NCERT Solutions for Class 12 Maths<\/a> Chapter 9 Differential Equations Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-4\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.4<\/h2>\n <\/p>\n
Class 12 Maths Chapter 9 Exercise 9.4\u00a0 Question 1.<\/strong> \n\\(\\frac { dy }{ dx } =\\frac { 1-cosx }{ 1+cosx } \\) \nSolution: \n \ny = 2tan\\(\\frac { x }{ 2 }\\) – x + C, is the required solution.<\/p>\nExercise 9.4class 12 NCERT Solutions Question 2.<\/strong> \n\\(\\frac{d y}{d x}=\\sqrt{4-y^{2}} \\quad(-2<y<2)\\) \nSolution: \n \n\u2234 y = 2sin (x + C), is the general solution.<\/p>\nExercise 9.4 Class 12 NCERT Solutions Question 3.<\/strong> \n\\(\\frac { dy }{ dx } +y=1(y\\neq 1)\\) \nSolution: \n \n1 – y = Ce-x<\/sup> \u21d2 y = 1 – Ce-x<\/sup> \n\u21d2 y = + Ae-x<\/sup>, where A = – C is the general solution.<\/p>\n <\/p>\n
Ex 9.4 Class 12 NCERT Solutions Question 4.<\/strong> \nsec\u00b2 x tany dx+sec\u00b2 y tanx dy = 0 \nSolution: \nsec\u00b2 x tany dx+sec\u00b2 y tanx dy = 0 \nsec\u00b2 x tan x dy = – sec\u00b2 x tan y dx \n \nis the required general solution.<\/p>\nEx 9.4 Class 12 Maths NCERT Solutions Question 5.<\/strong> \n\\(\\left( { e }^{ x }+{ e }^{ -x } \\right) dy-\\left( { e }^{ x }-{ e }^{ -x } \\right) dx=0\\) \nSolution: \n\\(\\left( { e }^{ x }+{ e }^{ -x } \\right) dy-\\left( { e }^{ x }-{ e }^{ -x } \\right)\\)dx = 0 \n(ex<\/sup> + e-x<\/sup>)dy = (ex<\/sup> – e-x<\/sup>)dx \ni.e., dy = \\(\\left(\\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\\right)\\)dx \nIntegrating both sides we get, \n\\(\\int d y=\\int \\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x\\) \ni.e., y = log|ex<\/sup> + e-x<\/sup>| + C is the general solution.<\/p>\nEx 9.4 Class 12 Learn Cbse NCERT Solutions Question 6.<\/strong> \n\\(\\frac { dy }{ dx } =\\left( { 1+x }^{ 2 } \\right) \\left( { 1+y }^{ 2 } \\right) \\) \nSolution: \n\\(\\frac { dy }{ { 1+y }^{ 2 } } =\\left( { 1+x }^{ 2 } \\right) dx \\) \nIntegrating on both side we get \n\\({ tan }^{ -1 }y={ x+\\frac { 1 }{ 3 } }x^{ 3 }+c \\) \nwhich is the required general solution.<\/p>\nQuestion 7. \ny logy dx – x dy = 0 \nSolution: \ny logy dx – x dy = 0 \ny logy dx = x dy \ni.e., \\(\\frac{d x}{x}=\\frac{d y}{y \\log y}\\) \nIntegrating both sides, we get \n \ny = eCx<\/sup> is the general solution.<\/p>\nQuestion 8. \n\\({ x }^{ 5 }\\frac { dy }{ dx } =-{ y }^{ 5 }\\) \nSolution: \n <\/p>\n
Question 9. \n\\(\\frac { dy }{ dx } ={ sin }^{ -1 }x\\) \nSolution: \n\\(\\frac { dy }{ dx } ={ sin }^{ -1 }x\\) \u2234 dy = sin-1<\/sup>x dx \nIntegrating both sides, we get \n \n\u2234 y = x sin-1<\/sup>x + \\(\\sqrt{1-x^{2}}\\) + C, is the general solution.<\/p>\nQuestion 10. \n\\({ e }^{ x }tany\\quad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0\\) \nSolution: \n <\/p>\n
Question 11. \n\\(\\left(x^{3}+x^{2}+x+1\\right) \\frac{d y}{d x}=2 x^{2}+x\\) y = 1 when x = 0 \nSolution: \n \nPut x = – 1, we get 2A = 1 \u2234 A = \\(\\frac { 1 }{ 2 }\\) \nEquating coefficients of x\u00b2, we get \nA + B = 2 \n\u2234 B = \\(\\frac { 3 }{ 2 }\\) \nPut x = 0, we get A + C = 0 \u2234 C = – \\(\\frac { 1 }{ 2 }\\) \n <\/p>\n
