Differential Equations PDF

60 394 Differential Equations E3 Many of the practical problems in physics and engineering can be converted into differential equations. The solution of differential equations is, therefore of paramount importance. This chapter deals with some elementary aspects of differential equations. These are...

60 394 Differential Equations E3 Many of the practical problems in physics and engineering can be converted into differential equations. The solution of differential equations is, therefore of paramount importance. This chapter deals with some elementary aspects of differential equations. These are addressed through simple application of differential and integral calculus. ID There are two important aspects of differential equation, which have just been touched in this chapter. How to formulate a problem as a differential equation is the one, and the other is how to solve it. U 8.1 Definition. An equation involving independent variable x, dependent variable y and the differential Examples : D YG dy d 2 y , ,........ is called differential equation. dx dx 2 coefficients (i) dy 1 x y dx (ii) 3  d4y  dy (iii)  4   4  4 y  5 cos 3 x dx  dx  dy  xy  cot x dx 2 (iv) x 2 d 2y  dy   1   0 2 dx  dx  U (1) Order of a differential equation : The order of a differential equation is the order of the highest derivative occurring in the differential equation. For example, the order of above differential equations are 1,1,4 and 2 respectively. ST Note :  The order of a differential equation is a positive integer. To determine the order of a differential equation, it is not needed to make the equation free from radicals. (2) Degree of a differential equation : The degree of a differential equation is the degree of the highest order derivative, when differential coefficients are made free from radicals and fractions. In other words, the degree of a differential equation is the power of the highest order derivative occurring in differential equation when it is written as a polynomial in differential coefficients. Note :  The definition of degree does not require variables x, y, t etc. to be free from radicals and fractions. The degree of above differential equations are 1, 1, 3 and 2 respectively. Differential Equations 395 2 The order and degree of the differential equation y  x (a) 1, 2 Solution: (a) (b) 2, 1 (c) 1, 1 (d) 2, 2 dy , having order 1. dx equation as a polynomial Clearly, highest order derivative involved is Expressing 2 dy  dy   a 2    b 2 are dx  dx  the above differential in derivative, we have 60 Example: 1 2 dy    dy  y  x   a2    b 2 dx    dx  2 The order and degree of the differential equation (a) 2, 3 Solution: (a) d 2y dx 2 (b) 3, 3 1 1  dy  3     x 4  0 , are respectively  dx  (c) 2, 6 d 2y (d) 2, 4 which is the 2nd order derivative. Hence order of the dx 2 differential equation is 2. Making the above equation free from radical, as far as the derivatives are concerned, we have The highest order derivative involved is 3 3 ID Example: 2 E3 dy  dy  i.e., (x 2  a 2 )    2 xy  y2  b2  0 dx  dx  In this equation, the power of highest order derivative is 2. So its degree is 2. U 1  1   d 2y  d 2y dy dy   4   4    x i.e.  x 0  dx 2   dx 2  dx dx     Example: 3 The degree of the differential equation (a) 1 Solution: (d) dx 2 will be 3. Hence degree of the differential equation is 3. 2  d 2y  d 2y  dy  2    3  x log    dx 2  is dx 2  dx    (b) 2 (c) 3 Solution: (c) The order of the differential equation whose solution is x 2  y 2  2 gx  2 fy  c  0, is (a) 1 (b) 2 (c) 3 (d) 4 To eliminate the arbitrary constants g, f and c, we need 3 more equations, that by differentiating the ST equation 3 times. Hence highest order derivative will be Example: 5 Solution: (a) (d) None of these  d 2y  The above equation cannot be written as a polynomial in derivatives due to the term x 2 log  2 .  dx  Hence degree of the differential equation is ‘not defined’. U Example: 4 d 2y D YG The exponent of highest order derivative d 3y. Hence order of the differential dx 3. equation will be 3. The order of the differential equation of all circles of radius r, having centre on y-axis and passing through the origin is (a) 1 (b) 2 (c) 3 (d) 4 2 The equation of the circle will be x 2  y  r   r 2. Hence the constant to be eliminated is r.  order of differential equation =1 Y Y O (0, r) O X X (0, – r) 396 Differential Equations Example: 6 The order of the differential equation, whose general solution is y  c1e x  c 2 e 2 x  c3 e 3 x  c4 e x  c 5 , where c1 , c2 , c3 , c4 , c5 are arbitrary constants is Solution: (c) (a) 5 (b) 4 Rewriting the given general solution, we have (c) 3 (d) None of these y  c1 e x  c 2 e 2 x  c 3 e 3 x  c 4 e x.e c5  (c 1  c 4.e c5 ) e x  c 2 e 2 x  c 3 e 3 x  c1 e x  c 2 e 2 x  c 3 e 3 x 60 where c 1  c 1  c 4.e c5. So there are 3 arbitrary constant associated with different terms. Hence the order of the differential equation formed, will be 3. The degree of the differential equation satisfying (a) 1 Solution: (a) (b) 2 (c) 3 2 dy will be1. Hence degree is dx 3 1  dy  1  dy  dy    ....... is    3 !  