{"id":169077,"date":"2024-03-28T16:33:29","date_gmt":"2024-03-28T11:03:29","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=169077"},"modified":"2024-05-04T14:58:54","modified_gmt":"2024-05-04T09:28:54","slug":"ml-aggarwal-class-12-maths-solutions-section-a-chapter-9-ex-9-1","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-12-maths-solutions-section-a-chapter-9-ex-9-1\/","title":{"rendered":"ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1"},"content":{"rendered":"

The availability of ML Aggarwal Class 12 ISC Solutions<\/a> Chapter 9 Differential Equations Ex 9.1 encourages students to tackle difficult exercises.<\/p>\n

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1<\/h2>\n

Very short answer type questions :<\/p>\n

Determine the order and the degree (when defined) of each of the following (1 to 20) differential equations:<\/p>\n

Question 1.
\n(i) \\(\\frac{d y}{d x}\\) – cos x = 0 (NCERT)
\n(ii) \\(\\left(\\frac{d y}{d x}\\right)^2+\\frac{d y}{d x}\\) – sin y = 0.
\nSolution:
\nGiven differential equation be,
\n\\(\\frac{d y}{d x}\\) – cos x = 0
\nWhich is polynomial in derivatives.
\nThe highest ordered derivative existing in the given diff. eqn. be \\(\\frac{d y}{d x}\\) so its order is 1.
\nAlso, the exponent of \\(\\frac{d y}{d x}\\) is 1.
\n\u2234 degree of given differential equation be 1.<\/p>\n

(ii) Given differential equation be,
\n\\(\\left(\\frac{d y}{d x}\\right)^2+\\frac{d y}{d x}\\) – sin y = 0
\nwhich is polynomial in derivatives.
\nThe higher order derivative existing in given the given diff. eqn. be \\(\\frac{d y}{d x}\\) so its order is 1.
\nand the highest power of \\(\\frac{d y}{d x}\\) in given diff. eqn. be 2 so its degree is 2.<\/p>\n

Question 2.
\n(i) y’ + y = ex<\/sup>
\n(ii) \\(\\frac{d^2 y}{d x^2}\\) = sin 3x + cos 3x. (NCERT)
\nSolution:
\nGiven differential eqn. be
\ny’ + y = ex<\/sup> i.e. \\(\\frac{d y}{d x}\\) + y = ex<\/sup>
\nwhich is polynomial in derivatives.
\nThe highest order derivative present in given differential eqn. be \\(\\frac{d y}{d x}\\) and its exponent be 1.
\nSo order of given diff. eqn. be 1 and its degree be also 1.<\/p>\n

(ii) Given diff. eqn. be,
\n\\(\\frac{d^2 y}{d x^2}\\) = sin 3x + cos 3x
\nwhich is polynomial in derivatives.
\nThe highest order derivative existing in given diff. eqn. be \\(\\frac{d^2 y}{d x^2}\\).
\nSo its order be 2.
\nThe highest power of \\(\\frac{d^2 y}{d x^2}\\) be 1.
\n\u2234 degree of given differential eqn. be 1.<\/p>\n

\"ML<\/p>\n

Question 3.
\n(i) (x2<\/sup>y – 3x) dy + (x3<\/sup> – 3y2<\/sup>) dx = 0
\n(ii) \\(\\sqrt{1-y^2} d x+\\sqrt{1-x^2} d y\\) = 0
\nSolution:
\n(i) Given differential eqn. can be written as,
\n\\(\\frac{d y}{d x}+\\frac{x^3-3 y^2}{x^2 y-3 x}\\) = 0
\nwhich is polynomial in derivatives.
\nThe highest order derivative present in given differential eqn. be \\(\\frac{d y}{d x}\\).
\nso its order be 1.
\nAlso the degree of given diff. eqn. be the highest exponent of \\(\\frac{d y}{d x}\\) which is 1.
\nThus its degree be 1.<\/p>\n

(ii) Given differential eqn. can be written as,
\n\\(\\frac{d y}{d x}+\\frac{\\sqrt{1-y^2}}{\\sqrt{1-x^2}}\\) = 0
\nwhich is polynomial in derivatives.
\nThe highest order derivatives present in given diff. eqn. be \\(\\frac{d y}{d x}\\) and its order be 1.
\nThe degree of given diff. eqn. be the highest power of \\(\\frac{d y}{d x}\\) which is 1.
\nSo its degree be 1.<\/p>\n

