Differential Equation Notes for Class XII / CBSE / NCERT/ISC / JEE Main, Study notes of Mathematics

Download Differential Equation Notes for Class XII / CBSE / NCERT/ISC / JEE Main and more Study notes Mathematics in PDF only on Docsity! DIFFERENTIAL EQUATION A. DEFINITION An equation that involves independent and dependent variables and the derivatives of the dependent variables is called a DIFFERENTIAL EQUATION. There are two types of differential equation : (i) Ordinary Differential Equation : A differential equation is said to be ordinary, if the differential coefficients have reference to a single independent variable only e.g. dx dy2 dx yd 2 2  + cos x = 0 (ii) Partial Differential Equation : A differential equation is said to be partial, if there are two or more independent variables, e.g. z u y u x u         = 0 is a partial differential equation. We are concerned with ordinary differential equations only. (a) Solution (primitive) of differential equation : Finding the unknown function which satisfies given differential equation is called SOLVING OR INTEGRATING the differential equation. The solution of the differential equation is also called its PRIMITIVE, because the differential equation can be regarded as a relation derived from it. (b) Order of differential equation : The order of a differential equation is the order of the highest differential coefficient occurring in it. (c) Degree of differential equation : The degree of a differential equation which can be written as a polynomial in the derivatives is the degree of the derivative of the highest order occurring in it, after it has been expressed in a form which is free from radicals and fractions so far as derivatives are concerned, thus the differential equation f (x, y) q 1m 1mp m m dx )y(d )y,x( dx yd                    + ....= 0 is order m and degree p. Note that : In the differential equation ey’’ – xy’’ + y = 0 order is three but degree doesn’t apply. Ex.1 Find the order and degree of the following differential equation : (i) 3 2 2 3 dx dy dx yd  (ii) 2 2 dx yd = sin       dx dy (iii) 5x3 dx dy  Sol. (i) The given differential equation can be re-written as 23 2 2 3 dx dy dx yd                Hence order is 2 and degree is 3. (ii)The given differential equation has the order 2. Since the given differential equation cannot be written as a polynomial in the differential coefficients, the degree of the equation is not defined. (iii)Its order is obviously 1 and degree 1. DIFFERENTIAL EQUATION DIFFERENTIAL EQUATI 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com Page # 17 Ex.2 The order and degree of the differential equation 2 32 2 d s ds 3 dtdt             + 4 = 0 are Sol. Clearly order is 2 and degree is 2. (from the definition of order and degree of differential equation). B. FORMATION OF A DIFFERENTIAL EQUATION If an equation with independent and dependent variables having some arbitrary constant is given, then a differential equation is obtained as follows : (a)Differentiate the given equation w.r.t. the independent variable (say x) as many times as the number of arbitrary constants in it. (b)Eliminate the arbitrary constants. The eliminant is the required differential equation. Note : A differential equation represents a family of curves all satisfying some common properties. This can be considered as the geometrical interpretation of the differential equation. Ex.3 Find the differential equation of all parabolas whose axes is parallel to the x-axis and having latus rectum a. Sol. Equation of parabola whose axes is parallel to x-axis and having latusrectum ‘a’ is (y – )2 = a (x –) Differentiating both sides, we get 2(y – ) dx dy = a  2(y – ) 2 2 2 dx dy 2 dx yd        =0  a. 3 2 2 dx dy 2 dx yd        =0 Ex.4 Find the differential equation whose solution represents the family : c (y + c)2 = x3 Sol. c (y + c)2 = x3 ...(i) Differentiating, we get, c.