{"id":12223,"date":"2024-03-10T14:32:35","date_gmt":"2024-03-10T09:02:35","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=12223"},"modified":"2024-03-13T17:26:37","modified_gmt":"2024-03-13T11:56:37","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-8-ex-8a","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-8-ex-8a\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions<\/a> Chapter 8 Differential Equations Ex 8(a) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)<\/h2>\n

I.
\nQuestion 1.
\nFind the order of the differential equation obtained by eliminating the arbitrary constants b and c from xy = cex<\/sup>+ be-x<\/sup> + x2<\/sup>.
\nSolution:
\nGiven equation is xy = cex<\/sup> + be-x<\/sup> + x2<\/sup> ………….(1)
\nDifferentiating (1) w.r.t x, we get
\nxy1<\/sub> + y = cex<\/sup> – be-x<\/sup> + x2<\/sup>
\nAgain differentiating w.r.t x, we get
\nxy2<\/sub> + y1<\/sub> + y1<\/sub> = cex<\/sup> + be-x<\/sup> + 2
\n= (xy – x2<\/sup>) + 2
\n\u2234 xy2<\/sub> + 2y1<\/sub> – (xy – x2<\/sup>) – 2 = 0 ………………(2)
\nArbitrary constants a and b are eliminated in the differential equation (2).
\nThe order of the differential equation (2) is 2.<\/p>\n

Question 2.
\nFind the order of the differential equation of the family of all circles with their cen\u00actres at the origin.
\nSolution:
\nThe equation of circle with centre 0 is given by x2<\/sup> + y2<\/sup> = a2<\/sup> where a is any constant.
\nDifferentiating w.r.t x we get 2x + 2yy1<\/sub> = 0
\n\u21d2 x + yy1<\/sub> = 0
\nWhich is the required differential equation of family of all circles with their centres at origin.
\nThe order of the above differential equation is 1.<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nForm the differential equations of the following family of curves where parameters are given in brackets.
\n(i) y = c (x – c)2<\/sup>; (c)
\nGiven y = c(x – c)2<\/sup> ……………..(1)
\nDifferentiating w.r.t ‘x’ we have
\ny1<\/sub> = 2c(x – c) ………….(2)<\/p>\n

\"TS<\/p>\n

\u2234 y . y1<\/sub>3<\/sup> = (xy1<\/sub> – 2y) 4y2<\/sup>
\n\u21d2 y1<\/sub>3<\/sup> = (xy1<\/sub> – 2y) 4y
\n\u21d2 y1<\/sub>3<\/sup> = 4xyy1<\/sub> – 8y2<\/sup>
\n\u21d2 y1<\/sub>3<\/sup> – 4xyy1<\/sub> + 8y2<\/sup> = 0
\n\u21d2 \\(\\left(\\frac{d y}{d x}\\right)^3\\) – 4xy \\(\\frac{d y}{d x}\\) + 8y2<\/sup> = 0
\nThis the differential equation in which c is eliminated.<\/p>\n

ii) xy = aex<\/sup> + be-x<\/sup> ; (a, b)
\nSolution:
\nGiven xy = aex<\/sup> + be-x<\/sup> and ………….(1)
\nDifferentiating (1) w.r.t x
\nxy1<\/sub> + y = aex<\/sup> – be-x<\/sup>
\nAgain differentiating w.r.t x,
\nxy2<\/sub> + y1<\/sub> + y1<\/sub> = aex<\/sub> + be-x<\/sub> = xy
\n= xy2<\/sub> + 2y1<\/sub> – xy = 0
\nwhich is the required equation obtained on the elimination of a and b.<\/p>\n