dx  dx 2 !  dx  ID The order and degree of y  1  (a) 1, 2 (b) 1, 1 (c) Order 1, degree not defined (d) The given differential equation can be re-written as y  e dy dx  ln y  U Solution: (b) (d) None of these To eliminate a the above equation is differentiated once and exponent of 1 Example: 8 1  x 2 + 1  y 2 = a( x  y ) is E3 Example: 7 None of these dy dx This is a polynomial in derivative. Hence order is 1 and degree 1. The order and degree of d 2y 2  dy   sin   x is  dx  D YG Example: 9 dx (a) 2, 1 (b) Order 2, degree not defined (c) 2, 0 Solution: (b) As the highest order derivative involved is (d) None of these 2 d y dx 2. Hence order is 2. The given differential equation cannot be written as a polynomial in derivatives, the degree is not defined. U 8.2 Formation of Differential Equation. ST Formulating a differential equation from a given equation representing a family of curves means finding a differential equation whose solution is the given equation. If an equation, representing a family of curves, contains n arbitrary constants, then we differentiate the given equation n times to obtain n more equations. Using all these equations, we eliminate the constants. The equation so obtained is the differential equation of order n for the family of given curves. Consider a family of curves f (x , y, a1 , a 2....., a n )  0......(i) where a1 , a 2 ,...... a n are n independent parameters. Equation (i) is known as an n parameter family of curves e.g. y = mx is a one-parameter family of straight lines. x 2  y 2  ax  by  0 is a two parameters family of circles. If we differentiate equation (i) n times w.r.t. x, we will get n more relations between x , y, a1 , a 2 ,.....an and derivatives of y w.r.t. x. By eliminating a1 , a 2 ,...., an from these n relations and equation (i), we get a differential equation. Differential Equations 397 Clearly order of this differential equation will be n i.e. equal to the number of independent parameters in the family of curves. Algorithm for formation of differential equations Step (i) : Write the given equation involving independent variable x (say), dependent variable y (say) and the arbitrary constants. 60 Step (ii) : Obtain the number of arbitrary constants in step (i). Let there be n arbitrary constants. Step (iii) : Differentiate the relation in step (i) n times with respect to x. Differential equation whose general solution is y  c1 x  (a) Solution: (d) d 2 y x 2 dy   0 y dx dx 2 y  c1 x  (b) d 2y y dy   0 dx 2 x 2 dx c2 for all values of c1 and c 2 is x (c) d 2y 1 dy  0 dx 2 2 x dx ID Example: 10 E3 Step (iv) : Eliminate arbitrary constants with the help of n equations involving differential coefficients obtained in step (iii) and an equation in step (i). The equation so obtained is the desired differential equation. c2 x (d) d 2 y 1 dy y   0 dx 2 x dx x 2.....(i) D YG U There are two arbitrary constants. To eliminate these constants, we need to differentiate (i) twice. Differentiating (i) with respect to x, dy c  c1  22.....(ii) dx x Again differentiating with respect to x, d 2 y 2c2  3 dx 2 x From (iii), c2   c1 ......(iii) dy c2 x 3 d 2y  and from (ii), c1  ; dx x 2 2 dx 2 dy x d 2 y  dx 2 dx 2 U  dy x d 2 y  x 2 d 2y d 2y dy   2  x   2  y  x2 From (i), y   x 2 dx 2 2 dx dx dx dx   d 2 y 1 dy y   0 dx 2 x dx x 2 ST  Example: 11 y x is a solution of the differential equation x 1 (a) y 2 Solution: (b) dy  x2 dx We have y  (b) x 2 dy  y2 dx (c) y dy x dx x 1 x 1 1   1 x 1 y x x Differentiating w.r.t. x, 1 dy 1  2 0 2 y dx x  x2 Example: 12 dy  y2 dx The differential equation of all parabolas whose axes are parallel to y axis is (d) x dy y dx 398 Differential Equations (a) Solution: (a) d 3y 0 dx 3 (b) d2x c dy 2 (c) d 3y d 2y  0 dx 3 dx 2 ……..