Chapter ™He who seeks for methods without having a definite problem in mind seeks for the most part in vain. – D. HILBERT • 9.1 Introduction In Class XI and in Chapter 5 of the present book, we discussed how to differentiate a given function f with respect to an independent variable, i.e., how to find f ′(x) for a given function f at each x in its domain of definition. Further, in the chapter on Integral Calculus, we discussed how to find a function f whose derivative is the function g, which may also be formulated as follows: For a given function g, find a function f such that dy = g(x), where y = f(x) ... (1)dxAn equation of the form (1) is known as a differential equation. A formal definition will be given later. These equations arise in a variety of applications, may it be in Physics, Chemistry, Biology,Anthropology, Geology, Economics etc. Hence, an indepth study of differential equations has assumed prime importance in all modern scientific investigations. In this chapter, we will study some basic concepts related to differential equation, general and particular solutions of a differential equation, formation of differential equations, some methods to solve a first order - first degree differential equation and some applications of differential equations in different areas. 9.2 Basic Concepts We are already familiar with the equations of the type: x2 – 3x + 3 =0 ... (1) sin x + cos x = 0 ... (2) x + y = 7 ... (3) Let us consider the equation: dyx +y = 0 ... (4)dx We see that equations (1), (2) and (3) involve independent and/or dependent variable (variables) only but equation (4) involves variables as well as derivative of the dependent variable y with respect to the independent variable x. Such an equation is called a differential equation. In general, an equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation. A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation, e.g., 2dy ⎛ dy ⎞3 22 +⎜ ⎟ = 0 is an ordinary differential equation .... (5)dx ⎝ dx ⎠ Of course, there are differential equations involving derivatives with respect to more than one independent variables, called partial differential equations but at this stage we shall confine ourselves to the study of ordinary differential equations only. Now onward, we will use the term ‘differential equation’ for ‘ordinary differential equation’. 9.2.1. Order of a differential equation Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation. Consider the following differential equations: dy = ex ... (6)dx2dy +y = 0 ... (7) dx2 332⎛ dy ⎞⎛ dy ⎞ ⎜ ⎟+ x2 ⎜⎟ = 0 ... (8)dx3 ⎝ dx2 ⎠⎝⎠ The equations (6), (7) and (8) involve the highest derivative of first, second and third order respectively. Therefore, the order of these equations are 1, 2 and 3 respectively. 9.2.2 Degree of a differential equation To study the degree of a differential equation, the key point is that the differential equation must be a polynomial equation in derivatives, i.e., y′, y″, y″′ etc. Consider the following differential equations: 232dy ⎛ dy ⎞ dy + 2⎜ ⎟−+ y = 0 ... (9) dx3 dx 2 dx⎝⎠⎛ dy ⎞2 ⎛ dy ⎞ ⎜ ⎟+⎜ ⎟−sin2 y = 0 ... (10) ⎝ dx ⎠⎝ dx ⎠dy ⎛ dy ⎞+sin ⎜⎟ = 0 ... (11) dx ⎝ dx ⎠We observe that equation (9) is a polynomial equation in y″′, y″ and y′, equation (10) is a polynomial equation in y′ (not a polynomial in y though). Degree of such differential equations can be defined. But equation (11) is not a polynomial equation in y′ and degree of such a differential equation can not be defined. By the degree