[2(y + c)] dx dy = 3x2 Writing the value of c from (i), we have 2 3 2 2 3 x3 dx dy cy x2 x3 dx dy )cy( )cy( x2     i.e. 3 dx dy cy x2          dx dy 3 x2 = y + c Hence c =       dx dy 3 x2 – y Substituting value of c in equation (i), we get 2 dx dy 3 x2 y dx dy 3 x2                   = x3, Which is the required differential equation. C. GENERAL AND PARTICULAR SOLUTIONS The solution of a differential equation which contains a number of independent arbitrary constants equal to the order of the differential equation is called the GENERAL SOLUTION (OR COMPLETE INTEGRAL OR COMPLETE PRIMITIVE). A solution obtainable from the general solution by giving particular values to the constants is called a PARTICULAR SOLUTION. DIFFERENTIAL EQUATION Page # 18 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com Equations reducible to the homogeneous form : If 222 1111 cybxa cybxa dx dy    ; where a1b2 – a2b1  0, i.e. 2 2 1 1 b a b a  then the substitution x = u + h, y = v + k transform this equation to a homogeneous type in the new variables u and v where h and k are arbitrary constants to be chosen so as to make the given equation homogeneous which can be solved by the method as given in (b). If (i) a1b2 – a2b1 = 0, then a substitution u = a1 x + b1 y transforms the differential equation to an equation with variables separable and (ii) In an equation of the form : yf(xy)dx + xg(xy)dy = 0 the variables can be separated by the substitution xy = v. Important note : (i) The function f(x, y) is said to be a homogeneous function of degree n if for any real number t(0), we have f(tx, ty) = tn f(x, y). For e.g. f(x, y) = ax2/3 + hx1/3 . y1/3 + by2/3 is a homogeneous function of degree 2/3. (ii) A differential equation of the form dx dy = f(x , y) is homogeneous if f(x, y) is a homogeneous function of degree zero i.e. f(tx, ty) = tº f(x, y) = f(x, y). The function f does not depend on x and y separately but only on their ratio y x or x y . Ex.10 Solve dy dx = x 2y 3 2x 3y 4     Sol. Put x = X + h, y = Y + k. We have dX dY = )4k3h2(Y3X2 )3k2h(Y2X   To determine h and k, we write h + 2k + 3 = 0, 2h + 3k + 4 = 0  h = 1, k = –2 so that dX dY = Y3X2 Y2X   Putting Y= VX, we get V + X dX dV = V32 V21    1V3 V32 2   dV = – X dX               )1V3(2 32 )1V3(2 32 dV = – X dX  32 32  log ( 3 V – 1) – 32 32  log ( 3 V + 1) = (– log X + c) 32 32  log ( 3 Y – X) – 32 32  log ( 3 Y + X) = A where A is another constant and X = x – 1, Y= y+ 2. Ex.11 Solve the differential equation (1 + 2ex/y) dx + 2ex/y (1 – x/y) dy = 0. Sol. The equation is homogeneous of degree 0. Put x = vy, dx = v dy + y dv, Then (1 + 2ev) (v dy + y dv) + 2ev (1 – v) dy = 0  (v + 2ev) dy + y (1 + 2 ev) dv = 0 y dy + v v e2v e21   dv = 0. Integrating and replacing v by x/y n y + n (v + 2ev) = nC and x + 2 yex/y = c DIFFERENTIAL EQUATIONPage # 21 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com (3) LINEAR DIFFERENTIAL EQUATIONS : A differential equation is said to be linear if the dependent variable & its differential coefficients occur in the first degree only and are not multiplied together. The nth order linear differential equation is of the form ; a0 (x) n n dx yd + a1(x) 1n 1n dx yd   +...........+ an (x) . y =  (x), where a0 (x), a1 (x) ... an (x) are called the coefficients of the differential equation. Note that : A linear differential equation is always of the first degree but every differential equation of the first degree need not be linear. e.g. the differential equation 2 2 dx yd + 3 dx dy       + y2 = 0 is not linear, though its degree is 1. Ex.12 Which of the following equation is linear ? (A) dy dx + xy2 = 1 (B) x2 dy dx + y = ex (C) dy dx + 3y = xy2 (D) x dy dx + y2 = sin x Sol. Clearly answer is (B) Ex.13 Which of the following equation is non-linear ? (A) dx dy = cos x (B) 2 2 dx yd + y = 0 (C) dx + dy = 0 (D) x dx dy + dx dy 3 = y2 Sol. Clearly answer is (D) (a) Linear differential equations of first order : The most general form of a linear differential equations of the first order is dx dy + Py = Q, where P & Q are functions of x. To solve such an equation multiply both sides by Pdx e . Note : (i) The factor Pdx e on multiplying by which the left hand side of the differential equation becomes the differential coefficient of some function of x & y, is called integrating factor of the differential equation popularly abbreviated as I.F. (ii) It is very important to remember that on multiplying by the integrating factor, the left hand side becomes the derivative of the product of y and the I.F. (iii) Some times a given differential equation becomes linear if we take y as the independent variable and x as the dependent variable. e.g. the equation ; (x + y + 1) dx dy = y2 + 3 can be written as (y2 + 3) dy dx = x + y + 1 which is a linear differential equation. DIFFERENTIAL EQUATION Page # 22 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com Ex.14 Solve (1 + y2) + (x – 1tan ye  ) dy dx = 0. Sol. Differential equation can be rewritten as (1 + y2) dy dx + x = ytan 1 e  or dy dx + 2y1 1  .x = 2 ytan y1 e 1   ....