(i) There are three arbitrary constants ,  and a. We need to differentiate (i) 3 times dy Differentiating (i) w.r.t. x, 2(x   )  4 a dx 60 Y Again differentiating w.r.t. x, d 2y d 2y 1   2 dx 2 2 a dx X O E3 2  4a (, ) Differentiating w.r.t. x, d 3y 0 dx 3 The differential equation of family of curves whose tangent form an angle of /4 with the hyperbola xy  c is (a) dy x 2  c 2  2 dx x  c2 (b) ID 2 dy x 2  c 2  dx x 2  c 2 (c) The slope of the tangent to the family of curves is m 1  U Solution: (b) d 2y dy 2 c dx dx 2 The equation of a parabola whose axis is parallel to y-axis may be expressed as (x   ) 2  4 a(y   ) Example: 13 (d) Equation of the hyperbola is xy  c 2  y  dy c2  2 dx x  Slope of tangent to xy  c 2 is m 2   dy dx c2 x c2 x2 dy c 2  2 dy m1  m 2 Now tan   1  dx 2 x  dx 4 1  m1m 2 c dy 1 2 x dx  2  2    1  c   1  c   x 2   x 2   dy x 2  c 2  dx x 2  c 2 U  (d) None of these D YG  dy c2  2 dx x ST 8.3 Variable Separable type Differential Equation. (1) Solution of differential equations : If we have a differential equation of order ‘n’ then by solving a differential equation we mean to get a family of curves with n parameters whose differential equation is the given differential equation. Solution or integral of a differential equation is a relation between the variables, not involving the differential coefficients such that this relation and the derivatives obtained from it satisfy the given differential equation. The solution of a differential equation is also called its primitive. For example y  e x is a solution of the differential equation dy y. dx (i) General solution : The solution which contains as many as arbitrary constants as the order of the differential equation is called the general solution of the differential equation. For Differential Equations 399 example, y  A cos x  B sin x is the general solution of the differential equation d 2y  y  0. But dx 2 y  A cos x is not the general solution as it contains one arbitrary constant. (ii) Particular solution : Solution obtained by giving particular values to the arbitrary constants in the general solution of a differential equation is called a particular solution. For d 2y y 0 dx 2 60 example, y  3 cos x  2 sin x is a particular solution of the differential equation E3 (2) Differential equations of first order and first degree : A differential equation of first dy order and first degree involves x, y and. So it can be put in any one of the following forms: dx dy dy    f (x , y) or f  x , y,   0 or f (x , y ) dx  g(x , y ) dy  0 where f (x , y ) and g( x , y ) are obviously the dx  dx  functions of x and y. ID (3) Geometrical interpretation of the differential equations of first order and first dy   degree : The general form of a first order and first degree differential equation is f  x , y,   0 dx   U.....(i) D YG We know that the direction of the tangent of a curve in Cartesian rectangular coordinates at dy any point is given by , so the equation in (i) can be known as an equation which establishes dx dy the relationship between the coordinates of a point and the slope of the tangent i.e., to the dx integral curve at that point. Solving the differential equation given by (i) means finding those curves for which the direction of tangent at each point coincides with the direction of the field. All the curves represented by the general solution when taken together will give the locus of the differential equation. Since there is one arbitrary constant in the general solution of the curves. U equation of first order, the locus of the equation can be said to be made up of single infinity of ST (4) Solution of first order and first degree differential equations : A first order and first degree differential equation can be written as f(x, y)dx + g(x, y)dy = 0 or dy f (x , y ) dy  or   (x , y ) dx g(x , y )' dx Where f (x , y ) and g( x , y ) are obviously the functions of x and y. It is not always possible to solve this type of equations. The solution of this type of differential equations is possible only when it falls under the category of some standard forms. (5) Equations in variable separable form : If the differential equation of the form f1 (x )dx  f2 (y )dy.....(i) 400 Differential Equations where f1 and f2 being functions of x and y only. Then we say that the variables are separable in the differential equation. Thus, integrating both sides of (i), we get its solution as  f1 (x )dx   f2 (y )dy  C , where c is an arbitrary constant. 60 There is no need of introducing arbitrary constants to both sides as they can be combined together to give just one. dy  f(x) dx E3 (i) Differential equations of the type To solve this type of differential equations we integrate both sides to obtain the general dy solution as discussed following :  f (x )  dy  f (x ) dx dx (ii) Differential equations of the type ID  dy   f (x ) dx  C or y   f (x) dx  C. Integrating both sides, we obtain, dy  f (y ) dx U To solve this type of differential equations we integrate both sides to obtain the general D YG solution as discussed following : dy dx 1 1  f (y )    dx  dy dx dy f (y ) f (y ) Integrating both sides, we obtain, 1  dx   f (y) dy  C or x  1  f (y) dy  C. (6) Equations reducible to variable separable form dy  f (ax  by  c) can be reduced to variable separable dx ax + by + c = Z (i) Differential equations of the form form by the substitution dy dZ  dx dx U  ab dZ  dZ 1  a   f (Z)   a  bf (Z).  