of a differential equation, when it is a polynomial equation in derivatives, we mean the highest power (positive integral index) of the highest order derivative involved in the given differential equation. In view of the above definition, one may observe that differential equations (6), (7), (8) and (9) each are of degree one, equation (10) is of degree two while the degree of differential equation (11) is not defined. Example 1 Find the order and degree, if defined, of each of the following differential order is one. It is a polynomial equation in y′ and the highest power raised to equations: (i) dy dx − cos x = 0 (ii) xy 2 dx2 d y + x ⎜⎛ ⎝ dy ⎞2 dx ⎠⎟ − y dy dx = 0 (iii) y′′′ + y2 y′ + e = 0 Solution (i) The highest order derivative present in the differential equation is dy dx , so its dy dx is one, so its degree is one. 2dy(ii) The highest order derivative present in the given differential equation is 2 , sodx 2dy dyits order is two. It is a polynomial equation in 2 and dx and the highestdx2dypower raised to 2 is one, so its degree is one.dx(iii) The highest order derivative present in the differential equation is y′′′ , so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined. Determine order and degree (if defined) of differential equations given in Exercises 1 to 10. 4 42dy ⎛ ds ⎞ ds1. 4 + sin( y′′′ ) = 0 2. y′ + 5y = 0 3. ⎜ ⎟+ 3s 2 = 0 dx ⎝ dt ⎠ dt 2 2⎛ dy ⎞2 ⎛ dy ⎞ dy4. ⎜ ⎟+ cos ⎜ ⎟= 0 5. = cos3 x + sin3 x ⎝ dx2 ⎠⎝ dx ⎠ dx2 y′′′6. ()2 + (y″)3 + (y′)4 + y5 = 0 7. y′′′ + 2y″ + y′ = 0 8. y′ + y = ex 9. y″ + (y′)2 + 2y = 0 10. y″ + 2y′ + sin y = 0 11. The degree of the differential equation ⎛23dy ⎞⎛⎞ dy 2 ⎛⎞ dy++sin +1 =0 is⎜ 2 ⎟⎜⎟ ⎜⎟ ⎠dx ⎝⎠ dx⎝dx ⎝⎠ (A) 3 (B) 2 (C) 1 (D) not defined 12. The order of the differential equation 2 2 dy dy y 02x −3 +=is dx2 dx (A) 2 (B) 1 (C) 0 (D) not defined 9.3. General and Particular Solutions of a Differential Equation In earlier Classes, we have solved the equations of the type: x2 + 1 =0 ... (1) sin2 x – cos x = 0 ... (2) Solution of equations (1) and (2) are numbers, real or complex, that will satisfy the given equation i.e., when that number is substituted for the unknown x in the given equation, L.H.S. becomes equal to the R.H.S.. 2dyNow consider the differential equation 2 +=0y ... (3)dx In contrast to the first two equations, the solution of this differential equation is a function φ that will satisfy it i.e., when the function φ is substituted for the unknown y (dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S.. The curve y = φ (x) is called the solution curve (integral curve) of the given differential equation. Consider the function given by y = φ (x) = a sin (x + b), ... (4) where a, b ∈ R. When this function and its derivative are substituted in equation (3), L.H.S. = R.H.S.. So it is a solution of the differential equation (3). πLet a and b be given some particular values say a = 2 and b = , then we get a 4 ⎛ π⎞function y = φ1(x) = 2sin ⎜x +⎟ ... (5)⎝ 4 ⎠ When this function and its derivative are substituted in equation (3) again L.H.S. = R.H.S.. Therefore φ1 is also a solution of equation (3). Function φ consists of two arbitrary constants (parameters) a, b and it is called general solution of the given differential equation. Whereas function φ1 contains no arbitrary constants but only the particular values of the parameters a and b and hence is called a particular solution of the given differential equation. The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation. The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation. Example 