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. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3<\/h2>\n <\/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
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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>\nQuestion 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>\nQuestion 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 <\/p>\n
Question 6. \nForm the differential equation of the family of circles touching the y axis at origin \nSolution: \n \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>\nQuestion 7. \nForm the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis. \nSolution: \n \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 <\/p>\n
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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 \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 \n\u21d2 (x\u00b2 – 9)(y’)\u00b2 + x\u00b2 = 0 is the required differential equation.<\/p>\n
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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 <\/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 <\/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. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3\/ 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. …<\/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":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n