Chapter 11 Higher Engineering Mathematics PDF

Unit IV: Differential Equations 11 Differential Equations of First Order 1. Definitions. 2. Practical approach to differential equations. 3. Formation of a differential equation. 4. Solution of a d...

Unit IV: Differential Equations 11 Differential Equations of First Order 1. Definitions. 2. Practical approach to differential equations. 3. Formation of a differential equation. 4. Solution of a differential equation—Geometrical meaning—5. Equations of the first order and first degree. 6. Variables separable. 7. Homogeneous equations. 8. Equations reducible to homogeneous form. 9. Linear equations. 10. Bernoulli’s equation. 11. Exact equations. 12. Equations reducible to exact equations. 13. Equations of the first order and higher degree. 14. Clairut’s equation. 15. Objective Type of Questions. 11.1 DEFINITIONS (1) A differential equation is an equation which involves differential coefficients or differentials. d2 x Thus (i) ex dx + ey dy = 0 (ii) + n2x = 0 dt2 2 3/2 dy x ⎡ ⎛ dy ⎞ ⎤ d2 y (iii) y = x + (iv) ⎢1 + ⎜ ⎟ ⎥ =c dx dy dx ⎣⎢ ⎝ dx ⎠ ⎦⎥ dx2 dx dy ∂u ∂u (v) wy = a cos pt, + wx = a sin pt (vi) x + y = 2u dt dt ∂x ∂y ∂2 y ∂2 y (vii) = c2 are all examples of differential equations. ∂t 2 ∂x2 (2) An ordinary differential equation is that in which all the differential coefficients have reference to a single independent variable. Thus the equations (i) to (v) are all ordinary differential equations. A partial differential equation is that in which there are two or more independent variables and partial differential coefficients with respect to any of them. Thus the equations (vi) and (vii) are partial differential equations. (3) The order of a differential equation is the order of the highest derivative appearing in it. The degree of a differential equation is the degree of the highest derivative occurring in it, after the equa- tion has been expressed in a form free from radicals and fractions as far as the derivatives are concerned. Thus, from the examples above, (i) is of the first order and first degree ; (ii) is of the second order and first degree ; 2 dy ⎛ dy ⎞ + x is clearly of the first order but of second degree ; (iii) written as y = x⎜ ⎟ dx ⎝ dx ⎠ 2 3 2 ⎡ ⎛ dy ⎞ ⎤ 2 2⎛ d y⎞ and (iv) written as ⎢1 + ⎜ ⎟ ⎥ = c ⎜⎜ 2 ⎟⎟ is of the second order and second degree. ⎣⎢ ⎝ dx ⎠ ⎦⎥ ⎝ dx ⎠ 426  DIFFERENTIAL EQUATIONS OF FIRST ORDER 427 11.2 PRACTICAL APPROACH TO DIFFERENTIAL EQUATIONS Differential equations arise from many problems in oscillations of mechanical and electrical systems, bending of beams, conduction of heat, velocity of chemical reactions etc., and as such play a very important role in all modern scientific and engineering studies. The approach of an engineering student to the study of differential equations has got to be practical unlike that of a student of mathematics, who is only interested in solving the differential equations without knowing as to how the differential equations are formed and how their solutions are physically interpreted. Thus for an applied mathematician, the study of a differential equation consists of three phases : (i) formulation of differential equation from the given physical situation, called modelling. (ii) solutions of this differential equation, evaluating the arbitrary constants from the given conditions, and (iii) physical interpretation of the solution. 11.3 FORMATION OF A DIFFERENTIAL EQUATION An ordinary differential equation is formed in an attempt to eliminate certain arbitrary constant from a relation in the variables and constants. It will, however, be seen later that the partial differential equations may be formed by the elimination of either arbitrary constants or arbitrary functions. In applied mathematics, every geometrical or physical problem when translated into mathematical symbols gives rise to a differential equation. Example 11.1. Form the differential equation of simple harmonic motion given by x = A cos (nt + α). Solution. To eliminate the constants A and α differentiating it twice, we have dx d2 x = nA sin (nt + α) and = n2A cos (nt + α) = n2 x dt dt 2 d2 x Thus + n2 x = 0 dt2 is the desired differential equation which states that the acceleration varies as the distance from the origin. Example 11.2. Obtain the differential equation of all circles of radius a and centre (h, k). Solution. Such a circle is (x h)2 + (y k)2 = a2...