2 Verify that the function y = e–3x is a solution of the differential equation 2dy dy +−6y =0 dx2 dx Solution Given function is y = e–3x. Differentiating both sides of equation with respect to x , we get dy −3x=−3e ... (1)dx Now, differentiating (1) with respect to x, we have 2dy =9 e – 3x dx2 2dy dy Substituting the values of 2 ,and y in the given differential equation, we getdx dx L.H.S. = 9 e–3x + (–3e–3x) – 6.e–3x = 9 e–3x – 9 e–3x = 0 = R.H.S.. Therefore, the given function is a solution of the given differential equation. Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution 2dyof the differential equation y 0+= dx2 Solution The given function is y = a cos x + b sin x ... (1) Differentiating both sides of equation (1) with respect to x, successively, we get dy =– a sinx + b cos xdx2dy 2 =– a cos x – b sin xdx2dySubstituting the values of 2 and y in the given differential equation, we getdxL.H.S. = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R.H.S. Therefore, the given function is a solution of the given differential equation. In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: 1. y = ex + 1 : y″ – y′ = 0 2. y = x2 + 2x + C : y′ – 2x – 2 = 0 3. y = cos x + C : y′ + sin x = 0 xy4. y = 1+ x2: y′ = 21+ x 5. y = Ax : xy′ = y (x ≠ 0) 6. y = x sin x : xy′ = y + xx2 − y2 (x ≠ 0 and x > y or x < – y) y2 7. xy = log y + C : y′ = (xy ≠ 1)1−xy 8. y – cos y = x :(y sin y + cos y + x) y′ = y 9. x + y = tan–1y : y2 y′ + y2 + 1 = 0 dy10. y = 2 − x2 x ∈ (–a, a): x + y = 0 (y ≠ 0)a dx11. The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A)0 (B)2 (C)3 (D)4 12. The number of arbitrary constants in the particular solution of a differential equation of third order are: (A)3 (B)2 (C)1 (D)0 9.4 Formation of a Differential Equation whose General Solution is given We know that the equation x2 + y2 + 2x – 4y + 4 = 0 ... (1) represents a circle having centre at (–1, 2) and radius 1 unit. Differentiating equation (1) with respect to x, we get dy x +1 = (y ≠ 2) ... (2) dx2−y which is a differential equation. You will find later on [See (example 9 section 9.5.1.)] that this equation represents the family of circles and one member of the family is the circle given in equation (1). Let us consider the equation x2 + y2 = r2 ... (3) By giving different values to r, we get different members of the family e.g. x2 + y2 = 1, x2 + y2 = 4, x2 + y2 = 9 etc. (see Fig 9.1). Thus, equation (3) represents a family of concentric circles centered at the origin and having different radii. We are interested in finding a differential equation that is satisfied by each member of the family. The differential equation must be free from r because r is different for different members of the family. This equation is obtained by differentiating equation (3) with respect to x, i.e., dy dy2x + 2y = 0 or x + y = 0 ... (4) Fig 9.1dxdxwhich represents the family of concentric circles given by equation (3). Again, let us consider the equation y = mx + c ... (5) By giving different values to the parameters m and c, we get different members of the family, e.g., y = x (m = 1, c = 0) y = 3 x (m = 3, c = 0) y = x + 1 (m = 1, c = 1) y = – x (m = – 1, c = 0) y = – x – 1 (m = – 1, c = – 1) etc. ( see Fig 9.2). Thus, equation (5) represents the family of straight lines, where m, c are parameters. We are now interested in finding a differential equation that is satisfied by each member of the family. Further, the