Question 4.
\n(i) \\(\\frac{1}{x} \\cdot \\frac{d^2 y}{d x^2}+5 x \\frac{d y}{d x}\\) = sin 2x
\n(ii) \\(\\left(\\frac{d y}{d x}\\right)^4+3 y \\frac{d^2 y}{d x^2}\\) = 0
\nSolution:
\n(i) Given differential eqn. be,
\n\\(\\frac{1}{x} \\cdot \\frac{d^2 y}{d x^2}+5 x \\frac{d y}{d x}\\) = sin 2x
\nwhich is polynomial in derivatives.
\nThe highest ordered derivative present in given diff. eqn. be \\(\\frac{d^2 y}{d x^2}\\)
\nSo its order is 2.
\nThe degree of given diff. eqn. be the highest exponent of \\(\\frac{d^2 y}{d x^2}\\) which is 1.
\nSo its degree be 1.<\/p>\n

\"ML<\/p>\n

Question 5.
\n(i) (y’)2<\/sup> + y2<\/sup> – 1 = 0
\n(ii) y” + 5x (y’)2<\/sup> – 6y = log x
\nSolution:
\n(i) Given diff. eqn. be,
\ny’2<\/sup> + y2<\/sup> – 1 = 0
\nwhich is polynomial in derivatives.
\nThe highest ordered derivative present in given diff. eqn. be \\(\\frac{d y}{d x}\\) i.e. y’.
\nso its order be 1.
\nThe degree of the given diff. eqn. is the highest exponent of which is \\(\\frac{d y}{d x}\\).
\nSo its degree be 2.<\/p>\n

(ii) Given diff. eqn. be,
\ny” + 5xy’2<\/sup> – 6y = log x
\nwhich is polynomial in derivatives.
\nThe highest ordered derivative in the given diff. eqn. be y”.
\nso its order be 2.
\nThe degree of the given diff. eqn. be the highest exponent of \\(\\frac{d^2 y}{d x^2}\\) which is 1.
\nThus its degree be 1.<\/p>\n

Question 6.
\n(i) x3<\/sup> (\\(\\frac{d^2 y}{d x^2}\\))2<\/sup> + x (\\(\\frac{d y}{d x}\\))4<\/sup> = 0
\n(ii) y(iv)<\/sup> + sin y(i)<\/sup> = 0 (NCERT)
\nSolution:
\n(i) Here the highest ordered derivative existing in given differential eqn. be \\(\\frac{d^2 y}{d x^2}\\)
\nso its order be 2.
\nFurther given differential eqn. can be expressed as polynomial in derivatives.
\nSo the exponent of highest ordered derivative \\(\\frac{d^2 y}{d x^2}\\) which is 2 gives the degree of given differential eqn.<\/p>\n

(ii) Here the highest ordered derivative existing in given differential equation be y(iv)<\/sup> so its order be 4.
\nSince the given differential eqn. can\u2019t be expressed as polynomial in derivatives as it contains term like sin y(1)<\/sup> which contains infinite number of terms.
\nSo degree of given differential eqn. is not defined.<\/p>\n

\"ML<\/p>\n

Question 7.
\n(i) \\(\\left(\\frac{d^2 y}{d x^2}\\right)^3+\\frac{d^2 y}{d x^2}+\\sin \\left(\\frac{d y}{d x}\\right)\\) = 2x
\n(ii) \\(\\left(\\frac{d^2 y}{d x^2}\\right)^3+2 y \\frac{d y}{d x}+\\sin y=5 x^{\\frac{2}{3}}+\\log x\\)
\nSolution:
\n(i) Here, the highest ordered derivative existing in given diff. eqn. be \\(\\frac{d^2 y}{d x^2}\\) and its order be 2.
\nClearly given diff. eqn. cannot be expressed as polynomial in \\(\\frac{d y}{d x}\\)
\nso its degree is not defined [as it contains terms like sin (\\(\\frac{d y}{d x}\\))]<\/p>\n

(ii) Clearly the highest ordered derivative existing in given differential eqn. be \\(\\frac{d^2 y}{d x^2}\\)
\nso its order is 2.
\nSince each term in derivatives is a polynomial.
\nSo degree of given differential eqn. be highest exponent of \\(\\frac{d^2 y}{d x^2}\\) i.e. 3.<\/p>\n