(i) where h and k, the coordinates of the centre, and a are the constants. Differential it twice, we have 2 dy d 2 y ⎛ dy ⎞ x h + (y k) = 0 and 1 + (y k) +⎜ ⎟ =0 dx dx2 ⎝ dx ⎠ 1 + (dy / dx)2 Then y k= d 2 y / dx2 2 dy ⎡ ⎛ dy ⎞ ⎤ ⎢1 + ⎜ ⎟ ⎥ dx ⎣⎢ ⎝ dx ⎠ ⎦⎥ and x h= (y k) dy/dx = d 2 y / dx2 Substituting these in (i) and simplifying, we get [1 + (dy/dx)2]3 = a2 (d2y/dx2)2...(ii) as the required differential equation [1 + (dy / dx)2 ]3 / 2 Writing (ii) in the form = a, d 2 y / dx2 it states that the radius of curvature of a circle at any point is constant. Example 11.3. Obtain the differential equation of the coaxial circles of the system x2 + y2 + 2ax + c2 = 0 where c is a constant and a is a variable. (J.N.T.U., 2003)  428 HIGHER ENGINEERING MATHEMATICS Solution. We have x2 + y2 + 2ax + c2 = 0...(i) Differentiating w.r.t. x, 2x + 2ydy/dx + 2a = 0 ⎛ dy ⎞ or 2⎜x + y 2a = ⎟ ⎝ dx ⎠ Substituting in (i), x2 + y2 2 (x + y dy/dx)x + c2 = 0 or 2xy dy/dx = y2 x2 + c2 which is the required differential equation. 11.4 (1) SOLUTION OF A DIFFERENTIAL EQUATION A solution (or integral) of a differential equation is a relation between the variables which satisfies the given differential equation. For example, x = A cos (nt + α)...(1) d2 x is a solution of + n2x = 0 [Example 11.1]...(2) dt 2 The general (or complete) solution of a differential equation is that in which the number of arbitrary constants is equal to the order of the differential equation. Thus (1) is a general solution (2) as the number of arbitrary constants (A, α) is the same as the order of (2). A particular solution is that which can be obtained from the general solution by giving particular values to the arbitrary constants. For example, x = A cos (nt + π/4) is the particular solution of the equation (2) as it can be derived from the general solution (1) by putting α = π/4. A differential equation may sometimes have an additional solution which cannot be obtained from the general solution by assigning a particular value to the arbitrary constant. Such a solution is called a singular solution and is not of much engineering interest. Linearly independent solution. Two solutions y1(x) and y2(x) of the differential equation d2 y dy + a1 ( x)+ a2(x) y = 0...(3) 2 dx dx are said to be linearly independent if c1y1 + c2y2 = 0 such that c1 = 0 and c2 = 0 If c1 and c2 are not both zero, then the two solutions y1 and y2 are said to be linearly dependent. If y1(x) and y2(x) any two solutions of (3), then their linear combination c1y1 + c2y2 where c1 and c2 are constants, is also a solution of (3). Example 11.4. Find the differential equation whose set of independent solutions is [ex, xex]. Solution. Let the general solution of the required differential equation be y = c1ex + c2xex...(i) Differentiating (i) w.r.t. x, we get y1 = c1ex + c2 (ex + xex) ∴ y y1 = c 2 e x...(ii) Again differentiating (ii) w.r.t. x, we obtain y1 y2 = c 2 e x...(iii) Subtracting (iii) from (ii), we get Y y y1 (y1 y2) = 0 or y 2y1 + y2 = 0 A4 which is the desired differential equation. A3 (2) Geometrical meaning of a differential equation. Consider any differential equation of the first order and first degree A2 dy = f (x, y)...(1) A1 dx If P(x, y) be any point, then (1) can be regarded as an equation giving the A 0 value of dy/dx (= m) when the values of x and y are known (Fig. 11.1). Let the 0 X value of m at the point A0(x0, y0) derived from (1) be m0. Take a neighbouring Fig. 11.1  DIFFERENTIAL EQUATIONS OF FIRST ORDER 429 point A1(x1, y1) such that the slope of A0A1 is m0. Let the corresponding value of m at A1 be m1. Similarly take a neighbouring point A2(x2, y2) such that the slope of A1A2 is m1 and so on. If the successive points A0, A1, A2, A3... are chosen very near one another, the broken curve A0A1A2A3... approximates to a smooth curve C[y = φ(x)] which is a solution of (1) associated with the initial point A0 (x0, y0). Clearly the slope of the tangent to C at any point and the coordinates of that point satisfy (1). A different choice of the initial point will, in general, give a different curve with the same property. The equation of each such curve is thus a particular solution of the differential equation (1). The equation of the whole family of such curves is the general solution of (1). The slope of the tangent at any point of each member of this family and the co-ordinates of that point satisfy (1). Such a simple geometric interpretation of the solutions of a second (or higher) order differential equation is not available. PROBLEMS 11.1 Form the differential equations from the following equations : 1. y = ax3 + bx2. 