equation must be free from m and c because m and c are different for different members of the family. This is obtained by differentiating equation (5) with respect to x, successively we get 2dy dy= m , and 2 = 0 ... (6) X’dx dx The equation (6) represents the family of straight lines given by equation (5). Note that equations (3) and (5) are the general Y’ solutions of equations (4) and (6) respectively. Fig 9.2 9.4.1 Procedure to form a differential equation that will represent a given family of curves (a) If the given family F1 of curves depends on only one parameter then it is represented by an equation of the form F1 (x, y, a) = 0 ... (1) For example, the family of parabolas y2 = ax can be represented by an equation of the form f (x, y, a) : y2 = ax. Differentiating equation (1) with respect to x, we get an equation involving y′, y, x, and a, i.e., g (x, y, y′, a) =0 ... (2) The required differential equation is then obtained by eliminating a from equations (1) and (2) as F(x, y, y′) =0 ... (3) (b) If the given family F2 of curves depends on the parameters a, b (say) then it is represented by an equation of the from F2 (x, y, a, b) =0 ... (4) Differentiating equation (4) with respect to x, we get an equation involving y′, x, y, a, b, i.e., g (x, y, y′, a, b) =0 ... (5) But it is not possible to eliminate two parameters a and b from the two equations and so, we need a third equation. This equation is obtained by differentiating equation (5), with respect to x, to obtain a relation of the form h (x, y, y′, y″, a, b) =0 ... (6) The required differential equation is then obtained by eliminating a and b from equations (4), (5) and (6) as F (x, y, y′, y″) =0 ... (7) Example 4 Form the differential equation representing the family of curves y = mx, where, m is arbitrary constant. Solution We have y = mx ... (1) Differentiating both sides of equation (1) with respect to x, we get dy dx = m Substituting the value of m in equation (1) we get y dy dx x or x dy dx – y =0 which is free from the parameter m and hence this is the required differential equation. Example 5 Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants. Solution We have y = a sin(x + b) ... (1) Differentiating both sides of equation (1) with respect to x, successively we get dy = a cos (x + b) ... (2)dx2dy 2 =– a sin(x + b) ... (3)dxEliminating a and b from equations (1), (2) and (3), we get 2dy +y = 0 ... (4) dx2 which is free from the arbitrary constants a and b and hence this the required differential equation. Example 6 Form the differential equation representing the family of ellipses having foci on x-axis and centre at the origin. Solution We know that the equation of said family of ellipses (see Fig 9.3) is x2 y2 2 + 2 =1 ... (1)ab Fig 9.3 2x 2 y dy Differentiating equation (1) with respect to x, we get += 0 a2 b2 dx ydy ⎞−b2⎛ or ⎜⎟ = ... (2)x ⎝ dx ⎠ a2 Differentiating both sides of equation (2) with respect to x, we get dy2 xyy dy dx dy 2 2 =0 xdx xdx 2dy dy 2 dy or xy 2 x – y = 0 ... (3) dxdx dx which is the required differential equation. Example 7 Form the differential equation of the family of circles touching the x-axis at origin. Solution Let C denote the family of circles touching x-axis at origin. Let (0, a) be the coordinates of the centre of any member of the family (see Fig 9.4). Therefore, equation of family C is x2 + (y – a)2 = a2 or x2 + y2 = 2ay ... (1) X’ where, a is an arbitrary constant. Differentiating both sides of equation (1) with respect