{"id":27925,"date":"2019-10-19T11:12:30","date_gmt":"2019-10-19T05:42:30","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=27925"},"modified":"2019-10-25T11:51:30","modified_gmt":"2019-10-25T06:21:30","slug":"differential-equations-class-12-mathematics-extra-questions","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/differential-equations-class-12-mathematics-extra-questions\/","title":{"rendered":"Differential Equations Class 12 Mathematics Extra Questions"},"content":{"rendered":"

Differential Equations Class 12 Mathematics Extra Questions. <\/strong>myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in\u00a0myCBSEguide<\/a>\u00a0<\/strong>website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 9 Differential Equations Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus<\/a><\/strong>\u00a0and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.<\/p>\n

Class 12 Chapter 9 Maths Extra Questions<\/strong><\/p>\n

Download as PDF<\/a><\/strong><\/p>\n

Chapter 9 Mathematics Class 12 Important Questions<\/h2>\n

Chapter 9 Differential Equations<\/strong><\/p>\n

\n
    \n
  1. \n
    \n
    In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931).<\/div>\n<\/div>\n
    \n
    \n
    \n
      \n
    1. 9.93%<\/li>\n
    2. 7.93%<\/li>\n
    3. 6.93%<\/li>\n
    4. 8.93%<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
    5. \n
      \n
      General solution of{tex}co{s^2}x\\frac{{dy}}{{dx}} + y = \\;\\tan x\\;\\left( {0 \\leqslant x < \\frac{\\pi }{2}} \\right){\/tex}<\/span> is<\/div>\n<\/div>\n
      \n
      \n
        \n
      1. {tex}y = \\left( {\\tan x – 1} \\right) + C{e^{ – \\tan x}}{\/tex}<\/span><\/li>\n
      2. {tex}y = \\left( {\\tan x + 1} \\right) + C{e^{ – \\tan x}}{\/tex}<\/span><\/span><\/li>\n
      3. {tex}y = \\left( {\\tan x + 1} \\right) – C{e^{ – \\tan x}}{\/tex}<\/span><\/span><\/li>\n
      4. {tex}y = \\left( {\\tan x – 1} \\right) – C{e^{ – \\tan x}}{\/tex}<\/span><\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/li>\n
      5. \n
        \n
        The number of arbitrary constants in the general solution of a differential equation of fourth order are:<\/div>\n<\/div>\n
        \n
        \n
        \n
          \n
        1. 3<\/li>\n
        2. 2<\/li>\n
        3. 1<\/li>\n
        4. 4<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
        5. \n
          \n
          In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years {tex}\\left( {{e^{0.5}} = {\\text{ }}1.648} \\right).{\/tex}<\/span><\/div>\n<\/div>\n
          \n
          \n
          \n
            \n
          1. Rs 1848<\/li>\n
          2. Rs 1648<\/li>\n
          3. Rs 1748<\/li>\n
          4. Rs 1948<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
          5. \n
            \n
            What is the order of differential equation : {tex}\\frac{{{d^3}y}}{{d{x^3}}} + \\frac{{{d^2}y}}{{d{x^2}}} + {\\left( {\\frac{{dy}}{{dx}}} \\right)^2} = {e^x}{\/tex}<\/span>.<\/div>\n<\/div>\n
            \n
            \n
            \n
              \n
            1. 2<\/li>\n
            2. 3<\/li>\n
            3. 1<\/li>\n
            4. 0<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
            5. F(x, y)\u00a0=\u00a0{tex}\\frac{{\\sqrt {{x^2} + {y^2}} + y}}{x}{\/tex}<\/span> is a homogeneous function of degree ________.<\/li>\n
            6. The degree of the differential equation {tex}\\sqrt{1+\\left(\\frac{dy}{dx}\\right)^2}=x{\/tex}<\/span> is ________.<\/li>\n
            7. The order of the differential equation of all circles of given radius a is ________.<\/li>\n
            8. \n
              \n
              Verify that the function is a solution of the corresponding differential equation {tex}y = x\\sin x;\\,\\,x{y^,} = y + x\\sqrt {{x^2} – {y^2}} {\/tex}<\/span>.<\/div>\n<\/div>\n<\/li>\n
            9. \n
              \n
              Find order and degree. {tex}\\frac{{{d^4}y}}{{d{x^2}}} + \\sin \\left( {y”’} \\right) = 0{\/tex}<\/span>.<\/div>\n<\/div>\n<\/li>\n
            10. \n
              \n
              Write the solution of the differential equation {tex}\\frac { d y } { d x } = 2 ^ { – y }{\/tex}<\/span>.<\/div>\n<\/div>\n<\/li>\n
            11. \n
              \n
              Verify that the given function (explicit) is a solution of the corresponding differential equation: y = x2<\/sup> + 2x + C : y’ – 2x – 2 = 0.<\/div>\n<\/div>\n<\/li>\n
            12. \n
              \n
              Find the differential equation of all non-horizontal lines in a plane.<\/div>\n<\/div>\n<\/li>\n
            13. \n
              \n
              Verify that the function is a solution of the corresponding differential equation
              \n{tex}y = \\sqrt {1 + {x^2}} ;{y’} = \\frac{{xy}}{{1 + {x^2}}}{\/tex}<\/span>.<\/div>\n<\/div>\n<\/li>\n
            14. \n
              \n
              Solve the following differential equation.
              \n{tex} \\left( y + 3 x ^ { 2 } \\right) \\frac { d x } { d y } = x{\/tex}<\/span><\/div>\n
              <\/div>\n<\/div>\n<\/li>\n
            15. \n
              \n
              Solve the differential equation (1 + y2<\/sup>) tan-1<\/sup>x dx + 2y (1 + x2<\/sup>) dy = 0.<\/div>\n<\/div>\n<\/li>\n
            16. \n
              \n
              Find the particular solution of the differential equation (1 + e2x<\/sup>)dy + (1 + y2<\/sup>)ex<\/sup> dx = 0, given that y = 1, when x = 0.<\/div>\n<\/div>\n<\/li>\n
            17. \n
              \n
              Solve {tex}\\left( {1 + {e^{\\frac{x}{y}}}} \\right)dx + {e^{\\frac{x}{y}}}\\left( {1 – \\frac{x}{y}} \\right)dy = 0{\/tex}<\/span>.<\/div>\n<\/div>\n<\/li>\n<\/ol>\n