2. y = C1 cos 2x + C2 sin 2x (Bhopal, 2008) 3. xy = Aex + Be x + x2. (U.P.T.U., 2011) 4. y = ex (A cos x + B sin x). (P.T.U., 2003) 5. y = ae2x + be 3x x + ce. Find the differential equations of : 6. A family of circles passing through the origin and having centres on the x-axis. (J.N.T.U., 2006) 7. All circles of radius 5, with their centres on the y-axis. 8. All parabolas with x-axis as the axis and (a, 0) as focus. 9. If y1(x) = sin 2x and y2 (x) = cos 2x are two solutions of y″ + 4y = 0, show that y1 (x) and y2 (x) are linearly independent solutions. 10. Determine the differential equation whose set of independent solutions is [ex, xex, x2 ex] (U.P.T.U., 2002) 11. Obtain the differential equation of the family of parabolas y = x2 + c and sketch those members of the family which pass through (0, 0), (1, 1), (0, 1) and (1, 1) respectively. 11.5 EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE It is not possible to solve such equations in general. We shall, however, discuss some special methods of solution which are applied to the following types of equations : (i) Equations where variables are separable, (ii) Homogeneous equations, (iii) Linear equations, (iv) Exact equations. In other cases, the particular solution may be determined numerically (Chapter 31). 11.6 VARIABLES SEPARABLE If in an equation it is possible to collect all functions of x and dx on one side and all the functions of y and dy on the other side, then the variables are said to be separable. Thus the general form of such an equation is f(y) dy = φ(x) dx Integrating both sides, we get ∫ f ( y) dy = ∫ φ (x) dx + c as its solution. x ( 2 log x + 1) Example 11.5. Solve dy/dx =. (V.T.U., 2012) sin y + y cos y Solution. Given equation is x (2 log x + 1) dx = (sin y + y cos y) dy Integrating both sides, 2 ∫ (log x. x + x) dx = ∫ sin y dy + ∫ y cos y dy + c ⎡⎛ x2 1 x2 ⎞ x2 ⎤ or 2 ⎢⎜⎜ log x. ⎢⎣⎝ 2 ∫. x 2 dx ⎟⎟ + ⎠ 2 ⎥⎦ ⎥= cos y + ⎢⎡ y sin y ⎣ ∫ sin y. 1 dy + c⎦⎥⎤  430 HIGHER ENGINEERING MATHEMATICS x2 x2 or 2x2 log x + = cos y + y sin y + cos y + c 2 2 Hence the solution is 2x2 log x y sin y = c. dy Example 11.6. Solve = e3x 2y + x2 e 2y. dx dy 2y (e3x + Solution. Given equation is =e x2 ) or e2y dy = (e3x + x2) dx dx ∫e ∫ (e 2y 3x Integrating both sides, dy = + x2 ) dx + c e2 y e3 x x3 or = + +c or 3e2y = 2(e3x + x3) + 6c. 2 3 3 dy Example 11.7. Solve = sin (x + y) + cos (x + y). (V.T.U., 2005) dx Solution. Putting x + y = t so that dy/dx = dt/dx 1 dt The given equation becomes 1 = sin t + cos t dx or dt/dx = 1 + sin t + cos t dt Integrating both sides, we get dx = ∫ 1 + sin t + cos t + c. 2dθ or x= ∫ 1 + sin 2θ + cos 2θ +c [Putting t = 2θ] 2 dθ sec 2 θ = ∫ 2 cos2 θ + 2 sin θ cos θ +c= ∫ 1 + tan θ dθ + c = log (1 + tan θ) + c Hence the solution is x = log ⎡⎣1 + tan 12 ( x + y) ⎤⎦ + c. Example 11.8. Solve dy/dx = (4x + y + 1) 2, if y (0) = 1. dy dt Solution. Putting 4x + y + 1 = t, we get = 4. dx dx dt dt ∴ the given equation becomes 4 = t2 or = 4 + t2 dx dx dt Integrating both sides, we get ∫ 4 + t2 = dx + c ∫ 1 t 1 1 ⎡1 ⎤ or tan 1 = x + c or tan ⎢⎣ 2 (4 x + y + 1) ⎥⎦ = x + c. 2 2 2 or 4x + y + 1 = 2 tan 2(x + c) 1 1 (1) When x = 0, y = 1 ∴ tan = c i.e. c = π/8. 2 Hence the solution is 4x + y + 1 = 2 tan (2x + π/4). y dy x 2 + y2 1 Example 11.9. Solve + = 0. (Calicut, 2013 ; V.T.U., 2003) 2 2 x dx 2( x + y ) + 1 dy dt y dy 1 dt Solution. Putting x2 + y2 = t, we get 2x + 2y = or = 1. dx dx x dx 2 x dx  DIFFERENTIAL EQUATIONS OF FIRST ORDER 431 1 dt t 1 Therefore the given equation becomes 1+ =0 2 x dx 2t + 1 1 dt t 1 t+2 2t + 1 or =1 = or 2x dx = dt 2 x dx 2t + 1 2t + 1 t+2 ⎛ 3 ⎞ or 2x dx = ⎜ 2 ⎟ dt ⎝ t + 2⎠ Integrating, we get x2 = 2t 3 log (t + 2) + c or x2 + 2y2 3 log (x2 + y2 + 2) + c = 0 [∵ t = x2 + y2] which is the required solution. PROBLEMS 11.2 Solve the following differential equations : dy 1. y √(1 x2 ) dy + x √(1 y2 ) dx = 0. 2. (x2 yx2) + y2 + xy2 = 0. dx y dy 3. sec2 x tan y dx + sec2 y tan x dy = 0. (P.T.U., 2003) 4. = √(1 + x 2 + y2 + x 2 y2 ). (V.T.U., 2011) x dx dy x2 5. ex tan y dx + (1 ex) sec2 y dy = 0. (V.T.U., 2009) 6. = xe y , if y = 0 when x = 0. (J.N.T.U., 2006) dx dy 7. x + cot y = 0 if y = π/4 when x = 2. 8. (xy2 + x) dx + (yx2 + y) dy = 0. dx dy dy ⎛ dy ⎞ 9. = e2x 3y + 4x2 e 3y. 10. y x = a ⎜ y2 + ⎟. dx dx ⎝ dx ⎠ dy dy 11. (x + 1) + 1 = 2e y. (Madras, 2000 S) 12. (x y)2 = a2. dx dx dy 13. (x + y + 1)2 = 1. (Kurukshetra, 2005) 14. sin 1 (dy/dx) = x + y (V.T.U., 2010) dx dy dy 15. = cos (x + y + 1) (V.T.U., 2003) 16. x tan (y x) = 1. dx dx dy 17. x4 + x3y + cosec (xy) = 0. dx 11.7 HOMOGENEOUS EQUATIONS dy f ( x, y) are of the form = dx φ ( x, y) where f (x, y) and φ (x, y) are homogeneous functions of the same degree in x and y (see page 205). dy dv To solve a homogeneous equation (i) Put y = vx, then =v+ x , dx dx (ii) Separate the variables v and x, and integrate. Example 11.10. Solve (x2 y2) dx xy dy = 0. dy x 2 y2 Solution. Given equation is = which is homogeneous in x and y....