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Question 12. \n\\(x\\left(x^{2}-1\\right) \\frac{d y}{d x}=1\\)y = 0 when x = 2 \nSolution: \n \n1 = A(x + 1)(x – 1) + Bx(x – 1) + Cx(x + 1) \nWhen x = 0, A = – 1 \nWhen x = 1, C = \\(\\frac { 1 }{ 2 }\\) \nWhen x = – 1, B = \\(\\frac { 1 }{ 2 }\\) \n <\/p>\n
Question 13. \ncos(\\(\\frac { dy }{ dx }\\)) = a(a \u2208 R); y = 1, when x = 0. \nSolution: \ncos(\\(\\frac { dy }{ dx }\\)) = a \n\u2234 \\(\\frac { dy }{ dx }\\) = cos-1<\/sup>(a) \ni.e., dy = cos-1<\/sup>(a)dx \nIntegrating both sides we get \n\u222bdy = \u222bcos-1<\/sup>(a)dx \ni.e., y = cos-1<\/sup>(a) x + C …. (1) \nWhen x = 0, y = 1 \n\u2234 1 = 0 + C \ni.e., C = 1 \nSubstituting the value of C in (1), we get \ny = x cos-1<\/sup>(a) + 1 \ny – 1 = x cos-1<\/sup>(a) \ncos(\\(\\frac { y-1 }{ x }\\)) = a is the required particular solution.<\/p>\nQuestion 14. \n\\(\\frac { dy }{ dx }\\) = y tan x; y = 1, when x = 0. \nSolution: \n\\(\\frac { dy }{ dx }\\) \u2234\\(\\frac { dy }{ y }\\) = tanx dx \nIntegrating, we get \u222b\\(\\frac { dy }{ dx }\\) = \u222btan x dx \n\u21d2 log|y| = log|secx| + log|C| \nWhen x = 0, y = 1 \n\u21d2 log 1 = log sec0 + C \u21d2 0 = log1 + C \n\u21d2 C = 0 \n\u2234 logy = log sec x \n\u21d2 y = sec x is the required particular solution.<\/p>\n
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Question 15. \nFind the equation of the curve passing through the point (0,0) and whose differential equation y’ = ex<\/sup> sin x. \nSolution: \ny’ = ex<\/sup> sin x \ni.e., \\(\\frac { dy }{ dx }\\) = ex<\/sup> sin x \ndy = ex<\/sup> sinx dx \nIntegrating both sides, we get \n\u222bdy = \u222bex<\/sup> sin x dx \ni.e., y = \\(\\frac{e^{x}}{2}[\\sin x-\\cos x]+C\\) … (1) \n(1) passess through the point (0, 0) \nwe get 0 = \\(\\frac { 1 }{ 2 }\\) [0 – 1] + C \n\u2234 C = \\(\\frac { 1 }{ 2 }\\) \n\u2234 (1) \u2192 y = \\(\\frac{e^{x}}{2}[\\sin x-\\cos x]\\) + \\(\\frac { 1 }{ 2 }\\) \ni.e., 2y = ex<\/sup>[sinx – cosx] + 1 \n2y – 1 = ex<\/sup> [sinx – cosx], is the equation of the curve.<\/p>\nQuestion 16. \nFor the differential equation xy \\(\\frac { dy }{ dx }\\) = (x + 2) (y + 2), find the solution curve passing through the point (1, – 1) \nSolution: \n \ny – 2 log |y + 2| = x + 2 log |x| + C \ny – x = log (y + 2)\u00b2 + logx\u00b2 + C \ny – x = log (x(y + 2))\u00b2 + C … (1) \n(1) passess through (1, – 1), we get \n– 1 – 1 = log 1 + C \n– 2 = C \u21d2 C = – 2 \n(1) \u2192 y – x = log x(x(y + 2))\u00b2 – 2 \ni.e., y – x + 2 = log(x(y + 2))\u00b2, is the equation of the curve.<\/p>\n
Question 17. \nFind the equation of a curve passing through the point (0, -2) given that at any point (pc, y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point \nSolution: \nAccording to the question \\(y\\frac { dy }{ dx } =x\\) \n\\(\\Rightarrow \\int { ydy } =\\int { xdx } \\Rightarrow \\frac { { y }^{ 2 } }{ 2 } =\\frac { { x }^{ 2 } }{ 2 } +c\\) \n0, – 2) lies on it.c = 2 \n\u2234 Equation of the curve is : x\u00b2 – y\u00b2 + 4 = 0.<\/p>\n