\"TS<\/p>\n

(iii) y = (a + bx) ekx<\/sup>; (a, b)
\nSolution:
\nGiven y = (a + bx) ekx<\/sup> …………(1)
\nand Differentiating (1) w.r.t x, we get
\ny1<\/sub> = (a + bx) kekx<\/sup> + ekx<\/sup> . b
\n= ky + ekx<\/sup> . b
\n\u2234 y1<\/sub> – ky = bekx<\/sup> ………….(2)
\nAgain differentiating w.r.t x,
\ny2<\/sub> – ky1<\/sub> = kbekx<\/sup>
\n= k(y1<\/sub> – ky)
\n\u21d2 y2<\/sub> – 2ky1<\/sub> + k2<\/sup>y = 0
\n\u21d2 \\(\\frac{d^2 y}{d x^2}-2 \\mathrm{k} \\frac{d y}{d x}\\) + k2<\/sup>y = 0
\nis the required equation obtained on the elimination of a, b.<\/p>\n

v) y = a cos (nx + b); (a, b)
\nSolution:
\nGiven equation is y = a cos (nx + b)
\n\u2234 y1<\/sub> = – an sin (nx + b)
\n= – an2<\/sup> cos (nx + b)
\n= – n2<\/sup>y
\n\u2234 y1<\/sub> + n2<\/sup>y = 0
\n\u21d2 \\(\\frac{d^2 \\mathrm{y}}{d x^2}\\) + n2<\/sup>y = 0
\nis the required differential equation obtained on elimination of a and b.<\/p>\n

\"TS<\/p>\n

Question 2.
\nObtain the differential equation which corresponds to each of the following family of curves.
\n(i) The rectangular hyperbolas which have the coordinate axes as asymptotes.
\nSolution:
\nEquation of rectangular hyperbolas which have the coordinate axes as asymptotes is
\nxy = c2<\/sup>.
\nDifferentiating w.r.t x,
\nxy1<\/sub> + y = 0
\n\u21d2 x\\(\\frac{d y}{d x}\\) + y = 0 is the required equation.<\/p>\n

(ii) The ellipses with centres at the origin and having coordinate axes as axes.
\nSolution:
\nEquation of ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) = 1
\nDifferentiating w.r.t x\u2019 we get,<\/p>\n

\"TS<\/p>\n

\u21d2 \\(\\frac{2}{\\mathrm{~b}^2}\\) [yy1<\/sub> – xyy2<\/sub> – xy1<\/sub>2<\/sup>] = 0
\n\u21d2 yy1<\/sub> – xyy2<\/sub> – xy1<\/sub>2<\/sup> = 0
\n\u21d2 xyy2<\/sub> + xy1<\/sub> – yy1<\/sub> = 0
\n\u21d2 \\(x y \\frac{d^2 y}{d x^2}+x\\left(\\frac{d y}{d x}\\right)^2-y \\frac{d y}{d x}=0\\) is the required differential equation.<\/p>\n

\"TS<\/p>\n

III.
\nQuestion 1.
\nForm the differential equations of the following family of curves whose parameters are given in brackets.
\n(i) y = ae3x<\/sup> + be4x<\/sup>; (a, b)
\nSolution:
\nGiven y = ae3x<\/sup> + be4x<\/sup>
\nDifferentiating w.r.t. \u2018X\u2019
\ny1<\/sub> = 3ae3x<\/sup> + 4be4x<\/sup>
\n\u21d2 y1<\/sub> – 3ae3x<\/sup> = 4be4x<\/sup>
\n= 4 [y – ae3x<\/sup>]
\n\u21d2 y1<\/sub> – 4y = ae3x<\/sup> ………….(1)
\nAgain differentiating w.r.t. x,
\ny2<\/sub> – 4y1<\/sub> = – 3ae3x<\/sup>
\n\u21d2 y2<\/sub> – 4y1<\/sub> = 3 (y1<\/sub> – 4y)
\n\u21d2 y2<\/sub> – 7y1<\/sub> + 12y = 0 is the required differential equation.<\/p>\n

(ii) y = ax2<\/sup> + bx, (a, b)
\nSolution:
\nGiven equation is
\ny = ax2<\/sup> + bx …………..(1)
\nand dill erentiating w.r.t. x
\ny1<\/sub> = 2ax + b …………(2)
\nAgain differentiating w.r.t. x,
\ny2<\/sub> = 2a
\n\u21d2 x2<\/sup>y2<\/sub> = 2ax2<\/sup> …………..(3)
\nAlso from (2)
\n– 2xy1<\/sub> = – 4x2<\/sup>a – 2bx …………..(4)
\nFrom (1)
\n2y = 2ax2<\/sup> + 2bx …………(5)
\nAdding (3), (4), (5) we get
\nx2<\/sup>y2<\/sub> – 2xy1<\/sub> + 2y = 2ax2<\/sup> – 4ax2<\/sup> – 2bx + 2ax2<\/sup> + 2bx = 0
\n\u2234 \\(x^2 \\frac{d^2 y}{d x^2}-2 x \\frac{d y}{d x}+2 y=0\\) is the required differential equation in which a, b are eliminated.<\/p>\n

