-tanx<\/sup> is the general solution.<\/p>\nQuestion 6. \n\\(x \\frac{d y}{d x}+2 y=x^{2} \\log x\\) \nSolution: \n \ny = \\(\\frac{x^{2}}{16}(4 \\log x-1)+C x^{-2}\\) is the general solution.<\/p>\n
Question 7. \n\\(xlogx\\frac { dy }{ dx } + y=\\frac { 2 }{ x } logx\\) \nSolution: \n <\/p>\n
Question 8. \n(1 + x\u00b2)dy + 2xy dx = cotx dx(x \u2260 0) \nSolution: \n <\/p>\n
Question 9. \n\\(x\\frac { dy }{ dx } +y-x+xy\\quad cotx=0(x\\neq 0)\\) \nSolution: \n <\/p>\n
<\/p>\n
Question 10. \n\\((x+y)\\frac { dy }{ dx }\\) = 1 \nSolution: \n <\/p>\n
Question 11. \ny dx + (x – y\u00b2)dy = 0 \nSolution: \n \nis the general solution.<\/p>\n
Question 12. \n\\(\\left( { x+3y }^{ 2 } \\right) \\frac { dy }{ dx } =y(y>0)\\) \nSolution: \n <\/p>\n
Question 13. \n\\(\\frac { dy }{ dx } +2ytanx=sinx,y=0\\quad when\\quad x=\\frac { \\pi }{ 3 } \\) \nSolution: \n\\(\\frac { dy }{ dx }\\) + 2y tanx = sin x is a linear differential equation. \n\u2234 p = 2 tanx, Q = sin x \n <\/p>\n
Question 14. \n\\(\\left( 1+{ x }^{ 2 } \\right) \\frac { dy }{ dx } +2xy=\\frac { 1 }{ 1+{ x }^{ 2 } } ,y=0\\quad when\\quad x=1\\) \nSolution: \n <\/p>\n
<\/p>\n
Question 15. \n\\(\\frac { dy }{ dx } -3ycotx=sin2x,y=2\\quad when\\quad x=\\frac { \\pi }{ 2 } \\) \nSolution: \n\\(\\frac { dy }{ dx }\\) – 3y cot x = sin 2x is a linear differential equation. \n\u2234 p = – 3 cot x, Q = sin 2x \n \ny cosec\u00b3x = – 2 cosec + C … (1) \nWhen x = \\(\\frac { \u03c0 }{ 2 }\\), y = 2 \n(1) \u2192 2 x 1 = – 2 + C \u2234 C = 4 \nHence the particular solution is \ny cosec\u00b3x = – 2 cosec + 4 \ny = \\(\\frac{-2}{cosec^{2} x}\\) + \\(\\frac{4}{cosec^{3} x}\\) \ny = 4 sin\u00b3x – 2 sin\u00b2x<\/p>\n
Question 16. \nFind the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point \nSolution: \n\\(\\frac { dy }{ dx }\\) = x + y \n\\(\\frac { dy }{ dx }\\) – y = x, is a linear differential equation \n\u2234 P = – 1, Q = x \n \nThe curve is passing through the origin \n\u2234 x = 0, y = 0 \n(1) \u2192 C = 1 \ni.e., x + y + 1 = ex<\/sup> is the required equation of the curve.<\/p>\n <\/p>\n
Question 17. \nFind the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5 \nSolution: \nx + y = \\(\\frac { dy }{ dx }\\) \n\\(\\frac { dy }{ dx }\\) – y = x – 5, is a linear differential equation \n\u2234 P = – 1, Q = x – 5 \n\u222bP dx = \u222b-1 dx = – x \nI.F = e\u222bpdx<\/sup> = e-x<\/sup> \n\u2234 Solution of the differential equation is \n \nGiven that the cure pass through the point (0, 2) \n(1) \u2192 2 = 4 + C \u2234 C = – 2 \ny = 4 – x – 2ex<\/sup> is the required equation of the curve.<\/p>\nQuestion 18. \nThe integrating factor of the differential equation \\(x\\frac { dy }{ dx } -y={ 2x }^{ 2 }\\) \n(a) \\({ e }^{ -x }\\) \n(b) \\({ e }^{ -y }\\) \n(c) \\(\\frac { 1 }{ x } \\) \n(d) x \nSolution: \n <\/p>\n
Question 19. \nThe integrating factor of the differential equation \\(\\left( { 1-y }^{ 2 } \\right) \\frac { dx }{ dy } +yx=ay\\)(-1<y<1) is \n(a) \\(\\frac { 1 }{ { y }^{ 2 }-1 } \\) \n(b) \\(\\frac { 1 }{ \\sqrt { { y }^{ 2 }-1 } } \\) \n(c) \\(\\frac { 1 }{ 1-{ y }^{ 2 } } \\) \n(d) \\(\\frac { 1 }{ \\sqrt { { 1-y }^{ 2 } } } \\) \nSolution: \n(d) \\(\\frac { 1 }{ \\sqrt { { 1-y }^{ 2 } } } \\) \n <\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-6\/ NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.6 Question 1. + 2y = sin x Solution: + 2y = sin x is a linear differential equation. …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.6<\/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 9.6 - 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