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Question 18. \nAt any point (x, y) of a curve the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3) find the equation of the curve given that it passes through (- 2,1). \nSolution: \nThe slope of the line points (x, y) and (- 4, 3) is \\(\\frac { y+3 }{ x+4 }\\) \nAlso the slope of the tangent at any point on the curve is \\(\\frac { dy }{ dx }\\) \n \nSince it passes through the point (-2, 1), we get log(1 + 3) = log (-2 + 4)\u00b2 + C \nlog 4 = log 4 + C \n\u21d2 C = 0 \n\u2234 log|y + 3| = log(x + 4)\u00b2 \n\u21d2 y + 3 = (x + 4)\u00b2 is the equation of the curve.<\/p>\n
Question 19. \nThe volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and offer 3 seconds it is 6 units. Find the radius of balloon after t seconds. \nSolution: \n <\/p>\n
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Question 20. \nIn a bank principal increases at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years \nSolution: \nLet Pt<\/sub> be the principal after t years. \nThe initial principal P0<\/sub> = \u20b9 100 \nRate of interest = r %, time t = 10 years \nP10<\/sub> = 200 \nGiven that \\(\\frac{d P}{d t}=\\frac{r}{100} \\mathrm{P} \\Rightarrow \\frac{d P}{P}=\\frac{r}{100} d t\\) \nIntegrating both sides, we get \n <\/p>\nQuestion 21. \nIn a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years \nSolution: \nLet Pt<\/sub> be the principal after t years. \nThe initial principal P0<\/sub> = 1000 \nRate of interest = 5 %, time t = 10 years \nGiven that \\(\\frac{d P}{d t}=\\frac{5}{100} \\mathrm{P} \\Rightarrow \\frac{d P}{P}=\\frac{1}{20} d t\\) \nIntegrating both sides, we get \n <\/p>\nQuestion 22. \nIn a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present \nSolution: \nLet N1<\/sub> be the count of bacteria at any time t. Initial count of bacteria be N0<\/sub>. \nGiven that \\(\\frac { dN }{ dt }\\) \u221d N \n\u21d2 \\(\\frac { dN }{ dt }\\) = kN \n\\(\\frac { dN }{ N }\\) = k dt \nIntegrating both sides, we get \u222b\\(\\frac { dN }{ N }\\) = \u222bk dt \n\u21d2 log N = kt + C … (1) \nWhen t = 0, N = N0<\/sub> = 100000 \ni.e., C = log 100000 \n(1) \u21d2 logN = kt + log 100000 \nlog N – log 100000 = kt \nlog(\\(\\frac { N }{ 100000 }\\)) = kt … (2) \nWhen t = 2, N = N2<\/sub> \n <\/p>\n <\/p>\n
Question 23. \nThe general solution of a differential equation \\(\\frac{d y}{d x}=e^{x+y} \\text { is }\\) \n(a) \\({ e }^{ x }+{ e }^{ -y }=c \\) \n(b) \\({ e }^{ x }+{ e }^{ y }=c \\) \n(c) \\({ e }^{ -x }+{ e }^{ y }=c \\) \n(d) \\({ e }^{ -x }+{ e }^{ -y }=c \\) \nSolution: \n(a) \\(\\frac { dy }{ dx } ={ e }^{ x }.{ e }^{ y }\\Rightarrow \\int { { e }^{ -y }dy } =\\int { { e }^{ x }dx } \\) \n\\(\\Rightarrow { e }^{ -y }={ e }^{ x }+k\\Rightarrow { e }^{ x }+{ e }^{ -y }=C \\) is the general solution.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-4\/ NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.4 Class 12 Maths Chapter 9 Exercise 9.4\u00a0 Question 1. Solution: y = 2tan – x + C, is …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4<\/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 9.4 - 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