radius 3 units
\nSolution:
\nConsider a circle of radius 3 unit and centre on (0, a). The equation of the circle is
\n(x – 0)\u00b2 + (y – a)\u00b2 = 3\u00b2
\nx\u00b2 + (y – a)\u00b2 = 9 … (1)
\nDifferentiating both sides w.r.t. x, we get
\n2x + 2(y – a)y’ = 0
\nx + yy’ – ay’ = 0
\nay’ = x + yy’
\n\"NCERT
\n\u21d2 (x\u00b2 – 9)(y’)\u00b2 + x\u00b2 = 0 is the required differential equation.<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nWhich of the following differential equation has y = \\({ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }\\) as the general solution?
\n(a) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\\)
\n(b) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\\)
\n(c) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +1=0\\)
\n(d) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -1=0\\)
\nSolution:
\n(b) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\\)
\n\"NCERT<\/p>\n

Question 12.
\nWhich of the following differential equations has y = x as one of its particular solution ?
\n(a) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\\frac { dy }{ dx } +xy=x\\)
\n(b) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\\frac { dy }{ dx } +xy=x\\)
\n(c) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\\frac { dy }{ dx } +xy=0\\)
\n(d) \\(\\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\\frac { dy }{ dx } +xy=0\\)
\nSolution:
\n(c) y = x
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3 Question 1. = 1 Solution: = 1 i.e., bx + ay = ab … (1) Differentiating (1) w.r.t. x, …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3<\/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 9.3 - 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-9-3\/\" \/>\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 9.3 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3 Question 1. = 1 Solution: = 1 i.e., bx + ay = ab … (1) Differentiating (1) w.r.t. x, … NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 Read More »","og_url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3\/","og_site_name":"NCERT Solutions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2021-08-18T09:16:46+00:00","article_modified_time":"2021-08-18T09:19:23+00:00","og_image":[{"url":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"7 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/ncertsolutions.guru\/#organization","name":"NCERT Solutions","url":"https:\/\/ncertsolutions.guru\/","sameAs":["https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","https:\/\/www.instagram.com\/ncertsolutionsguru\/","https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/","https:\/\/in.pinterest.com\/ncertsolutionsguru\/","https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training","https:\/\/twitter.com\/ncertsolguru"],"logo":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#logo","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},"image":{"@id":"https:\/\/ncertsolutions.guru\/#logo"}},{"@type":"WebSite","@id":"https:\/\/ncertsolutions.guru\/#website","url":"https:\/\/ncertsolutions.guru\/","name":"NCERT Solutions","description":"NCERT Solutions for Class 1 to 12","publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/ncertsolutions.guru\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","width":170,"height":17,"caption":"NCERT Solutions"},{"@type":"WebPage","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3\/#webpage","url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3\/","name":"NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 - 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