(i) I. F = ytan dy y1 1 12 ee     so solution is dy y1 ee xe 2 ytanytan ytan 11 1      Let ytan 1 e  = t  2 ytan y1 e 1   dy = dt    dttxe ytan 1 [Putting ytan 1 e  = t] or 2 c 2 t xe 2 ytan 1    .ceex2 ytan2ytan 11   Ex.15 The solution of differential equation (x2 – 1) dy dx + 2xy = 2 1 x 1 is Sol. The given differential equation is (x2 – 1) dx dy + 2 xy = 1x 1 2   dx dy + 1x x2 2  y = 22 )1x( 1  This is linear differential equation of the form dx dy + Py = Q, where P = 1x x2 2  and Q = 22 )1x( 1   I.F. = )1x(eee 2)1xlog(dx)1x/(x2dxP 22   multiplying both sides of (i) by I.F. = (x2 – 1), we get (x2 – 1) dx dy + 2 xy = 1x 1 2  integrating both sides we get y(x2 – 1) =  1x 1 2 dx + C [Using : y (I.F.) =  .)F.I.(Q dx + C]  y(x2 – 1) = 2 1 log 1x 1x   + C. This is the required solution. DIFFERENTIAL EQUATIONPage # 23 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com E. TRAJECTORIES A curve which cuts every member of a given family of curves according to a given law is called a Trajectory of the given family. A curve making at each of its points a right angle with the curve of the family passing through that point is called an orthogonal trajectory of that family. ORTHOGONAL TRAJECTORIES : We set up the differential equation of the given family of curves. Let it be of the form F(x, y, y ' ) = 0 The differential equation of the orthogonal trajectories is of the form F        'y 1 ,y,x = 0 The general integral of this equation 1(x, y, C) = 0 gives the family of orthogonal trajectories. Ex.20 Find the value of k such that the family of parabolas y = cx2 + k is the orthogonal trajectory of the family of ellipses x2 + 2y2 – y = c. Sol. Differentiate both sides of x2 + 2y2 – y = c w.r.t. x, we get 2x + 4y dx dy – dx dy = 0 or 2x + (4y – 1) dx dy = 0, is the differential equation of the given family of curves. Replacing dx dy by – dy dx to obtain the differential equation of the orthogonal trajectories, we get 2x + dx dy )y41(  = 0  dx dy = x2 1y4    1y4 dy =  x2 dx  4 1 n (4y – 1) = 2 1 n x + 2 1 n a, where a is any constant.  n(4y – 1) = 2 n x + 2 n a or , 4y – 1 = a2x2 or, y = 4 1 a2x2 + 4 1 , is the required orthogonal trajectory, which is of the form y = cx2 + k where c = 4 a2 , k = 4 1 DIFFERENTIAL EQUATION Page # 26 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com Ex.21 Solve (y log x – 1) ydx = xdy. Sol. The given differential equation can be written as x dx dy + y = y2 log x ........(i) Divide by xy2 . Hence 2y 1 dx dy + xy 1 = x 1 log x Let y 1 = v  – 2y 1 dx dy = dx dv so that dx dv – x 1 v = – x 1 log x ........(ii) (ii) is the standard linear differential equation with P = – x 1 , Q = – x 1 log x I.F. = x/1ee dxx/1pdx    The solution is given by v . x 1 =         dxxlog x 1 x 1 = –  2x xlog dx = x xlog –  x 1 . x 1 dx = x xlog + x 1 + c  v = 1 + log x + cx = log ex + cx or y 1 = log ex + cx or y (log ex + cx) = 1. Ex.22 The solution of 2(y 1) x ye  {xy2 dy + y3 dx} + {ydx – xdy} = 0, is Sol. Hence y )1y( x 2 e  . y2 {xdy + ydx} + {ydx – xdy} = 0  exy . y2. {xdy + ydx} + ex/y {ydx – xdy} = 0  exy . {xdy + ydx} + ex/y 2y }xdyydx{  = 0  exy . d (xy) + ex/y . d       y x = 0 or d(exy) + d(ex/y) = 0, Integrating both sides we get exy + ex/y + c = 0. Ex.23 For a certain curve y = f(x) satisfying 2 2 d y 6x 4 dx   f(x) has a local minimum value 5 when x = 1. Find the equation of the curve and also the global maximum and global minimum values of f(x) given that DIFFERENTIAL EQUATIONPage # 27 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com 0 x 2  . Sol. Integrating 4x6 dx yd 2 2  , we get Ax4x3 dx dy 2  When x = 1, 0 dx dy  , so that A = 1. Hence 1x4x3 dx dy 2  ............(i) Integrating, we get Bxx2xy 23  When x = 1, y = 5, so that B = 5. Thus we have 5xx2xy 23  . From (i), we get the critical points x = 1/3, x = 1. At the critical point 2 2 dx yd , 3 1 x  is negative. Therefore at x = 1/3, y has a local maximum. At 2 2 dx yd ,1x  is positive. Therefore at x = 1, y has a local minimum Also 7)2(f,5)0(f. 27 157 3 1 f,5)1(f        Hence the global maximum value = 7, and the global minimum value = 5. Ex.24 Determine the equation of the curve passing through the origin, in the form y = f(x), which satisfies the differential equation dy sin(10x 6y) dx   . Sol. We have )y6x10sin( dx dy  . ........(i) Let t = 10x + 6y tsin610 dx dy 610 dx dt  Cx2 tsin35 dt dx2 tsin35 dt       .....(ii) Putting z 2 t tan  , we get Cx2 5 z1 z6 )z1( dz2 2 2            from (ii) DIFFERENTIAL EQUATION Page # 28 394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-info@motioniitjee.com