Question 8.
\n(i) 2x2<\/sup> \\(\\frac{d^2 y}{d x^2}\\) – 3 \\(\\frac{d y}{d x}\\) + y = 0 (NCERT)
\n(ii) (y”)3<\/sup> + (y’)2<\/sup> + sin y’ + 1 = 0. (NCERT)
\nSolution:
\n(i) Given differential eqn. be
\n2x2<\/sup> \\(\\frac{d^2 y}{d x^2}\\) – 3 \\(\\frac{d y}{d x}\\) + y = 0,
\nwhich is polynomial in derivatives.
\nThe highest ordered derivative present in \\(\\frac{d^2 y}{d x^2}\\) given diff. eqn. be \\(\\frac{d^2 y}{d x^2}\\).
\nSo its order be 2.
\nThe degree of given diff. eqn. is the highest power of \\(\\frac{d^2 y}{d x^2}\\) which is 1.
\nThus, its degree be 1.<\/p>\n

(ii) Given differential eqn. be
\n(y”)3<\/sup> + (y’)2<\/sup> + sin y’ + 1 = 0
\nThe highest order derivative existing in given diff. eqn. be y” and its order be 2.
\nThus, the order of given diff. eqn. be 2.
\nSince the term sin y\u2019 is not polynomial in y\u2019.
\n\u2234 The degree of given diff. eqn. is not defined.<\/p>\n

Question 9.
\n(i) x \\(\\frac{d^2 y}{d x^2}\\) = (1 + (\\(\\frac{d y}{d x}\\))2<\/sup>)4<\/sup>
\n(ii) \\(\\left(\\frac{d^4 y}{d x^4}\\right)^2=\\left(x+\\left(\\frac{d y}{d x}\\right)^2\\right)^3\\)
\nSolution:
\n(i) Clearly the highest ordered derivative existing in given diff. eqn. be \\(\\frac{d^2 y}{d x^2}\\).
\nso its order be 2.
\nClearly the given diff. eqn. can be expressed as polynomial in derivatives.
\nHence degree be the highest exponent of \\(\\frac{d^2 y}{d x^2}\\) which is 1.<\/p>\n

(ii) Clearly the highest ordered derivative existing in given diff. eqn. be \\(\\frac{d^4 y}{d x^4}\\).
\nso its order be 4.
\nHere given differential eqn. can be expressed as polynomial in derivatives.
\nHence degree of given diff. eqn. be the highest exponent of \\(\\frac{d^4 y}{d x^4}\\) which is 2.<\/p>\n

\"ML<\/p>\n

Question 10.
\n(i) 3x \\(\\frac{d y}{d x}\\) + \\(\\frac{5}{\\frac{d y}{d x}}\\) = y3<\/sup>
\n(ii) y = x \\(\\frac{d y}{d x}\\) + a \\(\\sqrt{1+\\left(\\frac{d y}{d x}\\right)^2}\\)
\nSolution:
\n(i) Given differential eqn. be,
\n3x \\(\\frac{d y}{d x}\\) + \\(\\frac{5}{\\frac{d y}{d x}}\\) = y3<\/sup>
\n\u21d2 3x (\\(\\frac{d y}{d x}\\))2<\/sup> + 5 = y3<\/sup> \\(\\frac{d y}{d x}\\)
\nwhich is polynomial in derivatives.
\nThe highest ordered derivative present in given diff. eqn. be \\(\\frac{d y}{d x}\\).
\nso its order be 1.
\nThe degree of given diff. eqn. be the highest exponent of \\(\\frac{d y}{d x}\\) which is 2.
\nThus its degree be 2.<\/p>\n

(ii) Given differential eqn. be,
\ny = x \\(\\frac{d y}{d x}\\) + a \\(\\sqrt{1+\\left(\\frac{d y}{d x}\\right)^2}\\)
\n\u21d2 y – x \\(\\frac{d y}{d x}\\) = a \\(\\sqrt{1+\\left(\\frac{d y}{d x}\\right)^2}\\)
\nOn squaring, we have
\n(y – x \\(\\frac{d y}{d x}\\))2<\/sup> = a2<\/sup> [1 + (\\(\\frac{d y}{d x}\\))2<\/sup>]
\nThe highest ordered derivative exiosting in given diff. eqn. be \\(\\frac{d y}{d x}\\) and its power 2.
\n\u2234 it is of order 1 and degree 2.
\nClearly it is a non-linear differential equation.<\/p>\n

