{"id":32426,"date":"2022-03-29T17:00:46","date_gmt":"2022-03-29T11:30:46","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=32426"},"modified":"2022-03-29T17:22:31","modified_gmt":"2022-03-29T11:52:31","slug":"ncert-solutions-for-class-12-maths-chapter-9-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 9 Differential Equations Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-miscellaneous-exercise\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFor each of the differential equations given in Exercises 1 to 12, find the general solution.
\ni. \\(\\frac{d^{2} y}{d x^{2}}+5 x\\left(\\frac{d y}{d x}\\right)^{2}-6 y=\\log x\\)
\nii. \\(\\left(\\frac{d y}{d x}\\right)^{3}-4\\left(\\frac{d y}{d x}\\right)^{2}+7 y=\\sin x\\)
\niii. \\(\\frac{d^{4} y}{d x^{4}}-\\sin \\left(\\frac{d^{3} y}{d x^{3}}\\right)=0\\)
\nSolution:
\ni. \\(\\frac{d^{2} y}{d x^{2}}+5 x\\left(\\frac{d y}{d x}\\right)^{2}-6 y=\\log x\\)
\n\u2234 Order 2 and degree = 1<\/p>\n

ii. \\(\\left(\\frac{d y}{d x}\\right)^{3}-4\\left(\\frac{d y}{d x}\\right)^{2}+7 y=\\sin x\\)
\n\u2234 Order 1 and degree = 3<\/p>\n

iii. \\(\\frac{d^{4} y}{d x^{4}}-\\sin \\left(\\frac{d^{3} y}{d x^{3}}\\right)=0\\)
\n\u2234 Order 4 and degree not defined.<\/p>\n

Question 2.
\nFor each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
\ni. y = a ex<\/sup> + be-x<\/sup> + x\u00b2:
\n\\(\\frac{d^{2} y}{d x^{2}}+2 \\frac{d y}{d x}\\) – xy + x\u00b2 – 2 = 0<\/p>\n

ii. y = ex<\/sup>(a cos x + b sin x) :
\n\\(\\frac{d^{2} y}{d x^{2}}-2 \\frac{d y}{d x}\\) + 2y = 0<\/p>\n

iii. y = x sin 3x :
\n\\(\\frac{d^{2} y}{d x^{2}}\\) + 9y – 6 cos 3x = 0<\/p>\n

iv. x\u00b2 = 2y\u00b2 log y :
\n(x\u00b2 + y\u00b2) \\(\\frac { dy }{ dx }\\) – xy = 0
\nSolution:
\ni. y = a ex<\/sup> + be-x<\/sup> + x\u00b2:
\nDifferentiating w.r.t. x, we get
\n\"NCERT<\/p>\n

ii. y = ex<\/sup>(a cos x + b sin x) :
\nDifferentiating w.r.t. x, we get
\n\"NCERT<\/p>\n

iii. y = x sin 3x :
\nDifferentiating w.r.t. x, we get
\n\\(\\frac { dy }{ dx }\\) = 3x cos3x + sin 3x
\n\\(\\frac{d^{2} y}{d x^{2}}\\) = 3x(- 3sin 3x) + 3cos 3x + 3cos 3x
\n= – 9x sin 3x + 6cos 3x
\n\\(\\frac{d^{2} y}{d x^{2}}\\) + 9y – 6cos 3x
\n= – 9x sin 3x + 6cos 3x + 9x sin 3x – 6cos 3x = 0
\n\u2234 y = xsin 3x is the solution of the differential equation \\(\\frac{d^{2} y}{d x^{2}}\\) + 9y – 6cos 3x = 0<\/p>\n

iv. x\u00b2 = 2y\u00b2 log y :
\nDifferentiating w.r.t. x, we get
\n\"NCERT<\/p>\n

Question 3.
\nForm the differential equation representing the family of curves given by (x – a)\u00b2 + 2y\u00b2 = a\u00b2, where a is an arbitrary constant.
\nSolution:
\n(x – a)\u00b2 + 2y\u00b2 = a\u00b2 … (1)
\nDifferentiating w.r.t. x, we get
\n2(x – a) + 4yy’ = 0
\nx – a + 2yy’ = 0
\n\u2234 a = x + 2yy’
\n(1) \u2192 (- 2yy’)\u00b2 + 2y\u00b2 =[x + 2yy’]\u00b2
\n4(yy’)\u00b2 + 2y\u00b2 = x\u00b2 + 4 xyy’ + 4 (yy’)\u00b2
\n2y\u00b2 = x\u00b2 + 4xyy’
\n\u2234 y’ = \\(\\frac{2 y^{2}-x^{2}}{4 x y}\\) is the required differential equation.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nProve that x\u00b2 – y\u00b2 = c (x\u00b2 + y\u00b2)\u00b2 is the general solution of differential equation (x\u00b3 – 3xy\u00b2) dx = (y\u00b3 – 3x\u00b2y) dy, where c is a parameter.
\nSolution:
\n\"NCERT
\nx\u00b2 – y\u00b2 = C(x\u00b2 + y\u00b2)\u00b2 where C = C\u00b21<\/sub> is the general solution.<\/p>\n