(i) dx xy dy dv dv 1 v2 Put y = vx, then =v+ x. ∴ (i) becomes v + x = dx dx dx v dv 1 v2 1 2v2 or x = v=. dx v v  432 HIGHER ENGINEERING MATHEMATICS v dx Separating the variables, 2 dv = 1 2v x v dv dx Integrating both sides, ∫1 2v 2 = ∫ x +c 1 4v dx 1 or 4 ∫1 2v 2 dv = ∫ x +c or 4 log (1 2v2) = log x + c or 4 log x + log (1 2v2) = 4c or log x4 (1 2v2) = 4c [Put v = y/x] or x4(1 2y2/x2) = e 3c = c′ Hence the required solution is x2(x2 2y2) = c′. Example 11.11. Solve (x tan y/x y sec2 y/x) dx x sec2 y/x dy = 0. (V.T.U., 2013) Solution. The given equation may be rewritten as dy ⎛ y y y⎞ = ⎜ sec 2 tan 2 ⎟ cos y/x...(i) dx ⎝ x x x⎠ dv which is a homogeneous equation. Putting y = vx, (i) becomes v + x = (v sec2 v tan v) cos2 v dx dv or x =v tan v cos2 v v dx sec 2 v dx Separating the variables dv = tan v x Integrating both sides log tan v = log x + log c or x tan v = c or x tan y/x = c. Example 11.12. Solve (1 + ex/y) dx + ex/y (1 x/y) dy = 0. (P.T.U., 2006 ; Rajasthan, 2005 ; V.T.U., 2003) Solution. The given equation may be rewritten as dx e x / y (1 x / y) =−...(i) dy 1 + ex / y which is a homogeneous equation. Putting x = vy so that (i) becomes dv ev (1 v) dv ev (1 v) v + ev v+y =− or y =− v= dy 1 + ev dy 1 + ev 1 + ev Separating the variables, we get dy 1 + ev d(v + ev ) = v dv = y v+e v + ev Integrating both sides, log y = log (v + ev) + c or y (v + ev) = e c or x + yex/y = c′ (say) which is the required solution. PROBLEMS 11.3 Solve the following differential equations : 1. (x2 y2) dx = 2xy dy 2. (x2y 2xy2) dx (x3 3x2y) dy = 0. (D.T.U., 2011 ; Bhopal, 2008) 3. x2y dx (x3 + y3) dy = 0. (V.T.U., 2010) 4. y dx x dy = x2 + y2 dx. (Raipur, 2005) dy dy 5. y2 + x2 = xy. 6. (3xy 2ay2) dx + (x2 2axy) dy = 0. (S.V.T.U., 2009) dx dx  DIFFERENTIAL EQUATIONS OF FIRST ORDER 433 [Equations solvable like homogeneous equations : When a differential equation contains y/x a number of times, solve it like a homogeneous equation by putting y/x = v]. dy y y 7. = + sin. (V.T.U., 2000 S) 8. yex/y dx = (xex/y + y2) dy. (V.T.U., 2006) dx x x 9. xy (log x/y) dx + [y2 x2 log (x/y)] dy = 0. 10. x dx + sin2 (y/x) (ydx xdy) = 0. y y 11. x cos (ydx + xdy) = y sin (xdy ydx). x x dy 12. [x cos (y/x) + y sin (y/x)] y [y sin (y/x) x cos (y/x)x = 0. (V.T.U., 2013) dx 11.8 EQUATIONS REDUCIBLE TO HOMOGENEOUS FORM dy ax + by + c The equations of the form =...(1) dx a′x + b′ y + c′ can be reduced to the homogeneous form as follows : a b Case I. When ≠ a ′ b′ Putting x = X + h, y = Y + k, (h, k being constants) so that dx = dX, dy = dY, (1) becomes dY aX + bY + (ah + bk + c) =...(2) dX a′X + b′ Y + (a′h + b′ k + c′) Choose h, k so that (2) may become homogeneous. Put ah + bk + c = 0, and a′h + b′k + c′ = 0 h k 1 so that = = bc′ b′ c ca′ c′a ab′ ba′ bc′ b′ c ca′ c′a or h= ,k=...(3) ab′ b′ a ab′ ba′ aX + bY Thus when ab′ ba′ ≠ 0, (2) becomes dY = which is homogeneous in X, Y and can be solved by dX a′X + b′ Y putting Y = vX. a b Case II. When =. a ′ b′ i.e., ab′ b′a = 0, the above method fails as h and k become infinite or indeterminate. a b 1 Now = = (say) a′ b′ m ∴ a′ = am, b′ = bm and (1) becomes dy ( ax + by) + c =...(4) dx m(ax + by) + c′ dy dt Put ax + by = t, so that a + b = dx dx dy 1 ⎛ dt 1 ⎛ dt t+c or = ⎜ a ⎞⎟ ∴ (4) becomes ⎜ a ⎞⎟ = dx b ⎝ dx ⎠ b ⎝ dx ⎠ mt + c′ dt bt + bc (am + b)t + ac′ + bc or =a+ = dx mt + c′ mt + c′ so that the variables are separable. In this solution, putting t = ax + by, we get the required solution of (1). dy y + x 2 Example 11.13. Solve =. (Raipur, 2005) dx y x 4 dy y + x 2 ⎡Case a ≠ b ⎤ Solution. Given equation is = ⎢⎣...(i) dx y x 4 a′ b′ ⎥⎦  434 HIGHER ENGINEERING MATHEMATICS Putting x = X + h , y = Y + k, (h, k being constants) so that dx = dX, dy = dY, (i) becomes dY Y + X + (k + h 2) =...(ii) dX Y X + (k h 4) Put k + h 2=0 and k h 4 = 0 so that h = 1, k = 3. dY Y + X ∴ (ii) becomes = which is homogeneous in X and Y....(iii) dX Y X dY dv ∴ put Y = vX, then =v+ X dX dX dv v + 1 v+1 1 + 2v v2 ∴ (iii) becomes v+X = or X dv = v= dX v 1 dX v 1 v 1 v 1 dX or dv =. 1 + 2v v 2 X 1 2 2v dX Integrating both sides, 2 ∫ 1 + 2v 2 v dv = ∫ X + c. 1 or log (1 + 2v v2) = log X + c 2 ⎛ 2Y Y2 ⎞ 2 or log ⎜ 1 + ⎟ + log X = 2c ⎝ X X2 ⎠ or log (X2 + 2XY Y2) = 2c or X2 + 2XY Y2 = e 2c = c′...