Question 11.
\n(i) \\(\\sqrt{1+\\left(\\frac{d y}{d x}\\right)^2}\\) = 3x – \\(\\frac{d y}{d x}\\)
\n(ii) \\(5 \\frac{d^2 y}{d x^2}=\\left(1+\\left(\\frac{d y}{d x}\\right)^2\\right)^{\\frac{1}{4}}\\)
\nSolution:
\n(i) Given differential eqn. be,
\n\\(\\sqrt{1+\\left(\\frac{d y}{d x}\\right)^2}\\) = 3x – \\(\\frac{d y}{d x}\\)
\nOn squaring both sides ;
\n1 + (\\(\\frac{d y}{d x}\\))2<\/sup> = 9x2<\/sup> + (\\(\\frac{d y}{d x}\\))2<\/sup> – 6x \\(\\frac{d y}{d x}\\)
\n\u21d2 6x \\(\\frac{d y}{d x}\\) = 9x2<\/sup> + 1 = 0
\nThe highest ordered derivative existing in given diff. eqn. be \\(\\frac{d y}{d x}\\) so its order be 1.
\nFurther given duff. eqn. can be expressed as polynomial in derivative.
\nSo degree be the highest exponent of which is 1.<\/p>\n

(ii) Given differential eqn. can be written as
\n\\(5 \\frac{d^2 y}{d x^2}=\\left(1+\\left(\\frac{d y}{d x}\\right)^2\\right)^{\\frac{1}{4}}\\)
\nClearly the highest ordered derivative existing in given diff. eqn. be \\(\\frac{d^2 y}{d x^2}\\) and
\nhence its order be 2.
\nHere, given diff. eqn. can be expressed as polynomial in derivatives.
\nSo degree of given diff. eqn. be the highest power of which is 4.<\/p>\n

\"ML<\/p>\n

Question 11 (old).
\n\\(\\left(\\frac{d^2 x}{d t^2}\\right)^4-7 t\\left(\\frac{d x}{d t}\\right)^3\\) = log t.
\nSolution:
\nGiven differential eqn. be,
\n\\(\\left(\\frac{d^2 x}{d t^2}\\right)^4-7 t\\left(\\frac{d x}{d t}\\right)^3\\) = log t
\nwhich is polynomial in derivatives.
\nThe highest ordered derivative present in given diff. eqn. be \\(\\frac{d^2 y}{d x^2}\\) and its order be 2.
\nThe degree of given diff. eqn. be the highest exponent of \\(\\frac{d^2 y}{d x^2}\\) which is 4.
\nThus its degree be 4.<\/p>\n

Question 12.
\n(i) y” + (y’)2<\/sup> + 2y = 0 (NCERT)
\n(ii) y” + 2y’ + sin y = 0
\nSolution:
\n(i) Given differential eqn. be,
\ny” + (y’)2<\/sup> + 2y = 0
\nwhich is polynomial in derivatives.
\nThe highest order derivative present in given diff. eqn. be y” and its order be 2.
\nThe degree of given differential equation is the highest exponent of y” which is 1.
\nThus its degree be 1.<\/p>\n

(ii) Given differential eqn. be,
\ny” + 2y’ + sin y = 0
\nwhich is polynomial in derivatives.
\nThe highest order derivative present in given diff. eqn. be y\u201d and its order is 2.
\nThe degree of given diff. eqn. be the highest exponent of y\u201d which is 1.
\nThus its degree be 1.<\/p>\n

Question 12 (old).
\ny(iv)<\/sup> + sin y”’ = 0 (NCERT)
\nSolution:
\nGiven diff. eqn. be,
\ny(iv)<\/sup> + sin y”’ = 0
\nThe order of given diff. eqn. be the highest order derivative present in given diff. eqn. which is y(iv)<\/sup> and its order be 4.
\nHere the term sin y”’ is not polynomial in y”’.
\nThus, the degree of given differential eqn. is not defined.<\/p>\n

\"ML<\/p>\n

Question 13.
\n(i) (y”’)2<\/sup> + (y”)3<\/sup> + (y’)4<\/sup> + y5<\/sup> = 0 (NCERT)
\n(ii) (1 – (y’)2<\/sup>)3\/2<\/sup> = ky”
\nSolution:
\n(i) Given diff. eqn. be,
\n(y”’)2<\/sup> + (y”)3<\/sup> + (y’)4<\/sup> + y5<\/sup> = 0
\nwhich is polynomial in derivatives.
\nThe highest order derivative present in given diff. eqn. be y”’ so its order be 3.
\nThe degree of given diff. eqn. is the highest exponent of y”’, which is 2.
\nThus degree of given diff. eqn. be 2.<\/p>\n