              Chapter 9 Differential Equations<\/strong><\/p>\n

              \n

              Solution<\/b><\/p>\n

                \n
              1. \n
                  \n
                1. 6.93%
                  \nExplanation:<\/strong> Let P be the principal at any time t. then,
                  \n{tex}\\frac{{dP}}{{dt}} = \\frac{{rP}}{{100}} \\Rightarrow \\frac{{dP}}{{dt}} = \\frac{P}{{100}}{\/tex}<\/span>
                  \n{tex} \\Rightarrow \\int {\\frac{1}{P}dP = \\int {\\frac{r}{{100}}dt} } {\/tex}<\/span>
                  \n{tex} \\Rightarrow \\log P = \\frac{r}{{100}}t + \\log c{\/tex}<\/span>
                  \n{tex} \\Rightarrow \\log \\frac{P}{c} = \\frac{r}{{100}}t{\/tex}<\/span>
                  \n{tex}\\Rightarrow P = c{e^{\\frac{r}{{100}}}}{\/tex}<\/span>
                  \nWhen P = 100 and t = 0., then, c = 100, therefore, we have:
                  \n{tex}\\Rightarrow P=100 \\ {{e}^{{}^{r}\\!\\!\\diagup\\!\\!{}_{100}\\;}}{\/tex}<\/span>
                  \nNow, let t = T, when P = 100., then;
                  \n{tex}\\Rightarrow 200 = 100{e^{\\frac{T}{{100}}}}{\/tex}<\/span>
                  \n{tex} \\Rightarrow {e^{\\frac{T}{{100}}}} = 2{\/tex}<\/span>
                  \n{tex} \\Rightarrow T = 100\\log 2{\/tex}<\/span> = 100(0.6931) = 6.93%<\/li>\n<\/ol>\n
                    \n
                  1. {tex}y = \\left( {\\tan x – 1} \\right) + C{e^{ – \\tan x}}{\/tex}<\/span>
                    \nExplanation:<\/strong> {tex}\\frac{{dy}}{{dx}} + {\\sec ^2}x.y = \\tan x.{\\sec ^2}x \\Rightarrow P = {\\sec ^2}x,Q = \\tan x.{\\sec ^2}x{\/tex}<\/span>
                    \n{tex} \\Rightarrow I.F. = {e^{\\int_{}^{} {{{\\sec }^2}xdx} }} = {e^{\\tan x}}{\/tex}<\/span>
                    \n{tex}\\Rightarrow y.{e^{\\tan x}} = \\int_{}^{} {\\tan x{{\\sec }^2}x{e^{\\tan x}}dx \\Rightarrow y.{e^{\\tan x}} = (\\tan x – 1){e^{\\tan x}} + C} {\/tex}<\/span>
                    \n{tex} \\Rightarrow y = (\\tan x – 1) + C{e^{ – \\tan x}}{\/tex}<\/span><\/li>\n<\/ol>\n
                      \n
                    1. 4
                      \nExplanation:<\/strong> 4, because the no. of arbitrary constants is equal to order of the differential equation.<\/li>\n<\/ol>\n
                        \n
                      1. Rs 1648
                        \nExplanation:<\/strong> Here P is the principal at time t
                        \n{tex}\\frac{{dP}}{{dt}} = \\frac{{5P}}{{100}} \\Rightarrow \\frac{{dP}}{{dt}} = \\frac{P}{{20}}{\/tex}<\/span>
                        \n{tex} \\Rightarrow \\int_{}^{} {\\frac{1}{P}dP = \\int_{}^{} {\\frac{1}{{20}}dt} } {\/tex}<\/span>
                        \n{tex}\\Rightarrow \\log P = \\frac{1}{{20}}t + \\log c{\/tex}<\/span>
                        \n{tex} \\Rightarrow \\log \\frac{P}{c} = \\frac{1}{{20}}t{\/tex}<\/span>
                        \n{tex}\\Rightarrow P =c {e^{\\frac{1}{{100}}}}{\/tex}<\/span>
                        \nWhen P = 1000 and t = 0 ., then ,
                        \nc = 1000, therefore, we have :
                        \n{tex} \\Rightarrow P = 1000{e^{\\frac{T}{{100}}}}{\/tex}<\/span>
                        \n{tex}\\Rightarrow A = 1000{e^{\\frac{5}{{10}}}}{\/tex}<\/span>
                        \n{tex}\\Rightarrow {e^{\\frac{5}{{10}}}} = A{\/tex}<\/span>
                        \n{tex} \\Rightarrow A = 1000\\log 0.5{\/tex}<\/span>
                        \n= 1000(1.648)
                        \n= 1648<\/li>\n<\/ol>\n
                          \n
                        1. 3