dx b dx   ST  This is variable separable form. (ii) Differential equation of the form dy ax  by  c a b  , where   K (say) dx Ax  By  C A B  dy K ( Ax  By )  C  dx Ax  By  C Put Ax + By = Z  AB dy dZ dZ KZ  C  dZ  1 KZ  C ,     A    AB ZC dx dx dx ZC  dx B Differential Equations 401 This is variable separable form and can be solved. (b) tan 1 y  log(1  x 2 )  c (c) 2 tan 1 y  log(1  x 2 )  c  0 (d) None of these  1 x 2 x dx 2 2  1y x 3 cos 3 x  c 3 3 (b) y  We have dy  (x 2  sin 3 x )dx  Example: 16 The solution of  2 tan 1 y  log e (1  x 2 )  c dy  x 2  sin 3 x is dx x 3 cos 3 x  c 3 3  dy   (x dy 1  is dx y 2  sin y y3  cos y  c 3 2 [DSSE 1981] (c) y   sin 3 x )dx  y  x3  sin 3 x  c 3 (b) y + cosy = x + c (d) None of these x 3 cos 3 x  c 3 3 (c) x  y3  cos y  c 3 (d) None of these U (a) x  Given equation may be re-written as dx  (y 2  sin y)dy  dx   (y 2  sin y ) dy D YG Integrating,  x y3  cos y  c 3 The solution of the differential equation dy  (4 x  y  1)2 is dx (a) 4x – y + 1 = 2 tan(2x – 2c) (b) 4x – y – 1 = 2 tan(2x – (c) 4x + y + 1 = 2 tan(2x + 2c) (d) None of these 2c) Let 4x + y + 1 = z  4  U Solution: (c) 2 The solution of the differential equation Solution: (b) Example: 17 dy 60 Separating the variables, we can re-write the given differential equation as (a) y  Solution: (a) [AISSE 1983] (a) 2 tan 1 y  log(1  x 2 )  c xdx dy   1  x 2 1  y2 Example: 15 dy  x (1  y 2 ) is dx E3 Solution: (a) The solution of the differential equation (1  x 2 ) ID Example: 14 dy  (4 x  y  1)2 dx ST  dy dz dy dz   4  dx dx dx dx  dz dz dz z 1  4x  y 1   4  z2   z2  4  2  dx  tan 1  x  c  tan 1    2 x  2c 2 dx dx 2 2 z 4    4x + y + 1 = 2tan (2x + 2c) Example: 18 Solution of the differential equation dy x  y 7 is  dx 2 x  2y  3 (a) 6(x + y) + 11 log(3x + 3y + 10) = 9x + c (b) 6(x + y) – 11 log(3x + 3y + 10) = 9x + c 10   (c) 6(x  y )  11 log  x  y    9x  c 3   (d) None of these Solution: (b, c) Given equation may be re-written as Let x + y = z dy x  y 7  dx 2(x  y)  3 402 Differential Equations  1  dy dz dy dz    1 dx dx dx dx dz z 7 1  dx 2z  3  2 11 z log( 3 z  10 )  x  c1  6 z  11 log( 3 z  10 )  9 x  9c1 3 9  6(x  y )  11 log( 3 x  3 y  10 )  9 x  c [9 c1  c]  2 11 dz  3 9  3 dz  3 z  10  dx 60 2 11 (3 z  10 )  dz 2z  3 z 7 3 z  10 3 3 dz  dx    1 dz  dx   dx 3 z  10 2z  3 2z  3 3 z  10 10    6(x  y )  11 log  x  y    9x  k 3   E3 10  10     6(x  y )  11 log 3  x  y    9 x  (c  11 log 3)   9 x  c  6(x  y )  11 log  x  y  3  3    (k = c + 11 log 3) ID 8.4 Homogeneous Differential Equation. (1) Homogeneous differential equation : A function f(x, y) is called a homogeneous function of degree n if f (x , y )  n f (x , y). U For example, f (x , y)  x 2  y 2  3 xy is a homogeneous function of degree 2, because f (x , y )  2 x 2  2 y 2  3x. y  2 f (x , y). A homogeneous function f(x, y) of degree n can always be written D YG x y f (x , y)  x n f   or f (x , y )  y n f  . If a first-order first degree differential equation is x y f (x , y ) dy  expressible in the form where f(x, y) and g(x, y) are homogeneous functions of the dx g(x , y ) as same degree, then it is called a homogeneous differential equation. Such type of equations can be reduced to variable separable form by the substitution y  vx. The given differential equation dy x n f (y / x ) f (y / x ) dy dv y  n   F . If y  vx , then. Substituting the vx dx g y x x ( / ) dx dx x g(y / x )   U can be written as dv dv dx dy y  F(v)  . On integration,  F  , we get v  x dx F(v)  v x dx x ST value of 1  F(v)  v dv   dx C x where C is an arbitrary constant of integration. After integration, v will be replaced by y in x complete solution. (2) Algorithm for solving homogeneous differential equation Step (i) : Put the differential equation in the form dy  (x , y)  dx  (x , y) dy dv in the equation in step (i) and cancel out x from the v  x dx dx dv right hand side. The equation reduces to the form v  x  F(v). dx Step (ii) : Put y = vx and Step (iii) : Shift v on RHS and separate the variables v and x Differential Equations 403 Step (iv) : Integrate both sides to obtain the solution in terms of v and x. Step (v) : Replace v by y in the solution obtained in step (iv) to obtain the solution in terms x of x and y. (3) Equation reducible to homogeneous form 60 A first order, first degree differential equation of the form a x  b1 y  c1 a b dy , where 1  1  1 dx a 2 x  b 2 y  c 2 a2 b 2.....(i) This is non-homogeneous. E3 It can be reduced to homogeneous form by certain substitutions. Put x  X  h, y  Y  k Where h and k are constants, which are to be determined. dy dy dY dX dY ..  dx dY dX dx dX Substituting these values in (i), we have.....(iii).....