to x,we get Y’dy dy2x 2y =2a Fig 9.4dxdxor x y dy dx = a dy dx or a = dyxy dx dy ... (2) dx Substituting the value of a from equation (2) in equation (1), we get dyxy x2 + y2 =2y dydx dx dy dy22 2 or ( xy ) =2xy 2ydxdx dy 2xyor = dxx2– y2 This is the required differential equation of the given family of circles. Example 8 Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis. Solution Let P denote the family of above said parabolas (see Fig 9.5) and let (a, 0) be the focus of a member of the given family, where a is an arbitrary constant. Therefore, equation of family P is y2 =4ax ... (1) Differentiating both sides of equation (1) with respect to x, we get dy2y =4a ... (2)dxSubstituting the value of 4a from equation (2) in equation (1), we get y2 = ⎛ dy ⎞⎜ 2y ⎟ () x ⎝ dx ⎠ 2 dyor y −2xy =0dx which is the differential equation of the given family of parabolas. Fig 9.5 In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. x y1. +=1 2. y2 = a (b2 – x2) 3. y = ae3x + be– 2x ab 4. y = e2x (a + bx) 5. y = ex (a cos x + b sin x) 6. Form the differential equation of the family of circles touching the y-axis at origin. 7. Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis. 8. Form the differential equation of the family of ellipses having foci on y-axis and centre at origin. 9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin. 10. Form the differential equation of the family of circles having centre on y-axis and radius 3 units. 11. Which of the following differential equations has y = c1 ex + c2 e –x as the general solution? 2 222dy dy dy dy(A) +y =0 y 0 10 (D) −=(B) −= (C) += 102 222dx dx dx dx 12. Which of the following differential equations has y = x as one of its particular solution? 2 2dy dy dydy (A) −x2 +xy =x (B) +x +xy =x dx2 dx dx2 dx 2 2dy 2 dy dydy (C) x xy 0 (D) +x +xy =0 dx2 dx dx2 dx 9.5. Methods of Solving First Order, First Degree Differential Equations In this section we shall discuss three methods of solving first order first degree differential equations. 9.5.1 Differential equations with variables separable A first order-first degree differential equation is of the form dy =F(x, y) ... (1)dxIf F(x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x and h(y) is a function of y, then the differential equation (1) is said to be of variable separable type. The differential equation (1) then has the form dy = h (y) . g (x) ... (2)dxIf h(y) ≠ 0, separating the variables, (2) can be rewritten as 1 dy = g (x) dx ... (3)()hy Integrating both sides of (3), we get 1 dy = g()xdx ... (4)()∫ hy ∫ Thus, (4) provides the solutions of given differential equation in the form H(y) =G(x) + C 1Here, H (y) and G (x) are the anti derivatives of and g (x) respectively and()hyC is the arbitrary constant. Example 9 Find the general solution of the differential equation 1 2 dy x dx y + = − , (y ≠ 2) Solution We have dy dx = 1 2 x y + − ... (1) Separating the variables in equation (1), we get (2 – y) dy =(x + 1) dx ... (2) Integrating both sides of equation (2), we get (2 )=(x +1) − ydy dx∫∫ y2 x2 or 2y − = + x + C122 or x2 + y2 + 2x – 4y + 2 C1 =0 or x2 + y2 + 2x – 4y + C = 0, where C = 2C1 which is the general solution of equation (1). dy 1+y2 Example 10 Find the general solution of the differential equation = . dx 1+x2 Solution Since 1 + y2 ≠ 0, therefore separating the variables, the given differential equation can be written as dy dx 2 = 2 ... (1)1+y1+x Integrating both sides of equation (1), we get dy dx = 2∫1+y2 ∫1+x or tan–1 