                          \nExplanation:<\/strong> Order = 3. Since the third derivative is the highest derivative present in the equation. i.e.{tex}\\frac{{{d^3}y}}{{d{x^3}}}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                        2. Zero<\/li>\n
                        3. not defined<\/li>\n
                        4. 2<\/li>\n
                        5. {tex}y = x.\\sin x{\/tex}<\/span>…(1)
                          \n{tex}{y^,} = x.\\cos x + \\sin x.1{\/tex}<\/span>
                          \n{tex} \\Rightarrow x{y^,} = {x^2}\\cos x + x.\\sin x{\/tex}<\/span>
                          \n{tex}x{y^,} = {x^2}\\sqrt {1 – {{\\sin }^2}x} + x.\\sin x{\/tex}<\/span>
                          \n{tex}x{y^,} = {x^2}\\sqrt {1 – {{\\left( {\\frac{y}{x}} \\right)}^2}} + x.\\sin x{\/tex}<\/span> {tex}\\left[ {\\because \\frac{y}{x} = \\sin x} \\right.]{\/tex}<\/span>
                          \n{tex}x{y^,} = {x^2}\\frac{{\\sqrt {{x^2} – {y^2}} }}{x} + x.\\sin x{\/tex}<\/span>
                          \n{tex}x{y^,} = x\\sqrt {{x^2} – {y^2}} + y{\/tex}<\/span>
                          \nHence proved.<\/li>\n
                        6. order = 4 ,degree = not defined<\/li>\n
                        7. Given differential equation is
                          \n{tex}\\frac { d y } { d x } = 2 ^ { -y }{\/tex}<\/span>
                          \non separating the variables, we get
                          \n2y<\/sup>dy = dx
                          \nOn integrating both sides, we get
                          \n{tex}\\int 2 ^ { y } d y = \\int d x{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad \\frac { 2 ^ { y } } { \\log 2 } = x + C_1{\/tex}<\/span>
                          \n{tex}\\Rightarrow{\/tex}<\/span> 2y<\/sup> = x log 2 + C1<\/sub> log 2
                          \n{tex}\\therefore{\/tex}<\/span> 2y<\/sup> = x log 2 + C, where C = C1<\/sub> log 2<\/li>\n
                        8. Given: y = x2<\/sup> + 2x + C …(i)
                          \nTo prove: y is a solution of the differential equation y’ – 2x – 2 = 0 …(ii)
                          \nProof:From, eq. (i),
                          \ny’ = 2x + 2
                          \nL.H.S. of eq. (ii),
                          \n= y’ – 2x – 2
                          \n= (2x + 2) – 2x – 2
                          \n= 2x + 2 – 2x – 2 = 0 = R.H.S.
                          \nHence, y given by eq. (i) is a solution of y’ – 2x – 2 = 0.<\/li>\n
                        9. The general equation of all non-horizontal lines in a plane is ax + by = c, where {tex}a \\ne 0{\/tex}<\/span>.
                          \ndifferentiating both sides w.r.t. y on both sides,we get
                          \n{tex}a\\frac{{dy}}{{dx}} + b = 0{\/tex}<\/span>
                          \nAgain, differentiating both sides w.r.t. y, we get
                          \n{tex}a\\frac{{{d^2}x}}{{d{y^2}}} = 0 \\Rightarrow \\frac{{{d^2}x}}{{d{y^2}}} = 0.{\/tex}<\/span><\/li>\n
                        10. {tex}y = \\sqrt {1 + {x^2}} {\/tex}<\/span> ……(i)
                          \n{tex}{y’} = \\frac{1}{{2\\sqrt {1 + {x^2}} }}.2x{\/tex}<\/span> ……(ii)
                          \n{tex}(ii) \\div (i){\/tex}<\/span>,we get,
                          \n{tex}\\Rightarrow\\frac{{{y’}}}{y} = \\frac{{\\frac{x}{{\\sqrt {1 + {x^2}} }}}}{{\\sqrt {1 + {x^2}} }}{\/tex}<\/span>
                          \n{tex}\\Rightarrow\\frac{{{y’}}}{y} = \\frac{x}{{1 + {x^2}}}{\/tex}<\/span>
                          \n{tex}{y’} = \\frac{{xy}}{{1 + {x^2}}}{\/tex}<\/span>
                          \nHence given value of y is the solution of given differential equation.<\/li>\n
