{"id":20745,"date":"2020-01-22T14:53:59","date_gmt":"2020-01-22T09:23:59","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20745"},"modified":"2021-07-02T15:27:56","modified_gmt":"2021-07-02T09:57:56","slug":"2nd-puc-maths-question-bank-chapter-9-ex-9-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-9-ex-9-3\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 9 Differential Equations Ex 9.3"},"content":{"rendered":"

Students can Download Maths Chapter 9 Differential Equations Ex 9.3 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 9 Differential Equations Ex 9.3<\/h2>\n

2nd PUC Maths Differential Equations NCERT Text Book Questions and Answers Ex 9.3<\/h3>\n

In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constant a and b.<\/p>\n

Question 1.
\n\\(\\frac{x}{a}+\\frac{y}{b}=1\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\ny2<\/sup> = a (b2<\/sup> – x2<\/sup>)
\nAnswer:
\ny2<\/sup> = a (b2<\/sup> – x2<\/sup>)
\nDifferentiating w.r.t x, we get
\n\\(2 y \\frac{d y}{d x}=a(0-2 x) \\Rightarrow 2 y y_{1}=-2 a x\\)
\n=y y, = -ax ………(1)
\ndifferentiating again
\nyy2<\/sub> + yy1<\/sub> = -a …(2)
\nputting value of (-a) from (2) in (1)
\nyy1<\/sub>= (yy2<\/sub> + y1<\/sub>y1<\/sub>) x
\n\u21d2 y y, = x (y1<\/sub>2<\/sup> + yy2<\/sub>) which is required differential equation.
\nNote:-
\nTry to form differential equation with lowest order by removing all arbitrary constants<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\ny = a e3x<\/sup> + b e-2x
\n<\/sup>Answer:
\ny = a e3x<\/sup> + b e_2x<\/sup> ….(1)
\nDifferentiating w.r.t x , y, = ae3x<\/sup>.3 + be 2x<\/sup> (-2)\u00a0\u00a0\u00a0\u00a0 ….(2)
\nmultiply (1) by 3
\n\u21d2 3y = 3ae3x<\/sup> + 3be_2x
\n<\/sup> \u21d2 3ae3x<\/sup> = 3y -3be-2x<\/sup> …(3)
\nput (3) in (2), we get y, = 3y – 3be-2x<\/sup> – 2 be-2x
\n<\/sup>\u21d2 y – 3y = -5 be-2x<\/sup> …….. (4)
\ndifferentiating again
\n\u21d2 y2<\/sub> – 3y1<\/sub> = -5 be-2x<\/sup> (-2) ….(5)
\nmultiply (4) by 2, we get
\n\u21d2 2y,-6y = -10be 2x<\/sup> ….(6)
\nputting (6) in (5), we get
\n\u21d2 -(2y1<\/sub> – 6y) = y2<\/sub> -3y1<\/sub>
\n\u21d2 y2 <\/sub>– 3y1<\/sub> + 2 (y1<\/sub>\u00a03y) = 0
\n\u21d2 y2<\/sub> -yt<\/sub> – 6y = 0 is the required differential equation.<\/p>\n

Question 4.
\ny = e2x<\/sup> (a + bx)
\nAnswer:
\ny = e2x<\/sup> (a + bx) …(1)
\ndifferentiating w.r.t x, we get
\n\u21d2 y1<\/sub>= e2x<\/sup> (0 + b) + (a + bx) e2x<\/sup> 2
\n\u21d2 y1<\/sub> = e2x<\/sup>xb + (a + bx) e2x<\/sup> 2\u00a0\u00a0\u00a0\u00a0\u00a0 ….(2)
\nputting (1) in (2),
\ny1<\/sub> = be2x<\/sup> + 2y \u21d2 y1<\/sub> – 2y = b e2x<\/sup> ….(3)
\nagain differentiating w.r.t x, we get
\ny2<\/sub> – 2y1<\/sub> = b.e2x<\/sup>.2 ….(4)
\nputting (3) in (4) we get
\ny2<\/sub> – 2y1<\/sub>\u00a0= 2<\/sup>y1<\/sub> – 4y
\n\u21d2 y2<\/sub> – 4y1<\/sub> + 4y = 0 is
\nthe required differential equation.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\ny = ex<\/sup> (a cos x + b sin x)
\nAnswer:
\ny1<\/sub> = ex<\/sup> (a cos x + b sin x) ….(1)
\ndifferentiate w.r.t x
\ny1<\/sub> = ex<\/sup> (a cos x + b sin x) + ex<\/sup> (-a sin x + b cos x) ….(2)
\nputting (1) in (2); we get
\n\u21d2 y1<\/sub> = y + ex<\/sup> (b cos x – a sin x)
\n\u21d2 y1 <\/sub>– y = ex<\/sup> (b cos x – a sin x) ….(3)
\nagain differentiating w.r.t x, we get
\n\u21d2 y2<\/sub> – y1<\/sub> = ex<\/sup> (b cos x – a sin x) +ex<\/sup> (-b sin x – a cos x) ….(5)
\nputting (3) and (1) in (5), we get
\n\u21d2 y2<\/sub> – y1 <\/sub>= (y1<\/sub> – y) +(-y)
\n\u21d2 y2<\/sub> – y1 <\/sub>+ 2y = 0 is the required differential equation.<\/p>\n