(ii) a1 h  b 1 k  c 1  0  a 2 h  b 2 k  c 2  0  h k 1   b1 c 2  b 2 c1 c1 a 2  c 2 a1 a1 b 2  a 2 b1 D YG i.e. (a X  b1 Y )  a1 h  b1 k  c1 dY  1 dX (a 2 X  b 2 Y )  a 2 h  b 2 k  c 2 U Now h, k will be chosen such that ID .....(iv) a X  b1 Y dY  1 dX a 2 X  b 2 Y For these values of h and k the equation (ii) reduces to which is a homogeneous differential equation and can be solved by the substitution Y = vX. Replacing X and Y in the solution so obtained by x  h and y  k respectively, we can obtain the required solution in terms of x and y. The solution of the differential equation x U Example: 19 ST (a) y  xe cx Solution: (a) (b) y  xe cx  0 Given equation may be expressed as Let dy  y(log y  log x  1) is dx (c) [IIT 1986] y  ex  0 dy y   y    log    1 dx x   x   …….(i) dy dv y vx  v  y = vx  dx dx x  From (i), v  x dv dv dv dx  v(log v  1)  x  v log v    dx dx v log v x  1 d (log v)  log v  log (log v) = log x + log c  log (log v) = log (cx)  log v = cx  v  e cx  Example: 20 (d) None of these  dx x y  e cx ,  y  xe cx x  y 2  (y 2 / x 2 )   is The solution of differential equation y y   x  2   (y 2 / x 2 )  x (a)  (y 2 / x 2 )  cx 2 (b) x 2(y 2 / x 2 )  c 2 y 2 (c) x 2 (y 2 / x 2 )  c (d)  (y 2 / x 2 )  cy x 404 Differential Equations 2 Solution: (a) Given equation may be re-written as Let y = vx  y dy  y   ((y / x )2 )     x dx  x   ((y / x )2 ).....(i) dy y dv vx v and dx x dx dv  (v 2 )  (v 2 )(2v dv )  (v 2 ) dv  dx   From (i), v v  x  vx  2    v2  2 2 dx  (v ) x dx   (v 2 )  (v )  60 Integrating, ln((v 2 ))  2 ln x  ln c   (v 2 )  cx 2   (y 2 / x 2 )  cx 2 dy y 3  2 x 2 y is  3 dx x  2 xy 2 The solution of (b) (x 2  y 2 )3  Bx 2 y 2 (a) (x 2  y 2 )3  Bx 2 y 2 Given equation is homogeneous. Let y = vx (d) None of these dy dv vx  dx dx  2 2    dv v 3  2v dv (y / x )3  2(y / x ) y 3  2 x 2y dv dv v 2   1v    v  1  v     x vx vx vx 2 2 2 3 2 2 dx dx 1  2 v 1  2 v     dx dx 1  2v 1  2(y / x ) x  2 xy      B D  dx 1  2v 2 1  2v 2 dx dx A  dv dv       dv  2 x v(1  v)(1  v) x x v(1  v )  v 1v 1v  ID Solution: (a) (c) (x 2  y 2 )3  x 2 y 2 E3 Example: 21 Putting v = 0, v = 1, A=1 3 B 2 3 2 D YG v = – 1, D   U where A(1  v)(1  v)  Bv (1  v)  Dv(1  v)  1  2v 2 3 1  dx 1 3 1     dv  x v 2 1v 2 1v  Integrating both side, we get ln v  3 ln(1  v) 3 3 3  ln(1  v)  ln x  ln c  ln v  ln(1  v)  ln(1  v)  ln cx 2 1 2 2 2 3 2 2  y2  x 2y 2  y   v   v /{(1  v)1  v}3 / 2  cx     (1  v 2 )3   2    1  2   (x 2  y 2 )3  2 x  c  cx   cx   U  1   (x 2  y 2 )3  Bx 2 y 2 ,  2  B   c  The solution of dy x  3 y  2 is  dx 3 x  y  6 (a) y 2  6(x  2)y  (x  2)2  c (b) y 2  6(x  2)y  (x  2)2  c (c) y 2  6(y  2)x  x 2  c (d) None of these ST Example: 22 Solution: (b) Given equation is non-homogeneous Let x = X + h, y = Y + k dy dY   dx dX  dY (X  h)  3(Y  k )  2 X  3Y  (h  3k  2)   dX 3(X  h)  (Y  k )  6 3 X  Y  (3h  k  6) Let us select h and k so that h – 3k + 2 = 0 and 3h – k + 6 = 0 Solving, k = 0, h = – 2  X = x – h = x + 2, Y  y  k  y  dY X  3Y  , which is homogeneous dX 3X  Y Now, let Y = vX Differential Equations 405  dY dv X  3Y dv 1  3v dv 1  3(Y / X ) dv    vX vX vX vX dX dX 3X  Y dX 3 v dX 3  (Y / X ) dX  X (3  v)dv dX 2v  6 dX dv 1  3v v 2  6v  1  2  2  dv  2  v  X X dX 3 v 3 v v  6v  1 v  6v  1 Integrating, ln(v 2  6v  1)  2 ln X  ln c   ln(v 2  6v  1)  ln X 2  ln c X 2 (v 2  6v  1)  c  Y  6 XY  X  c 2 2 60  y 2  6(x  2)y  (x  2)2  c 8.5 Exact Differential Equation. where E3 (1) Exact differential equation : If M and N are functions of x and y, the equation Mdx + Ndy = 0 is called exact when there exists a function f(x, y) of x and y such that f f dx  dy  Mdx  Ndy d[f(x, y)] = Mdx + Ndy i.e., x y f  Partial derivative of f(x, y) with respect to x (keeping y constant) x Note ID f  Partial derivative of f(x, y) with respect to y (treating x as constant) y :  An exact differential equation can always be derived from its general solution D YG U directly by differentiation without any subsequent multiplication, elimination etc. (2) Theorem : The necessary and sufficient condition for the differential equation Mdx + M N  Ndy = 0 to be exact is i.e., partial derivative of M(x, y) w.r.t. y = Partial derivative of y x U N(x, y) w.r.t. x (3) Integrating factor : If an equation of the form Mdx + Ndy = 0 is not exact, it can always be made exact by multiplying by some function of x and y. Such a multiplier is called an integrating factor. (4) Working rule for solving an exact differential equation : Step (i) : Compare the given equation with Mdx + Ndy = 0 and find out M and N. Then find M N M N  out and. If , the given equation is exact. y x y x ST Step (ii) : Integrate M with respect to x treating y as a constant. Step (iii) : Integrate N with respect to y treating x as constant and omit those terms which have been already obtained by integrating M. Step (iv) : On adding the terms obtained in steps (ii) and (iii) and equating to an arbitrary constant, we get the required solution. In other words, solution of an exact differential equation is  Mdx  Regarding y as constant  Ndy c Only those terms not containing x (5) Solution by inspection : If we can write the differential equation in the form f ( f1 (x , y ))d ( f1 (x , y ))   ( f2 (x , y ))d ( f2 (x , y )) ......  