                        11. According to the question,we have to solve the differential equation ,
                          \n{tex} \\left( y + 3 x ^ { 2 } \\right) \\frac { d x } { d y } = x \\Rightarrow \\frac { d y } { d x } = \\frac { y } { x } + 3 x{\/tex}<\/span>
                          \n{tex} \\Rightarrow \\quad \\frac { d y } { d x } – \\frac { y } { x } = 3 x{\/tex}<\/span>
                          \nwhich is a linear differential equation of the form
                          \n{tex} \\frac { d y } { d x } + P y = Q{\/tex}<\/span>.
                          \nHere, {tex} P = \\frac { – 1 } { x }{\/tex}<\/span>and Q = 3x
                          \n{tex} \\therefore \\quad \\mathrm { IF } = e ^ { \\int P d x } = e ^ { \\int – \\frac { 1 } { x } d x } = e ^ { – \\log | x | } = e ^ { \\log x ^ { – 1 } } = x ^ { – 1 }{\/tex}<\/span>
                          \n{tex} \\Rightarrow \\quad \\mathrm { IF } = x ^ { – 1 } = \\frac { 1 } { x }{\/tex}<\/span>
                          \nThe solution of linear differential equation is given by
                          \n{tex} y \\times I F = \\int ( Q \\times I F ) d x + C{\/tex}<\/span>
                          \n{tex} \\Rightarrow \\quad y \\times \\frac { 1 } { x } = \\int \\left( 3 x \\times \\frac { 1 } { x } \\right) d x+C{\/tex}<\/span>
                          \n{tex} \\Rightarrow \\quad \\frac { y } { x } = \\int 3 d x+C \\Rightarrow \\frac { y } { x } = 3 x + C{\/tex}<\/span>
                          \n{tex} \\therefore \\quad y = 3 x ^ { 2 } + C x{\/tex}<\/span>
                          \nwhich is the required solution.<\/li>\n
                        12. Given differential equation is
                          \n(1 + y2<\/sup>) tan-1<\/sup>x dx + 2y (1 + x2<\/sup>) dy = 0
                          \n{tex}\\Rightarrow \\left( {1 + {y^2}} \\right){\\tan ^{ – 1}}xdx = – 2y\\left( {1 + {x^2}} \\right)dy{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\frac{{{{\\tan }^{ – 1}}xdx}}{{1 + {x^2}}} = – \\frac{{2y}}{{1 + {y^2}}}dy{\/tex}<\/span>
                          \nOn integrating both sides, we get
                          \n{tex}\\int {\\frac{{{{\\tan }^{ – 1}}x}}{{1 + {x^2}}}dx = – \\int {\\frac{{2y}}{{1 + {y^2}}}dy} } {\/tex}<\/span>
                          \nPut {tex}{\\tan ^{ – 1}}x = t\\;in\\,\\,LHS,{\/tex}<\/span> we get
                          \n{tex}\\frac{1}{{1 + {x^2}}}dx = dt{\/tex}<\/span>
                          \nand put 1 + y2<\/sup> = u in RHS, we get
                          \n2ydy = du
                          \n{tex} \\Rightarrow \\int {tdt = – \\int {\\frac{1}{u} \\Rightarrow \\frac{{{t^2}}}{2} = – \\log u + C} } {\/tex}<\/span>
                          \n{tex}\\Rightarrow \\frac{1}{2}{\\left( {{{\\tan }^{ – 1}}x} \\right)^2} = – \\log \\left( {1 + {y^2}} \\right) + C{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\frac{1}{2}{\\left( {{{\\tan }^{ – 1}}x} \\right)^2} + \\log \\left( {1 + {y^2}} \\right) = C{\/tex}<\/span><\/li>\n
                        13. Given differential equation is,
                          \n(1 + e2x<\/sup>)dy + (1 + y2<\/sup>)ex<\/sup> dx = 0
                          \nAbove equation may be written as
                          \n{tex}\\frac { d y } { 1 + y ^ { 2 } } = \\frac { – e ^ { x } } { 1 + e ^ { 2 x } } d x{\/tex}<\/span>
                          \nOn integrating both sides, we get
                          \n{tex}\\int \\frac { d y } { 1 + y ^ { 2 } } = – \\int \\frac { e ^ { x } } { 1 + e ^ { 2 x } } d x{\/tex}<\/span>
                          \nOn putting ex<\/sup> = t {tex}\\Rightarrow{\/tex}<\/span> ex<\/sup> dx = dt in RHS, we get