y = tan–1x + C which is the general solution of equation (1). dy 2Example 11 Find the particular solution of the differential equation =−4xy givendx that y = 1, when x = 0. Solution If y ≠ 0, the given differential equation can be written as dy y2 =–4x dx ... (1) Integrating both sides of equation (1), we get dy = −4∫x dx∫2y 1 or − =–2x2 + C y 1 or y = 2 ... (2)2x −C Substituting y = 1 and x = 0 in equation (2), we get, C = – 1. Now substituting the value of C in equation (2), we get the particular solution of the 1given differential equation as y = . 2x2 +1 Example 12 Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x2 + 1) dx (x ≠ 0). Solution The given differential equation can be expressed as 2x21dy* = dx* x ⎛ 1 ⎞ or dy = ⎜ 2x +⎟dx ... (1)⎝ x ⎠ Integrating both sides of equation (1), we get ∫dy = ∫⎜⎛ 2x + 1 ⎟⎞dx ⎝ x ⎠ or y = x2 + log | x | + C ... (2) Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we get C = 0. Now substituting the value of C in equation (2) we get the equation of the required curve as y = x2 + log | x|. Example 13 Find the equation of a curve passing through the point (–2, 3), given that 2xthe slope of the tangent to the curve at any point (x, y) is . y2 Solution We know that the slope of the tangent to a curve is given by dy . dx dy 2x so, = 2 ... (1)dxy Separating the variables, equation (1) can be written as y2 dy =2x dx ... (2) Integrating both sides of equation (2), we get 2 ∫∫ ydy =2x dx y3 or = x2 + C ... (3) 3 dy* The notation due to Leibnitz is extremely flexible and useful in many calculation and formaldx transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers. Bytreating dx and dy like separate entities, we can give neater expressions to many calculations. Refer: Introduction to Calculus and Analysis, volume-I page 172, By Richard Courant, Fritz John Spinger – Verlog New York. Substituting x = –2, y = 3 in equation (3), we get C = 5. Substituting the value of C in equation (3), we get the equation of the required curve as 1 y32 2= x +5 or y = (3x +15) 3 3 Example 14 In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself? Solution Let P be the principal at any time t. According to the given problem, dp ⎛ 5 ⎞ = ⎜ ⎟× P dt⎝100 ⎠ dp P or = ... (1)dt20 separating the variables in equation (1), we get dpdt = ... (2)P 20 Integrating both sides of equation (2), we get log P = t + C120 t 20 C1or P = e ⋅e t 20 C1or P = C e (where e = C ) ... (3) Now P = 1000, when t = 0 Substituting the values of P and t in (3), we get C = 1000. Therefore, equation (3), gives t P = 1000 e20 Let t years be the time required to double the principal. Then t 2000 = 1000 e20 ⇒ t = 20 loge 2 For each of the differential equations in Exercises 1 to 10, find the general solution: dy 1cos dy 2− x1. = 2. = 4 − y (2 < y−< 2) dx 1cos dx+ x dy y3. +=1( y ≠1) 4. sec2 x tan y dx + sec2 y tan x dy = 0dx dy =+225. (ex + e –x) dy – (ex – e –x) dx = 0 6. (1 x ) (1 +y )dx dy7. y log y dx – x dy = 0 8. x5 dx =−y5 dy9. dx =sin−1 x 10. ex tan y dx + (1 – ex) sec2 y dy = 0 For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition: 3 2 dy+ ++11. (xxx 1) = 2x2 + x; y = 1 when x = 0dx dy12. (2 −1) =1xx ; y = 0 when x = 2dx dy⎛⎞ 13. cos ⎜ ⎟=a (a ∈ R); y = 2 when x = 0dx⎝⎠dy14. =y tan x ; y = 1 when x = 0 dx 15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x. dy16. For the differential equation xy (x 2) ( y +2) , find the solution curve=+ dx passing through the point (1, –1). 17. Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point. 