(iv) Putting X = x h = x + 1, Y = y k = y 3, (iv) becomes (x + 1)2 + 2(x + 1) (y 3) (y 3)2 = c′ or 2 x + 2xy y 2 4x + 8y 14 = c′ which is the required solution. Example 11.14. Solve (3y + 2x + 4) dx (4x + 6y + 5) dy = 0. (V.T.U., 2013) dy (2 x + 3 y) + 4 Solution. Given equation is =...(i) dx 2(2 x + 3 y) + 5 dy dt 1 ⎛ dt t+4 Putting 2x + 3y = t so that 2 + 3 = ∴ (i) becomes ⎜ 2 ⎞⎟ = dx dx 3 ⎝ dx ⎠ 2t + 5 dt 3t + 12 7t + 22 2t + 5 or =2+ = or dt = dx dx 2t + 5 2t + 5 7t + 22 2t + 5 Integrating both sides, ∫ 7t + 22 dt = ∫ dx + c ⎛2 9 1 ⎞ 2 9 or ∫ ⎜⎝ 7 − 7. 7t + 22 ⎟⎠ dt = x + c or 7 t− 49 log (7t + 22) = x + c Putting t = 2x + 3y, we have 14(2x + 3y) 9 log (14x + 21y + 22) = 49x + 49c or 21x 42y + 9 log (14x + 21y + 22) = c′ which is the required solution. PROBLEMS 11.4 Solve the following differential equations : 1. (x y 2) dx + (x 2y 3) dy = 0. (Rajasthan, 2006) 2. (2x + y 3) dy = (x + 2y 3) dx. (V.T.U., 2009 S) 3. (2x + 5y + 1) dx (5x + 2y 1) dy = 0. dy ax + hy + g dy x+ y+1 4. + = 0. 5. =. dx hx + by + f dx 2 x + 2 y + 3 6. (4x 6y 1) dx + (3y 2x 2) dy = 0. 7. (x + 2y)(dx dy) = dx + dy. (Bhopal, 2002 S ; V.T.U., 2001)  DIFFERENTIAL EQUATIONS OF FIRST ORDER 435 11.9 LINEAR EQUATIONS A differential equation is said to be linear if the dependent variable and its differential coefficients occur only in the first degree and not multiplied together. Thus the standard form of a linear equation of the first order, commonly known as Leibnitz's linear equation,* is dy + Py = Q where, P, Q are the functions of x....(1) dx To solve the equation, multiply both sides by e∫ Pdx so that we get dy ∫ + y (e∫ P ) = Qe∫ ( ye∫ ) = Qe∫ Pdx Pdx Pdx d Pdx Pdx.e i.e., dx dx Integrating both sides, we get ye∫ ∫ Q e∫ Pdx Pdx = dx + c as the required solution. Obs. The factor e∫ Pdx on multiplying by which the left-hand side of (1) becomes the differential coefficient of a single function, is called the integrating factor (I.F.) of the linear equation (1). It is important to remember that I.F. = e ∫ Pdx and the solution is y (I.F.) = ∫ Q (I.F.) dx + c. dy Example 11.15. Solve (x + 1) y e3x (x + 1)2. (V.T.U., 2014) dx Solution. Dividing throughout by (x + 1), given equation becomes dy y − = e3x (x + 1) which is Leibnitz s equation....(i) dx x + 1 1 dx Here P= x+1 and ∫ Pdx = − ∫ x+1 = log (x + 1) = log (x + 1) 1 I.F. = e∫ Pdx −1 1 ∴ = elog ( x + 1) = x+1 3x Thus the solution of (1) is y(I.F.) = ∫ [e ( x + 1)] (I.F.) dx + c y 3x 1 3x 1 3x or x+1 = ∫e dx + c = 3 e +c or y= ( 3 e ) + c (x + 1). ⎛ e−2 x y ⎞ dx Example 11.16. Solve ⎜ − ⎟ = 1. ⎜ x x ⎟⎠ dy ⎝ dy y e 2 x Solution. Given equation can be written as + =...(i) dx x x I.F. = e∫ 1/2 x dx ∴ = e2 x 2 x e Thus solution of (i) is y (I.F.) = ∫ x (I.F.) dx + c 2 x 2 x e 2 x or ye =∫.e dx + c x or ye2 x = ∫x −1 / 2 dx + c or ye2 x = 2 x + c. * See footnote p. 139.  436 HIGHER ENGINEERING MATHEMATICS dy Example 11.17. Solve 3x(1 x 2) y 2 + (2x2 1) y3 = ax3 (Rajasthan, 2006) dx dy dz Solution. Putting y3 = z and 3y2 = , the given equation becomes dx dx 2 dz dz 2 x − 1 ax3 x( 1 x2 ) + (2x2 1) z = ax3, or + z=...(i) dx dx x−x 3 x − x3 which is Leibnitz s equation in z ⎛ 2 x2 − 1 ⎞ ∴ I.F. = exp ⎜ ⎜ ∫ x−x dx ⎟ ⎟ 3 ⎝ ⎠ 2 x2 − 1 ⎛ 1 1 1 1 1 ⎞ 1 1 Now ∫ x−x 3 dx = ∫ ⎜ ⎝ +. x 2 1 + x 2 1 − x ⎟⎠ − dx = log x 2 log (1 + x) 2 log (1 x) = log [ x (1 − x2 )] 2 log [ x (1 − x ) ] 2 1 ∴ I.F. = e = [ x (1 − x )] Thus the solution of (i) is 3 ax z(I.F.) = ∫ x−x 3 (I.F.) dx + c 3 z x 1 2 3/2 or =a∫ 2. dx + c = a ∫ x(1 x ) dx [ x (1 − x )] 2 x(1 x ) x (1 − x ) 2 a ( 2 x)(1 − x2 )− 3 / 2 dx + c = a (1 x 2 )− 1 / 2 + c 2∫ = Hence the solution of the given equation is y3 = ax + cx (1 − x2 ). [∵ z = y3 ] Example 11.18. Solve y (log y) dx + (x log y) dy = 0. dx x 1 Solution. We have + =... (i) dy y log y y which is a Leibnitz s equation in x 1 ∫ y log y dy ∴ I.F. = e = elog (log y) = log y 1 Thus the solution of (i) is x (I.F.) = ∫ y (I.F.) dy + c 1 1 x log y = ∫ y log y dy + c = (log y)2 + c 2 1 i.e., x= log y + c (log y) 1. 2 Example 11.19. Solve (1 + y2) dx = (tan 1 y x) dy. (Bhopal, 2008 ; V.T.U., 2008 ; U.P.T.U., 2005) Solution. This equation contains y2 and tan 1 y and is, therefore, not a linear in y; but since only x occurs, it can be written as dx dx x tan − 1 y (1 + y2) dy = tan 1 y x or + = dy 1 + y2 1 + y2 which is a Leibnitz s equation in x. 1 ∫ dy I.F. = e∫ Pdy 2 −1 ∴ = e 1+ y = etan y tan − 1 y Thus the solution is x (I.F.) = ∫ 1 + y2 (I.F.) dy + c  DIFFERENTIAL EQUATIONS OF FIRST ORDER 437 ⎡ Put tan − 1 y = t ⎤ tan − 1 y tan − 1 y tan − 1 y ⎢ ⎥ or xe = ∫ 1 + y2.e dy + c ⎢∴ dy = dt ⎥ ⎢ 1 y 2 ⎥ ⎣ + ⎦ t t t = ∫ te dt + c = t. e − ∫ 1. e dt + c (Integrating by parts) 1 t t 1 = t. e − e + c = (tan y 1) etan y +c −1 1 − tan y or x = tan y 1 + ce. Example 11.20. Solve r sin θ dθ + (r3 2r2 cos θ + cos θ) dr = 0. Solution. Given equation can be rewritten as dθ 1 + (1 2r2) cos θ = r2 sin θ...