                          \n{tex}\\tan ^ { – 1 } y = – \\int \\frac { 1 } { 1 + t ^ { 2 } } d t{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad \\tan ^ { – 1 } y = – \\tan ^ { – 1 } t + C{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad \\tan ^ { – 1 } y = – \\tan ^ { – 1 } \\left( e ^ { x } \\right) + C{\/tex}<\/span> …(i) [put t = ex<\/sup>]\nAlso, given that y = 1, when x = 0.
                          \nOn putting above values in Eq. (i), we get
                          \ntan-1<\/sup>1 = -tan-1<\/sup>(e0<\/sup>) + C
                          \n{tex}\\Rightarrow \\quad \\tan ^ { – 1 }1 = – \\tan ^ { – 1 } 1 + C \\quad \\left[ \\because e ^ { 0 } = 1 \\right]{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad 2 \\tan ^ { – 1 } 1 = C{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad 2 \\tan ^ { – 1 } \\left( \\tan \\frac { \\pi } { 4 } \\right) = C{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad C = 2 \\times \\frac { \\pi } { 4 } = \\frac { \\pi } { 2 }{\/tex}<\/span>
                          \nOn putting {tex}C = \\frac { \\pi } { 2 }{\/tex}<\/span> in Eq. (i), we get
                          \n{tex}\\tan ^ { – 1 } y = – \\tan ^ { – 1 } e ^ { x } + \\frac { \\pi } { 2 }{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\quad y = \\tan \\left[ \\frac { \\pi } { 2 } – \\tan ^ { – 1 } \\left( e ^ { x } \\right) \\right] = \\cot \\left[ \\tan ^ { – 1 } \\left( e ^ { x } \\right) \\right]{\/tex}<\/span>
                          \n{tex}= \\cot \\left[ \\cot ^ { – 1 } \\left( \\frac { 1 } { e ^ { x } } \\right) \\right] \\left[ \\because \\tan ^ { – 1 } x = \\cot ^ { – 1 } \\frac { 1 } { x } \\right]{\/tex}<\/span>
                          \n{tex}\\therefore \\quad y = \\frac { 1 } { e ^ { x } }{\/tex}<\/span>
                          \nwhich is the required solution.<\/li>\n
                        14. {tex}\\left( {1 + {e^{\\frac{x}{y}}}} \\right)dx + {e^{\\frac{x}{y}}}\\left( {1 – \\frac{x}{y}} \\right)dy = 0{\/tex}<\/span>
                          \n{tex}\\Rightarrow\\frac{{dx}}{{dy}} = – \\frac{{{e^{x\/y}}\\left( {1 – \\frac{x}{y}} \\right)}}{{1 + {e^{x\/y}}}}{\/tex}<\/span>
                          \n{tex}\\Rightarrow\\frac{{dx}}{{dy}} = \\frac{{{e^{x\/y}}\\left( {\\frac{x}{y} – 1} \\right)}}{{1 + {e^{x\/y}}}}{\/tex}<\/span>……..(1)
                          \nLet x = vy, then,
                          \n{tex}\\frac{{dx}}{{dy}} = v + y\\frac{{dv}}{{dy}}{\/tex}<\/span>
                          \nPut {tex}\\frac{{dx}}{{dy}}{\/tex}<\/span> in eq (1),we get,
                          \n{tex}v + y \\frac{{dv}}{{dy}} = \\frac{{{e^v}(v – 1)}}{{{e^v} + 1}}{\/tex}<\/span>
                          \n{tex}\\Rightarrow y\\frac{{dv}}{{dy}} = \\frac{{v{e^v} – {e^v}}}{{{e^v} + 1}} – v{\/tex}<\/span>
                          \n{tex}\\Rightarrow y\\frac{{dv}}{{dy}} = \\frac{{v{e^v} – {e^v} – v{e^v} – v}}{{{e^v} + 1}}{\/tex}<\/span>
                          \n{tex} \\Rightarrow- \\int {\\frac{{dy}}{y}} = \\int {\\frac{{{e^v} + 1}}{{v + {e^v}}}dv} {\/tex}<\/span>
                          \n{tex}\\Rightarrow\\log ({e^v} + v) = – \\log (y) + c{\/tex}<\/span>
                          \n{tex}\\Rightarrow\\log (({e^v} + v).y) = c{\/tex}<\/span>
                          \n{tex}\\Rightarrow({e^v} + v)y = {e^c}{\/tex}<\/span>
                          \n{tex}\\Rightarrow({e^v} + v)y = A{\/tex}<\/span> [Putting ec<\/sup> = A]\n{tex}\\Rightarrow\\left( {{e^{x\/y}} + \\frac{x}{y}} \\right)y = A{\/tex}<\/span>
                          \n{tex}\\Rightarrow y{e^{x\/y}} + x = A{\/tex}<\/span><\/li>\n<\/ol>\n