Question 5.
\nForm the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
\nSolution:
\n\"NCERT
\nThe equation of the required circles is
\n(x – r)\u00b2 + (y – r)\u00b2 = r\u00b2 … (1)
\nx\u00b2 + y\u00b2 – 2rx – 2ry + r\u00b2 = 0
\nDifferentiating (1) w.r.t. x, we get
\n2x + 2yy1<\/sub> – 2r – 2ry1<\/sub> = 0
\nr = \\(\\frac{x+y y_{1}}{1+y_{1}}\\)
\nSubstituting r in (1), we get
\n\"NCERT
\nis the required differential equation.<\/p>\n

Question 6.
\nFind the general solution of the differential equation \\(\\frac{d y}{d x}+\\sqrt{\\frac{1-y^{2}}{1-x^{2}}}=0\\).
\nSolution:
\n\"NCERT
\n\u2234 sin-1<\/sup>y + sin-1<\/sup>x = C, is the general solution.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nShow that the general solution of the differential equation \\(\\frac{d y}{d x}+\\frac{y^{2}+y+1}{x^{2}+x+1}\\) = 0 is given by
\n(x + y + 1) = A(1 – x – y – 2xy), where A is the parameter.
\nSolution:
\n\"NCERT<\/p>\n

Question 8.
\nFind the equation of the curve passing through the point (0, \\(\\frac { \u03c0 }{ 4 }\\)) whose differential equation is sin x cos y dx + cos x siny dy = 0.
\nSolution:
\nsin x cos y dx + cos x sin y dy = 0.
\nDivide by cos x cos y, we get
\n\\(\\frac{\\sin x}{\\cos x} d x+\\frac{\\sin y}{\\cos y} d y=0\\)
\ntan x dx + tan y dy = 0
\nIntegrating both sides, we get
\n\u222btan x dx + \u222btan y dy = \u222bo dx
\nlog|sec x| + log|sec y| = log|C|
\nlog|sec x secy| = log|C|
\nsec x sec y = C … (1)
\n(1) Passes through the point
\nwe get C = \\(\\sqrt{2}\\)
\n\u2234 (1) \u2192 sec x secy = \\(\\sqrt{2}\\)
\n\u21d2 cos y = \\(\\frac{\\sec x}{\\sqrt{2}}\\), is the equation of the curve.<\/p>\n

Question 9.
\nFind the particular solution of the differential equation
\n(1 + e2x<\/sup>) dy + (1 + y\u00b2)ex<\/sup> dx = 0. Given that y = 1 when x = 0.
\nSolution:
\n(1 + e2x<\/sup>) dy + (1 + y\u00b2)ex<\/sup> dx = 0
\nDivide by (1 + e2x<\/sup>)(1 + y\u00b2), we get
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nSolve the differential equation
\n\\(y e^{\\frac{x}{y}} d x=\\left(x e^{\\frac{x}{y}}+y^{2}\\right) d y\\) (y \u2260 0).
\nSolution:
\n\"NCERT
\ni.e., \\(e^{\\frac{x}{y}}\\) = y + C is the general solution.<\/p>\n

Question 11.
\nFind a particular solution of the differential equation (x – y) (dx + dy) = dx – dy. Given that y = – 1, when x = 0.
\nSolution:
\n(x – y)(dx + dy) = dx – dy
\nxdx – ydx + xdy – ydy = dx – dy
\n(x – y – 1)dx = (y – x – 1)dy
\n\"NCERT
\nt + log|t| = 2x + C
\nx – y + log|x – y|= 2x + C … (2)
\nWhen x = 0, y = – 1
\n\u2234 (2) \u2192 1 + log 1 = 0 + C
\nC = 1
\n\u2234 (2) \u2192 x – y + log |x – y| = 2x + 1
\nHence x + y + 1 = log|x – y| is the particular solution<\/p>\n

Question 12.
\nSolve the differential equation \\(\\left[\\frac{e^{-2 \\sqrt{x}}}{\\sqrt{x}}-\\frac{y}{\\sqrt{x}}\\right] \\frac{d x}{d y}=1(x \\neq 0)\\).
\nSolution:
\n\"NCERT
\ni.e., \\(y e^{2 \\sqrt{x}}=2 \\sqrt{x}+C\\) is the general solution.<\/p>\n