(i ) dr r Put cos θ = y so that sin θ dθ/dr = dy/dr dy ⎛ 1 dy ⎛ 1 Then (i) becomes + ⎜ − 2r ⎞⎟ y = r2 or + ⎜ 2r − ⎞⎟ y = r2 dr ⎝ r ⎠ dr ⎝ r⎠ I.F. = e∫ (2 r − 1 / r ) dr 2 1 r2 which is a Leibnitz s equation ∴ = er − log r = e r 1 r2 r2 1 Thus its solution is y ⎛⎜ e ⎞⎟ = ∫ r. e. dr + c 2 ⎝r ⎠ r r 2 1 r 2 1 r2 or y e / r = ∫ e 2r dr + c = e + c 2 2 2 2 2 or 2 er cos θ = rer + 2 cr or r (1 + 2 ce− r ) = 2 cos θ. PROBLEMS 11.5 Solve the following differential equations : dy dy 1. cos2 x + y = tan x. 2. x log x + y = log x2. (V.T.U., 2011) dx dx 3. 2y′ cos x + 4y sin x = sin 2x, given y = 0 when x = π/3. (V.T.U., 2003) dy 4. cosh x + y sinh x = 2 cosh2 x sinh x. (J.N.T.U., 2003) dx dy 5. (1 x2) xy = 1 (V.T.U., 2010) 6. (1 x2) dy + 2xy = x (1 − x 2 ) (Nagpur, 2009) dx dx dy x + y cos x 7. =−. 8. dr + (2r cot θ + sin 2θ) dθ = 0. (J.N.T.U., 2003) dx 1 + sin x dy 2 dy 9. + 2xy = 2 e− x (P.T.U., 2005) 10. (x + 2y3) = y. (Marathwada, 2008) dx dx 1 (1 − y ) dx = (sin y x) dy. 2 11. 12. yey dx = (y3 + 2xey) dy. − tan − 1 y 13. (1 + y2) dx + (x e ) dy = 0. (V.T.U., 2013) 14. e y sec2 y dy = dx + x dy. 11.10 BERNOULLI S EQUATION dy The equation dx + Py = Qyn...(1) where P, Q are functions of x, is reducible to the Leibnitz s linear equation and is usually called the Bernoulli s equation*. *Named after the Swiss mathematician Jacob Bernoulli (1654 1705) who is known for his basic work in probability and elasticity theory. He was professor at Basel and had amongst his students his youngest brother Johann Bernoulli (1667 1748) and his nephew Niklaus Bernoulli (1687 1759). Johann is known for his basic contributions to Calculus while Niklaus had profound influence on the development of Infinite series and probability. His son Daniel Bernoulli (1700 1782) is known for his contributions to kinetic theory of gases and fluid flow.  438 HIGHER ENGINEERING MATHEMATICS dy To solve (1), divide both sides by yn, so that y n + Py1 n =Q...(2) dx dy dz Put y1 n = z so that (1 n) y n =. dx dx 1 dz dz ∴ (2) becomes + Pz = Q or + P(1 n) z = Q(1 n), 1 − n dx dx which is Leibnitz s linear in z and can be solved easily. dy Example 11.21. Solve x dx + y = x3y6. −5 dy y Solution. Dividing throughout by xy6, y 6 + = x2...(i) dx x 5 6 dy dz 1 dz z Put y = z, so that 5y = dx dx ∴ (i) becomes + = x2 5 dx x dz 5 or − z= 5x2 which is Leibnitz s linear in z....(ii) dx x I.F. = e ∫ − (5 / x ) dx 5 log x 5 5 =e = elog x =x 2 ∴ the solution of (ii) is z (I.F.) = ∫ ( 5 x 2 ) (I.F.) dx + c or zx 5 = ∫ ( 5x ) x 5 dx + c x− 2 or y 5x 5. − 2 + c 5 = [∵ z = y 5] Dividing throughout by y 5x 5, 1 = (2.5 + cx2) x3y5 which is the required solution. dy Example 11.22. Solve xy (1 + xy2) = 1. (Nagpur, 2009) dx Solution. Rewriting the given equation as dx yx = y3x2 dy and dividing by x2, we have 2 dx 1 x yx = y3...(i) dy 1 2 dx dz Putting x = z so that x = (i) becomes dy dy dz + yz = y3 which is Leibnitz s linear in z. dy I.F. = e∫ y dy 2 Here = ey /2 3 ∴ the solution is z (I.F.) = ∫ (− y ) (I.F.) dy + c 2 2 y /2 2 1 2 y Put 12 y = t or ze =−∫ y. e2. ydy + c so that y dy = dt t = 2 ∫ t. e dt + c [Integrate by parts] 2 = 2 [t. et ∫ 1. e t dt ] + c = 2 [tet et] + c = (2 y2 ) e y /2 +c 1 2 1 2 − y − y or z = (2 y2) + ce 2 or 1/x = (2 y2) + ce 2. dy Note. General equation reducible to Leibnitz s linear is f ′ (y) + Pf(y) = Q...(A) dx where P, Q are functions of x. To solve it, put f (y) = z.  DIFFERENTIAL EQUATIONS OF FIRST ORDER 439 dy Example 11.23. Solve + x sin 2y = x3 cos2 y. (V.T.U., 2013 ; Marathwada, 2008 ; J.N.T.U., 2005) dx dy sin y cos y Solution. Dividing throughout by cos2 y, sec2 y + 2x = x3 dx cos2 y dy or sec2 y + 2x tan y = x3 which is of the form (A) above....(i) dx dy dz dz ∴ put tan y = z so that sec2 y = ∴ (i) becomes + 2xz = x3. dx dx dx I.F. = e∫ 2x dx x2 This is Leibnitz s linear equation in z. ∴ =e x 2 x 2 1 2 2 ∴ the solution is ze = ∫e x3 dx + c = ( x − 1) e x + c. 2 1 2 2 Replacing z by tan y, we get tan y = (x 1) + ce− x which is the required solution. 2 dz ⎛ z ⎞ z Example 11.24. Solve + log z = (log z)2. dx ⎝ x ⎠ x Solution. Dividing by z, the given equation becomes 1 dz 1 1 + log z = (log z)2...(i) z dx x x 1 dz dt Put log z = t so that =. ∴ (i) becomes z dx dx dt t t 2 1 dt 1 1 1 + = or +. =...