NCERT Solutions for Class 12 Maths<\/a> Chapter 9 Differential Equations Ex 9.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-5\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.5<\/h2>\n <\/p>\n
Question 1. \n(x\u00b2+xy)dy = (x\u00b2+y\u00b2)dx \nSolution: \n <\/p>\n
Question 2. \ny’ = \\(\\frac { x+y }{ x }\\) \nSolution: \nWe have x.\\(\\frac { dy }{ dx }\\) = x + y \nLHS as \\(\\frac { dy }{ dx }\\) and convert RHS as a function of \\(\\frac { y }{ x }\\) a homogeneous differential equation substitute v for \\(\\frac { y }{ x }\\) differential y w.r.t. x put the values of \\(\\frac { dy }{ dx }\\) and \\(\\frac { y }{ x }\\) in (1) variable separble equation of v and x Replace the value of v by \\(\\frac { y }{ x }\\) \n \nwhich the required general solution.<\/p>\n
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Question 3. \n(x – y)dy – (x + y)dx = 0 \nSolution: \n(x – y)dy – (x + y)dx = 0 \ni.e., (x – y)dy = (x + y)dx \n \nis a homogeneous differential equation \n <\/p>\n
Question 4. \n(x\u00b2 – y\u00b2)dx + 2xy dy = 0 \nSolution: \n(x\u00b2 – y\u00b2)dx + 2xy dy = 0 \n2xy dy = (x\u00b2 – y\u00b2)dx \n \nis a homogeneous differential equation \n \nwhich the required general solution.<\/p>\n
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Question 5. \nx\u00b2\\(\\frac { dy }{ dx }\\) = x\u00b2 – 2y\u00b2 + xy \nSolution: \n\\(\\frac { dy }{ dx }\\) = \\(\\frac{x^{2}-2 y^{2}+x y}{x^{2}}\\) \n\\(\\frac { dy }{ dx }\\) = \\(1-2\\left(\\frac{y}{x}\\right)^{2}+\\left(\\frac{y}{x}\\right)\\) … (1) \nis a homogeneous differential equation \n <\/p>\n
Question 6. \n\\(xdy-ydx=\\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx\\) \nSolution: \n <\/p>\n
Question 7. \n\\(\\left\\{ xcos\\left( \\frac { y }{ x } \\right) +ysin\\left( \\frac { y }{ x } \\right) \\right\\} ydx=\\left\\{ ysin\\left( \\frac { y }{ x } \\right) -xcos\\left( \\frac { y }{ x } \\right) \\right\\} xdy\\) \nSolution: \n \nis a function of \\(\\frac { y }{ x }\\) \n\u2234 The given differential equation is homogeneous. \n \nis the general solution of the differential equation.<\/p>\n
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Question 8. \n\\(x\\frac { dy }{ dx } -y+xsin\\left( \\frac { y }{ x } \\right) =0\\) \nSolution: \n\\(x\\frac { dy }{ dx } -y+xsin\\left( \\frac { y }{ x } \\right) =0\\) \ni.e., x\\(\\frac { dy }{ dx }\\) = y – xsin\\(\\frac { y }{ x }\\) \n\\(\\frac { dy }{ dx }\\) = \\(\\frac { y }{ x }\\) – sin(\\(\\frac { y }{ x }\\)) is a homogeneous differential equation. \nPut \\(\\frac { y }{ x }\\) = v \u21d2 y = vx and \\(\\frac { dy }{ dx }\\) = v + x\\(\\frac { dv }{ dx }\\) \n \nis the general solution.<\/p>\n
Question 9. \n\\(ydx+xlog\\left( \\frac { y }{ x } \\right) dy-2xdy=0\\) \nSolution: \n <\/p>\n
Question 10. \n\\(\\left( { 1+e }^{ \\frac { x }{ y } } \\right) dx+{ e }^{ \\frac { x }{ y } }\\left( 1-\\frac { x }{ y } \\right) dy=0\\) \nSolution: \n \nis a homogeneous differential equation. \nPut v = \\(\\frac { x }{ y }\\), \u2234 x = vy \nDifferentiating w.r.t. y we get \n \n\\(x+y e^{\\frac{x}{y}}=C\\), is the general solution of the differential equation.<\/p>\n
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Question 11. \n(x + y) dy+(x – y)dx = 0,y = 1 when x = 1 \nSolution: \n <\/p>\n
Question 12. \nx\u00b2dy + (xy + y\u00b2)dx = 0, y = 1 when x = 1 \nSolution: \nx\u00b2dy + (xy + y\u00b2)dx = 0 \nx\u00b2dy = – (xy + y\u00b2) dx \n <\/p>\n
Question 13. \n\\(\\left[x \\sin ^{2}\\left(\\frac{y}{x}\\right)-y\\right] d x+x d y=0 ; y=\\frac{\\pi}{2}\\) \nSolution: \n <\/p>\n
Question 14. \n\\(\\frac { dy }{ dx }\\) – \\(\\frac { y }{ x }\\) + cosec\\(\\frac { y }{ x }\\) = 0; y = 0 when x = 1 \nSolution: \n\\(\\frac { dy }{ dx }\\) – \\(\\frac { y }{ x }\\) + cosec\\(\\frac { y }{ x }\\) = 0 is a homogeneous differential equation. \n <\/p>\n
Question 15. \n\\(2xy+{ y }^{ 2 }-{ 2x }^{ 2 }\\frac { dy }{ dx } =0,y=2,when\\quad x=1\\) \nSolution: \n <\/p>\n
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Question 16. \nA homogeneous equation of the form \\(\\frac{d x}{d y}=h\\left(\\frac{x}{y}\\right)\\) can be solved by making the substitution, \n(a) y = vx \n(b) v = yx \n(c) x = vy \n(d) x = v \nSolution: \n\\(\\frac{d x}{d y}=h\\left(\\frac{x}{y}\\right)\\) is a homogeneous equation. \nHence the substitution is \\(\\frac { x }{ y }\\) = v or x = vy<\/p>\n
Question 17. \nWhich of the following is a homogeneous differential equation? \n(a) (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0 \n(b) \\((x y) d x-\\left(x^{3}+y^{3}\\right) d y=0\\) \n(c) \\(\\left(x^{3}+2 y^{2}\\right) d x+2 x y d y=0\\) \n(d) \\(y^{2} d x+\\left(x^{2}-x y-y^{2}\\right) d y\\) \nSolution: \n(d) \\(y^{2} d x+\\left(x^{2}-x y-y^{2}\\right) d y\\) \n \nHence it is a homogeneous differential equation.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-ex-9-5\/ NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.5 Question 1. (x\u00b2+xy)dy = (x\u00b2+y\u00b2)dx Solution: Question 2. y’ = Solution: We have x. = x + y …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5<\/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 9.5 - 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