0 , then each term can be easily integrated separately. For this the following results must be memorized. (i) d(x + y) = dx + dy (ii) d(xy) = xdy + ydx 406 Differential Equations  y  xdy  ydx (iv) d    x2 x  x 2  2 xydx  x 2 dy  (v) d   y2  y   y 2  2 xydy  y 2 dx (vi) d    x2  x   x 2  2 xy 2 dx  2 x 2 ydy (vii) d  2   y4 y   y 2  2 x 2 ydy  2 xy 2 dx (viii) d  2   x4 x   x  ydx  xdy (ix) d  tan 1   y x2  y2   y  xdy  ydx  (x) d  tan 1   x x2  y2    x   ydx  xdy (xii) d  ln     xy   y  xdy  ydx xy E3 (xi) d[ln( xy )]  60  x  ydx  xdy (iii) d    y2 y   y  xdy  ydx (xiv) d ln    xy   x  ID 1  xdx  ydy (xiii) d  ln( x 2  y 2 )  2 x2  y2    e x  ye x dx  e x dy (xvi) d    y2  y   1  xdy  ydx   (xv) d   x 2y 2  xy   e y  xe y dy  e y dx (xvii) d    2 (xviii) d (x m y n )  x m 1 y n 1 (mydx  nxdy ) U  x  x 1 x  y  x dy  y dx  (xx) d  log x  y  x2  y2 2 (xxi) D YG xdx  y dy (xix) d  x 2  y 2     x2  y2 d [ f (x , y )]1n f (x , y )  1n ( f (x , y ))n Example: 23 The general solution of the differential equation (x + y)dx + xdy = 0 is (a) x  y  c 2 Solution: (c) (b) 2 x  y  c 2 2 2 [MP PET 1994, 95] (c) x  2 xy  c (d) y  2 xy  c (c) xy 2  e  x  c (d) None of these 2 2 We have xdx + (ydx + xdy) = 0  xdx + d(xy) = 0 x2 c  xy  2 2 U Integrating,  x 2  2 xy  c Solution of y(2 xy  e x )dx  e x dy is ST Example: 24 (a) yx 2  e x  cy Solution: (a) (b) xy 2  e x  cx Re-writing the given equation, 2 xy 2dx  ye x dx  e x dy  2 x dx  Integrating, x 2   ex ye x dx  e x dy 2  d ( x )  d   0  y y2   0   ex c y  yx 2  e x  cy Example: 25 Solution of (x 2  4 xy  2y 2 )dx  (y 2  4 xy  2 x 2 )dy  0 is (a) x 3  y 3  6 xy (x  y)  c (b) x 3  y 3  6 xy (x  y)  c (c) x 3  y 3  6 xy (x  y)  c (d) x 3  y 3  6 xy (x  y)  c Differential Equations 407 Solution: (a) Comparing given equation with Mdx + Ndy = 0, We get, M  x 2  4 xy  2y 2 , N  y 2  4 xy  2 x 2 M  4 x  4 y y N  4 y  4 x x M N  y x 60  So the given differential equation is exact.  M dx   (x 2  4 xy  2 y 2 )dx  x3  2 x 2y  2y 2 x 3 Integrating N w.r.t. y, treating x as constant, 2  4 xy  2 x 2 )dy  y3 y3  2 xy 2  2 x 2 y  ; (omitting  2 xy 2  2 x 2 y which already occur in 3 3  Solution of the given equation is  x 3  y 3  6 xy (x  y)  c (3 = c) x3 y3  2 x 2 y  2 xy 2     x 3  y 3  6 xy (x  y)  3 3 3 U 8.6 Linear Differential Equation.  Mdx ) ID  Ndy   (y E3 Integrating m w.r.t. x, treating y as constant, D YG (1) Linear and non-linear differential equations : A differential equation is a linear differential equation if it is expressible in the form d ny d n 1 y d n2 y dy  P  P ...  Pn 1  Pn y  Q where 1 2 n n 1 n2 dx dx dx dx constants or functions of independent variable x. Po P0 , P1 , P2 ,..., Pn 1 , Pn and Q are either U Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no product of these, and also the coefficient of the various terms are either constants or functions of the independent variable, then it is said to be linear differential equation. Otherwise, it is a non linear differential equation. ST It follows from the above definition that a differential equation will be non-linear differential equation if (i) its degree is more than one (ii) any of the differential coefficient has exponent more than one. (iii) exponent of the dependent variable is more than one. (iv) products containing dependent variable and its differential coefficients are present. (2) Linear differential equation of first order : The general form of a linear differential equation of first order is dy  Py  Q dx.....(i) Where P and Q are functions of x (or constants) dy dy dy  xy  x 3 , x  2y  x 3 ,  2y  sin x etc. are linear differential equations. dx dx dx This type of differential equations are solved when they are multiplied by a factor, which is For example, 408 Differential Equations called integrating factor, because by multiplication of this factor the left hand side of the differential equation (i) becomes exact differential of some function. Pdx , we get e  Pdx On integrating both sides w. r. t. x, we get ; Pdx d   Pdx   dy   Pdx  Py   Q e    y e Qe dx    dx  ye Pdx  Qe  Pdx dx  C 60 Multiplying both sides of (i) by e  ……..(ii) which is the required solution, where C is the constant of integration. e  Pdx is called the  E3 integrating factor. The solution (ii) in short may also be written as y.