(ii) dx x x t 2 dx x t x This being Bernoulli s equation, put 1/t = v so that (ii) reduces to dv v 1 dv 1 1 + = or − v=− dx x x dx x x This is Leibnitz s linear in v. ∴ I.F. = e ∫ 1 / x dx = 1 / x 1 1 1 1 ∴ the solution is = − ∫. dx + c = + c v. x x x x Replacing v by 1/log z, we get (x log z) 1 = x 1 + c or (log z) 1 = 1 + cx which is the required solution. PROBLEMS 11.6 Solve the following equations : dy dr 1. = y tan x y2 sec x. (P.T.U., 2005) 2. r sin θ = r2. cos θ dθ (V.T.U., 2005) dx 3. 2xy′ = 10x3y5 + y. 4. (x3y2 + xy) dx = dy. (B.P.T.U., 2005) 2 2 dy x + y + 1 5. =. (Bhillai, 2005) 6. x(x y) dy + y2 dx = 0. (I.S.M., 2001) dx 2 xy dy tan y ⎛ dy + 1 ⎞ 7. dx − 1 + x = (1 + x) ex sec y. (Bhopal, 2009) 8. ey ⎜ ⎟ = ex. (V.T.U., 2009) ⎝ dx ⎠ dy dy 9. sec2 y dx + x tan y = x3. 10. tan y dx + tan x = cos y cos2 x. (Sambalpur, 2002) dy y 11. =. (V.T.U., 2011) 12. (y log x 2) ydx xdy = 0. (V.T.U., 2006) dx x ( xy)  440 HIGHER ENGINEERING MATHEMATICS 11.11 EXACT DIFFERENTIAL EQUATIONS (1) Def. A differential equation of the form M(x, y) dx + N (x, y) dy = 0 is said to be exact if its left hand member is the exact differential of some function u (x, y) i.e., du = Mdx + Ndy = 0. Its solution, therefore, is u (x, y) = c. (2) Theorem. The necessary and sufficient condition for the differential equation Mdx + Ndy = 0 to be exact is ∂M ∂N = ∂y ∂x Condition is necessary : The equation Mdx + Ndy = 0 will be exact, if Mdx + Ndy ≡ du...(1) where u is some function of x any y. But du = ∂u dx + ∂u dy...(2) ∂x ∂y ∂u ∂u ∴ equating coefficients of dx and dy in (1) and (2), we get M = and N = ∂x ∂y 2 2 ∂M ∂ u ∂N ∂ u ∴ = and =. ∂y ∂y∂x ∂x ∂x∂y ∂ 2u ∂ 2u But = (Assumption) ∂y∂x ∂x∂y ∂M ∂N ∴ = which is the necessary condition for exactness. ∂y ∂x ∂M ∂N Condition is sufficient : i.e., if = , then Mdx + Ndy = 0 is exact. ∂y ∂x Let ∫ Mdx = u, where y is supposed constant while performing integration. ⎧ ∂M ∂N Then ∂ ∂x ( ∫ Mdx ) = ∂∂ux , i.e., M = ∂u ∂x ⎪ ∂y = ⎪ ∂x (given)...(3) ⎨ 2 ∂M ∂ u ∂N ∂ 2u ∂ ⎛ ∂u ⎞ ⎪ ∂2u ∂2u ∴ = ∂y ∂y∂x or ∂x ∂x∂y ∂x ⎜⎝ ∂y ⎟⎠ = = ⎪⎩ and = ∂y∂x ∂x∂y Interating both sides w.r.t. x (taking y as constant). ∂u N= + f (y), where f (y) is a function of y alone....(4) ∂y ∂u ⎧ ∂u ⎫ ∴ Mdx + Ndy = dx + ⎨ + f ( y) ⎬ dy [By (3) and (4)] ∂x ⎩ ∂y ⎭ ⎧ ∂u ∂u ⎫ = ⎨ ⎩ ∂x dx + ∂y dy ⎬ + f (y) dy = du + f(y) dy = d [u + ⎭ ∫ f ( y) dy]...(5) which shows that Mdx + Ndy = 0 is exact. (3) Method of solution. By (5), the equation Mdx + Ndy = 0 becomes d[u + ∫ f ( y) dy] = 0 Integrating u+ ∫ f ( y) dy = 0. But u= ∫ y constant Mdx and f(y) = terms of N not containing x. ∴ The solution of Mdx + Ndy = 0 is ∫ ( yMcons.) dx + ∫ (terms of N not containing x) dy =c ∂M ∂N provided =. ∂y ∂x  DIFFERENTIAL EQUATIONS OF FIRST ORDER 441 2 2 Example 11.25. Solve (y2 e xy + 4x3) dx + (2xy e xy 3y2) dy = 0. (V.T.U., 2006) 2 2 xy Solution. Here M = y2 e xy + 4x3 and N = 2xy e 3y2 ∂M 2 2 ∂N ∴ = 2y e xy + y2 e xy. 2xy = ∂y ∂x Thus the equation is exact and its solution is ∫ ( y cons Mdx t.) + ∫ (terms of N not containing x) dy = c 2 2 2 y x 3 2 i.e., ∫ ( y const.) ( y e + 4 x ) dx + ∫ ( 3 y ) dy = c or e xy + x4 y3 = c. 1 Example 11.26. Solve y ⎛⎜ 1 + ⎞⎟ + cos y dx + (x + log x { } x sin y) dy = 0. ⎝ x⎠ (Marathwada, 2008 S ; V.T.U., 2006) Solution. Here M = y (1 + 1/x) + cos y and N = x + log x x sin y ∂M ∂N ∴ = 1 + 1/x sin y = ∂y ∂x Then the equation is exact and its solution is ∫ ( y const) Mdx + ∫ (terms of N not containing x) dy = c ⎛ 1 + 1 ⎞ y + cos y dx = c ∫ ( y const) {⎜ ⎝ x⎠ ⎟ } or (x + log x) y + x cos y = c. Example 11.27. Solve (1 + 2xy cos x2 2xy) dx + (sin x2 x2) dy = 0. Solution. Here M = 1 + 2xy cos x2 2xy and N = sin x2 x2 ∴ ∂M = 2x cos x2 2x = ∂N ∂y ∂x Thus the equation is exact and its solution is ∫ ( y const) Mdx + ∫ (terms of N not containing x) = c 2 i.e., ∫ ( y const) (1 + 2 xy cos x − 2 xy) dx = c or x + y ⎡ ∫ cos x2. 2 xdx − ⎣ ∫ 2 x dx ⎦⎤ =c or x + y sin x2 yx2 = c. dy y cos x + sin y + y Example 11.28. Solve + = 0. (Rohtak, 2011) dx sin x + x cos y + x Solution. Given equation can be written as (y cos x + sin y + y) dx + ( sin x + x cos y + x) dy = 0. Here M = y cos x + sin y + y and N = sin x + x cos y + x. ∂M ∂N ∴ = cos x + cos y + 1 =. ∂y ∂x Thus the equation is exact and its solution is ∫ ( y cons Mdx t.) + ∫ (terms of N not containing x) dy = c i.e., ∫ ((yyconst.) cos x + sin y + y) dx + ∫ (0) dx = c or y sin x + (sin y + y) x = c.  442 HIGHER ENGINEERING MATHEMATICS Example 11.29. Solve (2x2 + 3y2 7) xdx (3x2 + 2y2 8) ydy = 0. (U.P.T.U., 2005) Solution. Given equation can be written as 2 2 ydy 2 x + 3 y − 7 = 2 x dx 3 x + 2 y2 − 8 y dy + x dx 5( x2 + y2 − 3) or = [By componendo & dividendo] y dy − x dx − x2 + y2 + 1 x dx + y dy x dx − y dy or 2 2 =5. 