Question 13.
\nFind a particular solution of the differential equation \\(\\frac { dy }{ dx }\\) + y cot x = 4x cosec x (x \u2260 0).
\nGiven that y = 0 when x = \\(\\frac { \u03c0 }{ 2 }\\)
\nSolution:
\n\\(\\frac { dy }{ dx }\\) + y cot x = 4x cosec x is a linear differential equation.
\n\u2234 P = cot x and Q = 4x cosec x
\n\u222bPdx = \u222bcot x dx = log sinx
\n\u2234 I.F. = e\u222bp dx<\/sup> = elog sinx<\/sup> = sin x
\nThe solution is y(I.F) = \u222bQ(I.F)dx + C
\ny sinx = \u222b4x cosec x sin xdx + C
\n= \u222b4x dx + C
\ny sin x = 2x + C … (1)
\nWhen x = \\(\\frac { \u03c0 }{ 2 }\\), y = 0
\n(1) \u2192 o = \\(\\frac { \u03c0\u00b2 }{ 2 }\\) + C
\n\u2234 C = \\(\\frac { \u03c0\u00b2 }{ 2 }\\)
\nHence the particular solution is
\nysin x = 2x\u00b2 – \\(\\frac { \u03c0\u00b2 }{ 2 }\\)<\/p>\n

\"NCERT<\/p>\n

Question 14.
\nFind a particular solution of the differential equation (x + 1) \\(\\frac { dy }{ dx }\\) = 2e-y<\/sup> – 1, given that y = 0 when x = 0.
\nSolution:
\n\"NCERT
\nHence y = log\\(\\left|\\frac{2 x+1}{x+1}\\right|\\), is the particular solution.<\/p>\n

Question 15.
\nThe population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?
\nSolution:
\nLet P be the population at time t
\nGiven \\(\\frac { dP }{ dt }\\) \u221d P
\ni.e, \\(\\frac { dP }{ dt }\\) = kP, k is a constant
\ni.e., \\(\\frac { dP }{ dt }\\) = kdt
\nIntegrating both sides, we get
\n\u222b\\(\\frac { dP }{ dt }\\) = k \u222bdt
\nlog|P| = kt + C … (1)
\nThe initial population in 1999 is 20,000
\ni.e., when t = 0, P = 20,000
\nHence (1) \u2192 log 20,000 = k(0) + C
\n\u2234 C = log 20,000
\nAfter 5 years, i.e., in 2004, the population is 25,000
\ni.e., when t = 5, P = 25000
\nHence (1) \u2192 log 25000 = 5k + log 20000
\nlog 25000 – log 20000 = 5k
\n\"NCERT<\/p>\n

Question 16.
\nThe general solution of the differential equation \\(\\frac{y d x-x d y}{y}\\) = 0 is
\na. xy = C
\nb. x = Cy\u00b2
\nc. y = Cx
\nd. y = Cx\u00b2
\nSolution:
\nc. y = Cx
\n\\(\\frac{y d x-x d y}{y}\\) = 0
\n\u2234 ydx – xdy = 0
\nydx = xdy
\n\\(\\frac { dy }{ y }\\) = \\(\\frac { dx }{ x }\\)
\nIntegrating both sides, we get
\n\u222b\\(\\frac { dy }{ y }\\) = \u222b\\(\\frac { dx }{ x }\\)
\nlog|y| = log|x| + log|C|
\nlog|y| = log|Cx|
\ny = Cx<\/p>\n

Question 17.
\nThe general solution of a differential equation of the type \\(\\frac { dx }{ dy }\\) + P1<\/sub>, x = Q1<\/sub> is
\n\"NCERT
\nSolution:
\n\\(e^{\\int {P}_{1} d y}\\) is the integrating factor
\n\u2234 The general solution is
\n\\(x \\cdot e^{\\int {P}_{1} d y}=\\int\\left({Q}_{1} e^{\\int {P}_{1} d y}\\right) d y+ {C}\\)<\/p>\n

\"NCERT<\/p>\n

Question 18.
\nThe general solution of the differential equation ex<\/sup>dy + (yex<\/sup> + 2x) dx = 0 is
\na. x ey<\/sup> + x\u00b2 = C
\nb. x ey<\/sup> + y\u00b2 =C
\nc. y ex<\/sup> + x\u00b2 = C
\nd. y ey<\/sup> + x\u00b2 = C
\nSolution:
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-9-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise Question 1. For each of the differential equations given in Exercises 1 to 12, find the general solution. …<\/p>\n

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