18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1). 19. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds. 20. 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). 21. 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 (e0.5 = 1.648). 22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present? dy x+y23. The general solution of the differential equation =e is dx (A) ex + e –y = C (B) ex + ey = C (C) e –x + ey = C (D) e –x + e –y = C 9.5.2 Homogeneous differential equations Consider the following functions in x and y (x, y) = y2 + 2xy, F (x, y) = 2x – 3y,F1 2y⎛⎞F3 (x, y) = cos ⎜⎟, F4 (x, y) = sin x + cos yx⎝⎠If we replace x and y by λx and λy respectively in the above functions, for any nonzero constant λ, we get F1 (λx, λy) = λ2 (y2 + 2xy) = λ2 F1 (x, y) F2 (λx, λy) = λ (2x – 3y) = λ F2 (x, y) λyy⎛⎞ ⎛⎞ F3 (λx, λy) = cos ⎜ ⎟ =cos ⎜⎟ = λ0 F3 (x, y)λx ⎝⎠x⎝⎠ (λx, λy) = sin λx + cos λy ≠λn F(x, y), for any n ∈ NF4 4 Here, we observe that the functions F1, F2, F3 can be written in the form F(λx, λy) = λn F(x, y) but F4 can not be written in this form. This leads to the following definition: A function F(x, y) is said to be homogeneous function of degree n if F(λx, λy) = λn F(x, y) for any nonzero constant λ. We note that in the above examples, F1, F2, F3 are homogeneous functions of degree 2, 1, 0 respectively but F4 is not a homogeneous function. We also observe that ⎛y22 y ⎞ y22 ⎛⎞F1(x, y) = x ⎜2 + ⎟=xh 1 ⎜⎟⎝xx ⎠ x⎝⎠ 2xx2 ⎛⎞ 2 ⎛⎞ or F(x, y) = y ⎜1+ ⎟=yh 2 ⎜⎟ 1yy⎝⎠ ⎝⎠ 3yy1 ⎛⎞ 1 ⎛⎞ F2(x, y) = x ⎜2 − ⎟=xh 3 ⎜⎟ x ⎠ x⎝ ⎝⎠ 1 x ⎞ 1 x23 yh or F2(x, y) = y ⎛ −= ⎛⎞ ⎜⎟ 4 ⎜⎟ yy⎝ ⎠⎝⎠ 0 ⎛⎞ 0 ⎛⎞yyF3(x, y) = x cos ⎜⎟=xh5 ⎜⎟⎝⎠xx⎝⎠ n ⎛⎞yF(x, y) ≠ xh6 ⎜⎟, for any n ∈ N x x 4⎝⎠⎛⎞yhor F(x, y) ≠ n 7 ⎜⎟, for any n ∈ N4 y⎝⎠Therefore, a function F (x, y) is a homogeneous function of degree n if n ⎛⎞y nx⎛⎞F(x, y) = xg ⎜⎟ or yh ⎜⎟xy⎝⎠ ⎝⎠ dyA differential equation of the form = F (x, y) is said to be homogenous ifdx F(x, y) is a homogenous function of degree zero. To solve a homogeneous differential equation of the type dy yF, ⎜⎟=( )xy = g ⎛⎞... (1)⎝⎠dx x We make the substitution y = v . x ... (2) Differentiating equation (2) with respect to x, we get dy dv+= vx ... (3)dxdx dySubstituting the value of from equation (3) in equation (1), we getdxdv+vx = g (v)dx dv or x = g (v) – v ... (4)dxSeparating the variables in equation (4), we get () dv g v v− = dx x ... (5) Integrating both sides of equation (5), we get ∫ () dv g v v− = 1 Cdx x∫ + ... (6) Equation (6) gives general solution (primitive) of the differential equation (1) when we replace v by y . x dyExample 15 Show that the differential equation (x – y) = x + 2y is homogeneousdxand solve it. Solution The given differential equation can be expressed as dy dx = 2x y x y + − ... (1) Let F (x, y) = x x 2y y Now F(λx, λy) = ( 2 ) ( ) x y xy 0 (, ) F xy Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation. Alternatively, ⎛ 2y ⎞1+dy ⎜⎟ yx ⎛⎞=⎜ ⎟= g ⎜⎟... (2)⎝⎠dxy x⎜1−⎟ ⎝ x ⎠yR.H.S. of differential equation (2) is of the form g and so it is a homogeneousxfunction of degree zero. Therefore, equation (1) is a homogeneous differential equation. To solve it we make the substitution y = vx ... (3) Differentiating equation (3) with respect to, x we get dy dv+= vx ... (4)dxdx dySubstituting the value of y and in equation (1) we getdx dv 12+v vx =+dx 1−v dv 12+v or x = −vdx1−v dv v2 v 1 or x = dx1 v v 1 dx or dv = v2 v1 x Integrating both sides of equation (5), we get v 1 dx 2 dv = vv1 x 12v 13 or 