\"TS<\/p>\n

(iii) ax2<\/sup> + by2<\/sup> = 1; (a, b)
\nSolution:
\nGiven equation of the curve is
\nax2<\/sup> + by2<\/sup> = 1 ……………(1)
\nDifferentiating (1) w.r.t. \u2018x we get
\n2ax + 2by \\(\\frac{d y}{d x}\\) = 0 and
\nby2<\/sup> = 1 – ax2<\/sup> from (1)
\n\u21d2 2ax + 2byy1<\/sub> = 0 ……………(2)
\n\u21d2 b(2yy1<\/sub>) = – 2ax …………….(3)
\nFrom (3) + (2) we get<\/p>\n

\"TS<\/p>\n

\u21d2 – xay = y1<\/sub> (1 – ax2<\/sup>)
\n\u21d2 – axy = y1<\/sub> – ax2<\/sup>y1<\/sub>
\n\u21d2 y1<\/sub> = ax (xy1<\/sub> – y)
\n\u21d2 a = \\(\\frac{y_1}{x\\left(x y_1-y\\right)}\\)
\nDifferentiating w.r.t x,
\n0 = \\(\\frac{d}{d x}\\left[\\frac{y_1}{x\\left(x y_1-y\\right)}\\right]\\)
\n= \\(\\frac{y_2\\left(x^2 y_1-x y\\right)-y_1\\left(\\frac{d}{d x}\\left(x^2 y_1-x y\\right)\\right)}{x^2\\left(x y_1-y\\right)^2}\\)
\n\u21d2 (x2<\/sup>y1<\/sub> – xy)y2<\/sub> – y1<\/sub>(x2<\/sup>y2<\/sub> + 2xy1<\/sub> – xy1<\/sub> – y) = 0
\n\u21d2 x2<\/sup>y1<\/sub>y2<\/sub> – xyy1<\/sub> – x2<\/sup>y1<\/sub>y2<\/sub> – 2xy1<\/sub>2<\/sup> + xy1<\/sub>2<\/sup> + yy1<\/sub> = 0
\n\u21d2 xyy2<\/sub> + xy1<\/sub>2<\/sup> – yy1<\/sub> = 0
\n\u21d2 \\(x y \\frac{d^2 y}{d x^2}+x\\left(\\frac{d y}{d x}\\right)^2-y\\left(\\frac{d y}{d x}\\right)\\) = 0 is the required differential equation obtained on elimination of constants a and b.<\/p>\n

\"TS<\/p>\n

(iv) xy = ax2<\/sup> + \\(\\frac{b}{x}\\); (a, b)
\nSolution:
\nGiven equation is x2<\/sup>y = ax3<\/sup> + b ………….(1)
\nDifferentiating (1) w.r.t. ‘x\u2019
\n2xy + x2<\/sup>y1<\/sub> = 3ax2<\/sup> ………….(2)
\nAgain differentiating w.r.t x,
\nx2<\/sup>y2<\/sub> + 2xy1<\/sub> + 2xy1<\/sub> + 2y = 6ax
\n\u21d2 x2<\/sup>y2<\/sub> + 4xy1<\/sub> + 2y = 6ax
\n\u21d2 x3<\/sup>y2<\/sub> + 4x2y1<\/sub> + 2xy = 6ax2<\/sup>
\n= 2(3ax2<\/sup>)
\n= 2 [2xy + x2<\/sup>y1<\/sub>]
\n= 2x2<\/sup>y1<\/sub> + 4xy
\n\u21d2 x3<\/sup>y2<\/sup> + 2x2<\/sup>y1<\/sub> – 2xy = 0
\n\u21d2 x2<\/sup>y2<\/sub> + 2xy1<\/sub> – 2y = 0
\n\u21d2 \\(x^2 \\frac{d^2 y}{d x^2}+2 x \\frac{d y}{d x}\\) – 2y = 0 which is the required differential equation on elimination of constants a and b from (1).<\/p>\n