2 x + y −3 x − y2 − 1 Integrating both sides, we get 2 xdx + 2 ydy 2 xdx − 2 ydy 2 ∫ x + y −32 =5 x 2 − y2 − 1 +c ∫ or log (x2 + y2 3) = 5 log (x2 y2 1) + log c′ [Writing c = log c′] or x2 + y2 3 = c′ (x2 y2 1)5 which is the required solution. PROBLEMS 11.7 Solve the following equations : 1. (x2 ay) dx = (ax y2) dy. (Calicut, 2012) 2. (x2 + y2 a2) xdx + (x2 y2 b2) ydy = 0 (Kurukshetra, 2005) 3. (x2 4xy 2y2) dx + (y2 4xy 2x2) dy = 0. 4. (x4 2xy2 + y4) dx (2x2y 4xy3 + sin y) dy = 0 5. yexydx + (xexy + 2y) dy = 0 6. (5x4 + 3x2y2 2xy3) dx + (2x3y 3x2y2 5y4) dy = 0 (V.T.U., 2008) 2x y2 − 3 x 2 7. (3x2 + 6xy2) dx + (6x2y + 4y3) dy = 0 8. dx + dy = 0 3 y y4 9. y sin 2x dx (1 + y2 + cos2 x) dy = 0 (D.T.U., 2011) 10. (sec x tan x tan y ex ) dx + sec x sec2 y dy = 0 (Marathwada, 2008) 11. (2xy + y tan y) dx + x2 x tan2 y + sec2 y) dy = 0. (Nagpur, 2009) 11.12 EQUATIONS REDUCIBLE TO EXACT EQUATIONS Sometimes a differential equation which is not exact, can be made so on multiplication by a suitable factor called an integrating factor. The rules for finding integrating factors of the equation Mdx + Ndy = 0 are as follows : (1) I.F. found by inspection. In a number of cases, the integrating factor can be found after regrouping the terms of the equation and recognizing each group as being a part of an exact differential. In this connection the following integrable combinations prove quite useful : xdy + ydx = d (xy) xdy − ydx ⎛ y ⎞ xdy _ ydx ⎡ ⎛ y ⎞⎤ 2 = d ⎜ ⎟; = d ⎢ log ⎜ ⎟ ⎥ x x ⎝ ⎠ xy ⎣ ⎝ x ⎠⎦ xdy − ydx ⎛ x ⎞ xdy − ydx ⎛ y⎞ 2 = − d ⎜ ⎟; 2 2 = d ⎜ tan − 1 ⎟ y ⎝ y⎠ x +y ⎝ x⎠ xdy − ydx ⎛1 x + y⎞ 2 2 = d ⎜ log ⎟. x −y ⎝ 2 x − y⎠ Example 11.30. Solve y (2xy + ex) dx = ex dy. (Kurukshetra, 2005) Solution. It is easy to note that the terms yexdx and exdy should be put together. ∴ (yexdx exdy) + 2xy2 dx = 0  DIFFERENTIAL EQUATIONS OF FIRST ORDER 443 Now we observe that the term 2xy2 dx should not involve y2. This suggests that 1/y2 may be I.F. Multiply- ing throughout by 1/y2, it follows ye x dx − e x dy ⎛ ex ⎞ + 2xdx = 0 or d ⎜⎜ ⎟⎟ + 2xdx = 0 y2 ⎝ y ⎠ ex Integrating, we get + x2 = c which is the required solution. y (2) I.F. of a homogeneous equation. If Mdx + Ndy = 0 be a homogeneous equation in x and y, then 1/(Mx + Ny) is an integrating factor (Mx + Ny ≠ 0). Example 11.31. Solve (x2y 2xy2) dx (x3 3x2y) dy = 0. (Osmania, 2003 S) Solution. This equation is homogeneous in x and y. 1 1 1 ∴ I.F. = = 2 2 3 2 = 2 2 Mx + Ny ( x y −2 xy ) x − ( x − 3 x y) y x y Multiplying throughout by 1/x2y2, the equation becomes ⎛1 2⎞ ⎛ x 3⎞ ⎜ − ⎟ dx − ⎜ 2 − ⎟ dy = 0 which is exact. ⎝ y x⎠ ⎝ y y⎠ x ∴ the solution is ∫ ( y const) Mdx + ∫ (terms of N not containing x) dy = c or y 2 log x + 3 log y = c. (3) I.F. for an equation of the type f1(xy)ydx + f2(xy)xdy = 0. If the equation Mdx + Ndy = 0 be of this form, then 1/(Mx Ny) is an integrating factor (Mx Ny ≠ 0). Example 11.32. Solve (1 + xy) ydx + (1 xy) xdy = 0. (V.T.U., 2014 ; C.S.V.T.U., 2012) Solution. The given equation is of the form f1(xy) ydx + f2(xy) xdy = 0 Here M = (1 + xy) y, N = (1 xy) x. 1 1 1 ∴ I.F. = = = Mx − Ny (1 + xy) yx − (1 − xy) xy 2 x2 y2 Multiplying throughout by 1/2x2y2, it becomes ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎜ ⎟ dx + ⎜ ⎟ y ⎠ dy = 0, which is an exact equation. 2 + 2 − ⎝ 2x y 2 x ⎠ ⎝ 2 xy 2 ∴ the solution is ∫ ( y const) Mdx + ∫ (terms of N not containing x) dy = c 1 ⎛ 1⎞ 1 1 x 1 or ⎜ − ⎟ + log x − log y = c or log − = c′. 2y ⎝ x ⎠ 2 2 y xy (4) In the equation Mdx + Ndy = 0, ∂M ∂N − ∂y ∂x (a) if be a function of x only = f (x) say, then e∫ f ( x ) dx is an integrating factor. N ∂N ∂M − ∂x ∂y (b) if be a function of y only = F(y) say, then e∫ F ( y) dy is an integrating factor. M 3 Example 11.33. Solve ( xy 2 − e1/x ) dx x2 ydy = 0. (C.S.V.T.U., 2009 ; Mumbai, 2007) 3 Solution. Here M = xy2 e1/x and N = x2 y ∂M ∂N − ∂y ∂x 2 xy − (− 2 xy) 4 = 2 = − which is a function of x only. N −x y x  444 HIGHER ENGINEERING MATHEMATICS −4 ∴ I.F. = e∫ x dx =e 4 log x =x 4 ⎛ y2 1 3 ⎞ y Multiplying throughout by x 4, we get ⎜⎜ 3 − 4 e1 / x ⎟⎟ dx − 2 dy = 0 ⎝ x 4 ⎠ x which is an exact equation. ∴ the solution is ∫ ( y const) ( Mdx) + ∫ (terms of N not containing x) dy = c. ⎛ y2 1 1 / x3 ⎞ or ∫ ⎜⎜ 3 − 4 e ⎝ x x ⎟⎟ dx + 0 = c ⎠ y2 x − 2 1 1 x − 3 1 y2 ∫e x− 3 4) or + ( 3x dx = c or e − = c. 2 3 3 2 x2 Otherwise it can be solved as a Bernoulli s equation (§ 11.10) Example 11.34. Solve (xy3 + y) dx + 2 (x2y2 + x + y4) dy = 0. (D.T.U., 2013) Solution. Here M = xy3 + y, N = 2(x2y2 + x + y4) 1 ⎛ ∂N ∂M ⎞ 1 1 ∴ ⎜ − ⎟= (4xy2 + 2