(I.F.)  Q.(I.F.) dx  C (3) Algorithm for solving a linear differential equation : Step (i) : Write the differential equation in the form dy  Py  Q and obtain P and Q. dx Pdx. ID Step (ii) : Find integrating factor (I.F.) given by I.F.  e  Step (iii) : Multiply both sides of equation in step (i) by I.F.  y(I.F.)  Q(I.F.) dx  C D YG This gives the required solution. U Step (iv) : Integrate both sides of the equation obtained in step (iii) w. r. t. x to obtain (4) Linear differential equations of the form equation can be put in the form dx  Rx  S. Sometimes a linear differential dy dx  Rx  S where R and S are functions of y or constants. Note dy that y is independent variable and x is a dependent variable. U (5) Algorithm for solving linear differential equations of the form Step (i) : Write the differential equation in the form dx  Rx  S and obtain R and S. dy R dy ST Step (ii) : Find I.F. by using I.F.  e  dx  Rx  S dy Step (iii) : Multiply both sides of the differential equation in step (i) by I.F. Step (iv) : Integrate both sides of the equation obtained in step (iii) w. r. t. y to obtain the solution given by  x (I.F.)  S (I.F.) dy  C where C is the constant of integration. (6) Equations reducible to linear form (Bernoulli's differential equation) : The dy differential equation of type  Py  Qy n dx......(i) Differential Equations 409 Where P and Q are constants or functions of x alone and n is a constant other than zero or unity, can be reduced to the linear form by dividing by y n and then putting y n 1  v , as explained below. dy  Py n 1  Q dx 1 dv dv dy dv , we get  Pv  Q   (1  n)Pv  (1  n)Q   n  1 dx dx dx dx Putting y n 1  v so that (n  1)y n y n 60 Dividing both sides of (i) by y n , we get E3 which is a linear differential equation. Remark : If n  1 , then we find that the variables in equation (i) are separable and it can be easily integrated by the method discussed in variable separable from. dy  P (y)  Q (y) dx where P and Q are functions of x alone or constants. (7) Differential equation of the form :  (y) d   (y )  dv  v , so that    (y) dx  (y )  dx or dv 1 dy , where k is constant k dx  (y) dx U Now put 1 dy  (y)  PQ  (y) dx  (y) ID Dividing by  (y), we get dv  kP v  kQ dx Which is linear differential equation. Example: 26 D YG We get Which of the following is a linear differential equation 2 2  d 2y   dy  (a)  2   x 2    0  dx   dx  Solution: (c) (b) y  dy  dy   1  dx  dx  2 (c) dy y   log x dx x (d) y dy 4  x dx (a), (b), (d) do not fulfill the criteria of a linear differential equation but (c) do. dy y   log x is a linear differential equation. dx x Find the integral factor of equation (x 2  1) U Example: 27 ST (a) x 2  1 Solution: (a) 2x x 1 2 Given equation may be written as Comparing with P (c) [UPSEAT 2002] x2 1 x2 1 (d) None of these dy 2x x2 1  2 y 2 dx x  1 x 1 dy  Py  Q , dx 2x x2 1 I.F.  e Example: 28 (b) dy  2 xy  x 2  1 dx  Pdx e  The solution of 2 xdx 1 x 2  e ln(1 x 2 )  1  x2 dy  2 y tan x  sin x is dx (a) y sec 3 x  sec 2 x  c (b) y sec 2 x  sec x  c (c) y sin x  tan x  c (d) None of these 410 Differential Equations Solution: (b) dy  Py  Q , dx Comparing with P  2 tan x , Q  sin x  2 tan xdx  e 2 ln sec x  e ln sec 2 x  sec 2 x Multiplying given equation by I.F. and integrating, y sec 2 x   sin x. sec  y sec 2 x  sec x  c The solution of the differential equation (1  y 2 )  (x  e tan (c) xe tan Solution: (b) 1 1 We have (x  e tan 1  x  e tan dy dx  (1  y 2 )     1  y2 dx dy  1 y ) 1 S  Rdy e tan y 1  y2 e   e tan 1 Example: 30 y  e 2 tan Solution of 1 y U Let y 1  v   y 2 ST  tan x 1 y  (b) sec x  tan x  c y Example: 31 y  e tan 1 k y k tan y    dx  1 x  e 2  dy 1  y 1  y2  ……..(i) dx  R(y).x  S (y) dy  1 e tan y tan 1 y e dy  1  y2  1 1 (e tan y )2 dy (e tan y )2 k   2 2 1  y2 (c) y cos x = tan x + c (d) None of these dy  y 1 tan x   sec x dx ……..(i)  e ln sec x  sec x Multiplying (i) by sec x and integrating, v sec x   1 dy dv  dx dx dv  tan x. v  sec x dx I.F.  e y dy  y tan x  y 2 sec x is dx Re-writing the given equation, y  2  1 k (a) y sec x = tan x + c Solution: (b)  e 2 tan y Multiplying (i) by I.F. and integrating, xe tan 1 y dy 1 y 2 D YG  y 1 [AIEEE 2003] 1 U 1 , 1  y2 Integrating factor  e  2 xe tan dy  0 is dx (d) xe 2 tan This is a linear differential equation of the form R ) (b) 2 xe tan y  tan 1 y  k y y E3 (a) (x  2)  ke tan 1  sec x tan xdx ID Example: 29 xdx  2 60 I.F.  e  sec 2 xdx  tan x  c sec x  tan x  c y The solution of dz z z  log z  2 (log z )2 is dx x x  1   x  2  x 2c (a)   log z   1   x  2  x 2c (b)   log z   1   x  x 2 c (c)   log z   1  1  x   cx 2 (d)  2  log z  Differential Equations 411 Let dt 1 1 dz 1   t   log z (log z ) z dx dx   dt t 1   dx x x 2  dt t 1   2 dx x x  I.F.  e  dx x ……..