(ii) Given diff. eqn. can be written as
\n(1 – (y’)2<\/sup>)3<\/sup> = (ky”)2<\/sup>
\nClearly the highest ordered derivative existing in given duff. eqn. be y” so its order 2.
\nHere the given duff. eqn. can be expressed as polynomial in derivatives.
\nThus its degree be the highest exponent of y” which is 2.<\/p>\n

Question 14.
\nWrite the sum of the order and the degree of the differential equation \\(\\left(\\frac{d y}{d x}\\right)^5+3 x y\\left(\\frac{d^3 y}{d x^3}\\right)^2+y\\left(\\frac{d^2 y}{d x^2}\\right)^4\\) = 0.
\nSolution:
\nClearly the highest ordered derivative existing in given diff. eqn. be \\(\\frac{d^3 y}{d x^3}\\).
\nso its order be 3.
\nClearly the given duff. eqn. can be expressed as polynomial in derivatives.
\nSo degree of given diff. eqn. be the highest exponent of \\(\\frac{d^3 y}{d x^3}\\) which is 2.
\n\u2234 required sum = 3 + 2 = 5.<\/p>\n

Question 15.
\nFind the product of the order and degree of the following differential equation :
\nx (\\(\\frac{d^2 y}{d x^2}\\))2<\/sup> + (\\(\\frac{d y}{d x}\\))2<\/sup> + y2<\/sup> = 0.
\nSolution:
\nThe given differential eqn. be
\nx (\\(\\frac{d^2 y}{d x^2}\\))2<\/sup> + (\\(\\frac{d y}{d x}\\))2<\/sup> + y2<\/sup> = 0
\nwhich is polynomial in derivatives.
\nThe highest order derivative present in given diff. eqn. be \\(\\frac{d^2 y}{d x^2}\\).
\nSo its order be 2.
\nThe degree of given diff. eqn. be the highest exponent of \\(\\frac{d^2 y}{d x^2}\\) which is 2.
\nThus, its degree be 2.
\n\u2234 required product of order and degree of given differential equation 2 \u00d7 2 = 4.<\/p>\n

\"ML<\/p>\n

Question 20 (old).
\ny = px + \\(\\sqrt{a^2 p^2+b^2}\\), where p = \\(\\frac{d y}{d x}\\).
\nSolution:
\nGiven diff. eqn. can be written as
\n(y – px)2<\/sup> = a2<\/sup>p2<\/sup> + b2<\/sup>; where p = \\(\\frac{d y}{d x}\\)
\nClearly it is a polynomial in p.
\nThe highest order derivative existing in diff. eqn. be p i.e. \\(\\frac{d y}{d x}\\) and its power 2.
\nThus given diff. eqn. is of order 1 and degree 2.
\nClearly it is a non-linear differential equation.<\/p>\n

Question 21 (old).
\nWrite the sum of the order and the degree of the differential equation \\(\\left(\\frac{d^2 y}{d x^2}\\right)^2-\\left(\\frac{d y}{d x}\\right)^3\\) = y3<\/sup>.
\nSolution:
\nGiven differential equation be,
\n\\(\\left(\\frac{d^2 y}{d x^2}\\right)^2-\\left(\\frac{d y}{d x}\\right)^3\\) = y3<\/sup>
\nwhich is polynomial in derivatives.
\nThe highest order derivative present in given
\ndiff. eqn. be \\(\\frac{d^2 y}{d x^2}\\).
\nso its order be 2.
\nThe degree of given diff. eqn. be the highest power of \\(\\frac{d^2 y}{d x^2}\\) which is 2.
\nSo its degree be 2 and required sum = 2 + 2 = 4.<\/p>\n","protected":false},"excerpt":{"rendered":"

The availability of ML Aggarwal Class 12 ISC Solutions Chapter 9 Differential Equations Ex 9.1 encourages students to tackle difficult exercises. ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1 Very short answer type questions : Determine the order and the degree (when defined) of each of the following (1 …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/169077"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=169077"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/169077\/revisions"}],"predecessor-version":[{"id":169760,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/169077\/revisions\/169760"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=169077"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=169077"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=169077"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}