                          Chapter Wise Important Questions Class 12 Maths Part I and Part II<\/h2>\n
                            \n
                          1. Relations and Functions<\/a><\/li>\n
                          2. Inverse Trigonometric Functions<\/a><\/li>\n
                          3. Matrices<\/a><\/li>\n
                          4. Determinants<\/a><\/li>\n
                          5. Continuity and Differentiability<\/a><\/li>\n
                          6. Application of Derivatives<\/a><\/li>\n
                          7. Integrals<\/a><\/li>\n
                          8. Application of Integrals<\/a><\/li>\n
                          9. Differential Equations<\/a><\/li>\n
                          10. Vector Algebra<\/a><\/li>\n
                          11. Three Dimensional Geometry<\/a><\/li>\n
                          12. Linear Programming<\/a><\/li>\n
                          13. Probability<\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                            Differential Equations Class 12 Mathematics Extra Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in\u00a0myCBSEguide\u00a0website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1346,1432],"tags":[1867,1839,1838,1833,1832,1854],"class_list":["post-27925","post","type-post","status-publish","format-standard","hentry","category-cbse","category-mathematics-cbse-class-12","tag-cbse-class-12-mathematics","tag-extra-questions","tag-important-questions","tag-latest-exam-questions","tag-practice-questions","tag-practice-test"],"yoast_head":"\nDifferential Equations Class 12 Mathematics Extra Questions<\/title>\n<meta name=\"description\" content=\"Differential Equations Class 12 Mathematics Extra Questions These 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