Question 2.
\nObtain the differential equation which corresponds to each of the following family of curves.
\n(i) The circles which touch the Y – axis at the origin.
\nSolution:
\nThe cquation of circle which touch the Y-axis at the origin is x2<\/sup> + y2<\/sup> + 2gx = 0 ………..(1)
\nDifferentiating wr.t. x we get
\n2x + 2yy1<\/sub> + 2g = 0
\n\u21d2 g = – (x + yy1<\/sub>)
\nHence from (1)
\nx2<\/sup> + y2<\/sup> + 2x [- (x + yy1<\/sub>)] = 0]
\nx2<\/sup> + y2<\/sup> – 2x2<\/sup> – 2xyy1<\/sub> = 0
\n\u21d2 – y2<\/sup> – x2<\/sup> = 2xy . \\(\\frac{d y}{d x}\\) which is the required differential equation obtained on elimination of ‘g’ from (1).<\/p>\n

\"TS<\/p>\n

(ii) The parabola each of which has a laws rectum 4a and whose axis are parallel to X- axis.
\nSolution:
\nEquation of parabola which has latus rectum 4a and whose axes are parallel to X-axis is
\n(y – k)2<\/sup> = 4a(x – h) ………….(1)
\nDifferentiating w.r.t ‘x’
\n2 (y – k) y1<\/sub> = 4a
\n\u21d2 (y – k) y1<\/sub> = 2a ……………(2)
\nDifferentiating again w.r.t ‘x’
\n(y – k) y2<\/sub> + y1<\/sub>2<\/sup> = 0
\nFrom (2)
\ny – k = \\(\\frac{2 a}{y_1}\\)
\n\u2234 From (3)
\n\\(\\frac{2 a}{y_1}\\) y2<\/sub> + y1<\/sub>2<\/sup> = 0
\n\u21d2 2ay2<\/sub> + y1<\/sub>3<\/sup> = 0
\n\u21d2 2a \\(\\frac{d^2 y}{d x^2}+\\left(\\frac{d y}{d x}\\right)^3\\) = 0 which is the required differential equation obtained on elimination of constants h, k from (1).<\/p>\n

(iii) The parabolas having their focli at the origin and axis along the X-axis.
\nSolution:
\nEquation of parabola having focii at origin and axis is along X-axis is given by
\ny2<\/sup> = 4a(x + a) ……….(1)
\nDifferentiating w.r.t x
\n2yy1<\/sub> = 4a
\na = \\(\\frac{\\mathrm{yy}_1}{2}\\)
\n\u2234 From (1)
\ny2<\/sup> = 4a(x + a)
\n= 4 \\(\\frac{\\mathrm{yy}_1}{2}\\) (x + \\(\\frac{\\mathrm{yy}_1}{2}\\))
\n= 2yy1<\/sub> (x + \\(\\frac{\\mathrm{yy}_1}{2}\\))
\n= 2xyy1<\/sub> + y2<\/sup>y1<\/sub>2<\/sup>
\n\u21d2 y = 2xy1<\/sub> + yy1<\/sub>2<\/sup>
\n\u21d2 yy1<\/sub>2<\/sup> + 2xy1<\/sub> – y = 0
\n\u21d2 \\(y\\left(\\frac{d y}{d x}\\right)^2+2 x\\left(\\frac{d y}{d x}\\right)\\) – y = 0
\nwhich is the required differential equation obtained on elimination of ‘a’ from (1).<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a) I. Question 1. Find the order of the differential equation obtained by eliminating the arbitrary constants b and … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/12223"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=12223"}],"version-history":[{"count":2,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/12223\/revisions"}],"predecessor-version":[{"id":12229,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/12223\/revisions\/12229"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=12223"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=12223"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=12223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}