(i)  e  ln x  e Multiplying (i) by dz 1 1 1 1   z(log z )2 dx x log z x 2 ln 1 x  1 x t 1 and integrating,  x x  x 1 3 dx   1  1    x     cx 2 ln z   2 1 1 1 c   c x log z 2 x 2 2x 2 ID 8.7 Application of Differential Equation. 60 Dividing the given equation by z(log z )2 , E3 Solution: (d) D YG U Differential equation is applied in various practical fields of life. It is used to define various physical laws and quantities. It is widely used in physics, chemistry, engineering etc. Some important fields of application are ; (i) Rate of change (ii) Geometrical problems etc. Differential equation is used for finding the family of curves for which some conditions involving the derivatives are given. dy Equation of the tangent at a point (x, y) to the curve y = f(x) is given by Y  y  (X  x ) dx ……..(i) and equation of normal at (x, y) is Y  y   ……..(ii) 1  dy   dx    (X  x ) U     dy  y   The tangent meets X-axis at  x  , 0  and Y-axis at  0, y  x  dx   dy        dx    ST     x  dy    The normal meets X-axis at  x  y , 0  and Y-axis at  0, y  dx   dy         dx    dx  ( x  1) (x is the distance described). dt The time taken by a particle to transverse a distance of 99 metres 1 log 10 e (a) log 10 e (b) 2 log e 10 (c) log 10 e (d) 2 Example: 32 A particle moves in a straight line with a velocity given by Solution: (b) We have dx  dt x 1 412 Differential Equations Integrating,  99 0 dx  x 1  dt  [ln(x  1)] t 99 0 t 0  t  ln 100  log e (10 )2  2 log e 10 Example: 33 y y    cos 2   , then the The slope of the tangent at (x, y) to a curve passing through  1,  as given by x x  4 equation of the curve is [Kurukshetra CEE 2002] We have   e  (c) y  x tan 1 log     x  dy y y   cos 2   dx x x Let y = vx  (d) None of these dx dy dv dv dv  vx  tan v = – ln x + c vx   cos 2 v  sec 2 vdv    v  cos 2 v  x x dx dx dx dx E3 Solution: (c)   x  (b) y  x tan 1 log     e  60   e  (a) y  tan 1 log     x   tan (y/x) = – ln x + c For x = 1, y = /4  tan  / 4   ln 1  c  1  0  c ID c=1  tan( y / x )  1  ln x   e   y / x  tan 1 (1  ln x )  tan 1 (ln e  ln x )  tan 1 ln    x  U   e   y  x tan 1 ln    x  The equation of the curve which is such that the portion of the axis of x cut off between the origin and tangent at any point is proportional to the ordinate of that point (b is constant of proportionality) x (a) y  (b) log x  by 2  a (c) x 2  y(a  b log y) (d) None of these (a  b log x ) Solution: (d) Tangent at P(x, y) to the curve y = f(x) may be expressed as Y  y  D YG Example: 34 dy (X  x ) dx  dx   Q   x  y , 0 dy   As per question, OQ  y x dx dx dx  by   y  x y  b dy dy y dy U  x y P(x, y) O Q dx x  b dy y ST  (0, 0) Let x dx x dy dv dv dv dv  b  v  y  v b  v  y  b  y  b  dv  v  x = vy  vy y y dy dy y dy dy dy Integrating,  dv  b  dy x  v  b ln y  a   a  b ln y y y (a, an arbitrary constant)  x  y(a  b ln y ) 8.8 Miscellaneous Differential Equation. (1) A special type of second order differential equation : Equation (i) may be re-written as d 2y  f (x ) dx 2 d  dy   dy     f (x )  d    f (x )dx dx  dx   dx  ……..(i) Differential Equations 413 Integrating, dy  dx Where F(x )    f (x )dx  c1 i.e. dy  F(x )  c1 dx ……..(ii) f (x )dx  c 1 dx From (ii), dy  f (x )dx  c1 dx F(x )dx  c1 x  c 2 60  Integrating, y   y  H (x )  c1 x  c 2  c 1 and c 2 are arbitrary constants. F(x )dx E3 where H (x )  The solution of the equation x 2 Solution: (d) 1 (ln x )2  ln x 2 We have (b) 1 (ln x )2  ln x 2 (c)  1 (ln x )2  ln x 2 [Orissa JEE 2003] (d)  1 (ln x )2  ln x 2 U (a) dy d 2y  1 is  ln x when x = 1, y = 0 and 2 dx dx d 2 y ln x  dy  ln x  2  d    2 dx x dx 2 x  dx   1  ln xd   x    Integrating, dy  dx When x = 1, dy  1 dx ln x  x  1 ln x 1 dy 1  ln x dx    c   c x x dx x x2 D YG Example: 35 ID (2) Particular solution type problems : To solve such a problem, we proceed according to the type of the problem (i.e. variable-separable, linear, exact, homogeneous etc.) and then we apply the given conditions to find the particular values of the arbitrary constants. –1=–1+cc=0  dy 1  ln x 1  ln x  dx   dy    dy   dx x x    dx 1 1  ln x. dx   y  ln x  (ln x )2   2 x x y = 0 when x = 1 U  0  0  0 2      0   y  ln x  1 (ln x )2  ln x 2 ST  y Example: 36 Solution: (d) 1 (ln x )2 2 A continuously differentiable function (x) in (0, ) satisfying y   1  y 2 , y(0)  0  y( ) is (a) tan x (b) x(x – ) For (x) = y, y   1  y 2  dy  1  y2  dx (c) (x   )(1  e x )  1  y   dx dy 2  tan 1 y  x  c  y = tan (x + c) i.e., (x) = tan (x + c) As y(0) = 0, 0 = tan c and y() = 0  0 = tan ( + c) = tan c c=0  (x) = y = tan x. But tan x is not continuous in (0, ) (d) Not possible 414 Differential Equations Since tan  is not defined. 2 ST U D YG U ID E3 60 Hence there exists not a function satisfying the given condition.