Question 6.
\nForm the differential equation of the family of circles touching the y-axis at origin.
\nAnswer:
\nAny circle that touches \u2018y\u2019 axis at origin should have centre on \u2018x\u2019 axis and pass through origin. If centre is (h, 0) then radius is distance between (h, 0) and (0,0), so radius
\n\\(=\\sqrt{(h-0)^{2}+(0-0)^{2}}=h\\)
\nEquation of circle
\n= (x – h)2<\/sup> + (y – 0)2<\/sup> = radius2<\/sup> = h2
\n<\/sup>\u21d2 x2<\/sup> + h2<\/sup> – 2hx + y2<\/sup> = h2
\n<\/sup>\u21d2 x2<\/sup>-2hx + y2<\/sup> = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 …(1)
\ndifferentiating w.r.t x, we get
\n\u21d2 2x + 2y y, – 2h = 0\u00a0 …(2)
\nmultiply (2) with x and subtracting from (1), we get
\n\u21d22x2<\/sup> + 2y y1<\/sub> x – 2hx – x2<\/sup> + 2hx – y2<\/sup> = 0
\n\u21d2 x2<\/sup> – y2<\/sup> + 2y y1<\/sub> x = 0 is the required differential equation.<\/p>\n

Question 7.
\nForm the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
\nAnswer:
\nEquation of parabola required in question is:-
\nx2<\/sup> = 4 ay ………… (1)
\na is parameter differentiating w.r.t x, we get
\n2x = 4ay ……….. (2)
\nputting (1) in (2), we get x2
\n<\/sup>\\(2x=\\frac{x^{2}}{y} y_{1} \\Rightarrow 2y=x y_{1} \\Rightarrow x y_{1}-2 y=0\\)
\nrequired differential equation.<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nForm the differential equation of the family of ellipses having foci on y-axis and centre at origin.
\nAnswer:
\nEquation of ellipse required in question is
\n\\(\\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}=1\\)
\na=semi major axes (along \u2018y\u2019axis)
\nb = semi minor axes differentiating w.r.t x, we get
\n\"2nd<\/p>\n

Question 9.
\nForm the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.
\nAnswer:
\nEquation of hyperbola required in question is:-
\n\"2nd
\n\"2nd<\/p>\n

Question 10.
\nForm the differential equation of the family of circles having centre on y-axis and radius 3 units.
\nAnswer:
\nRequired equation of circle is :-
\n(x – 0)2<\/sup> + (y – k)2<\/sup> = 32<\/sup> ……….(1)
\n(k is any value) – as along \u2018y’ axis, centre will be (0, k)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nWhich of the following differential equations has y = c1 <\/sub>ex<\/sup> + c2<\/sub> e-x<\/sup> as the general solution?
\n\"2nd
\nAnswer:
\n\"2nd
\nbut (1) = (3)
\n\u21d2 y = y2
\n<\/sub>so y2<\/sub> – y = 0
\nso correct answer is (B).<\/p>\n

Question 12.
\nWhich of the following differential equations has y = x as one of its particular solution ?
\n\"2nd
\nAnswer:
\ny = x
\ny1 <\/sub>= 1 y2<\/sub>\u00a0= o
\nthis is satisfied by
\ny2<\/sub> – x2<\/sup> y, + xy = 0
\n0 – x2<\/sup> x 1 + x x x = 0
\nso answer is (C).<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 9 Differential Equations Ex 9.3 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 9 Differential Equations Ex 9.3 2nd PUC Maths Differential Equations …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20745"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=20745"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20745\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20745"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20745"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20745"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}