2 dv = – log |x | + C12 vv1 12v 1 31dv dv log x C1or 2 v2 v 12 v2 v 1 1 log v2 v 1 31 2 dv log xor 2 C12213 v 22 1 32 12v 1log v2 v 1 . tan log x C1or 2 23 3 1 2v 1 or log v2 v 1 1 log x2 3 tan 1 3 C1 (Why?)2 2 yReplacing v by , we getx 1 y2 y 1 2yx or 2log 1 log x2 3tan 1 C2 2xx 3x 1 1 ⎛y2 y ⎞ ⎛2y +x ⎞log ⎜2 ++1⎟x2 =3 tan −1 +C⎟ 1or 2 ⎜xx ⎝ 3x ⎠⎝⎠ −12yx ⎞⎛+log ( y2 xy x2)++ =2 3 tan +2C ⎟ 1or ⎜⎝ 3x ⎠ −1 ⎛x +2y ⎞(x2 ++ y2)xy log =2 3 tan +Cor ⎜ ⎟⎝ 3x ⎠ which is the general solution of the differential equation (1) ⎛⎞ydy y⎛⎞Example 16 Show that the differential equation x cos ⎜⎟ =y cos ⎜⎟+x is ⎝⎠xdx x⎝⎠ homogeneous and solve it. Solution The given differential equation can be written as y⎛⎞y cos⎜⎟+xdy x⎝⎠ = ... (1)dxy⎛⎞x cos ⎜⎟x⎝⎠ dyIt is a differential equation of the form =F( , xy). dx y⎛⎞y cos⎜⎟+x x⎝⎠Here F(x, y) = y⎛⎞x cos⎜⎟x⎝⎠ Replacing x by λx and y by λy, we get y⎛⎞ ⎜⎟xF(λx, λy) = λ[cos y ⎝⎠+x] =λ0 [F( x, y)] ⎛ y ⎞λx cos ⎜⎟⎝ x ⎠ Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it we make the substitution y = vx ... (2) Differentiating equation (2) with respect to x, we get dy +dv = vx ... (3)dxdx dySubstituting the value of y and in equation (1), we getdxdv vcosv +1 vx =+dx cosv dv vcos v +1 or x = −vdxcos v dv 1 or x = dxcosv dx or cos v dv = x Therefore ∫cosvdv = ∫1 dx x or sin v = log | x | + log |C| or sin v = log |Cx | yReplacing v by , we getx ysin ⎛⎞ = log |Cx |⎜⎟x⎝⎠which is the general solution of the differential equation (1). x ⎛ x ⎞ y ⎜ y ⎟Example 17 Show that the differential equation 2ye dx +− 2xe is⎝y ⎠dy =0 homogeneous and find its particular solution, given that, x = 0 when y = 1. Solution The given differential equation can be written as x dx 2xey −y= ... (1)dyx 2yey x 2xe y −yLet F(x, y) = x 2ye y ⎛ x ⎞ λ⎜2xe y −y ⎟ ⎜⎟ Then F(λx, λy) = ⎝ ⎠=λ0[F( , xy)] ⎛ x ⎞ λ⎜2ye y ⎟ ⎜⎟⎝⎠ Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it, we make the substitution x = vy ... (2) Differentiating equation (2) with respect to y, we get dx dv vy=+dydy 404 MATHEMATICS dxSubstituting the value of x and in equation (1), we getdydv 2vev −1 vy =+dy 2ev dv 2vev −1 or y = v − v dy2e dv 1 or y = − dy2ev −dyor 2ev dv = y vdy∫⋅ −∫or 2edv = y or 2 ev = – log |y| + C x and replacing v by , we getyx 2 ey + log | y| =C ... (3) Substituting x = 0 and y = 1 in equation (3), we get 2 e0 + log |1| = C ⇒ C = 2 Substituting the value of C in equation (3), we get x 2 ey + log | y | =2 which is the particular solution of the given differential equation. Example 18 Show that the family of curves for which the slope of the tangent at any x2 + y2 point (x, y) on it is , is given by x2– y2 = cx. 2xy dySolution We know that the slope of the tangent at any point on a curve is dx . dy x2 + y2 Therefore, = dx2xy y2 1+ 2dy x or = ... (1)dx2y x Clearly, (1) is a homogenous differential equation. To solve it we make substitution y = vx Differentiating y = vx with respect to x, we get dy dv+= vxdxdx dv 1+ v2 +or vx = dx 2v dv 1− v2 or x = dx 2v 2v dxdv = 1− v2 x 2v dx or dv = − v2 −1 x 2v 1Therefore dv = −∫ dx∫2v −1 x or log | v2 – 1| = – log | x | + log |C1| or log |(v2 – 1) (x) | = log |C1| or (v2 – 1) x =± C1 yReplacing v by , we getx⎛ y2 ⎞ ⎜ 2 −1⎟ x =± C1⎝ x ⎠or (y2 – x2) =± C1 x or x2 – y2 = Cx In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them. x +y′1. (x2 + xy) dy = (x2 + y2) dx 2. y = x 3. (x – y) dy – (x + y) dx = 0 4. (x2 – y2) dx + 2xy dy = 0 2 dy 225. x =x −2y +xy 6. x dy – y dx = dx ⎛⎞yy ⎫ ⎧⎛⎞ y⎧ ⎛⎞ y ⎛⎞ ⎫ 7. ⎨x cos ⎜⎟+y sin ⎜⎟⎬y dx =⎨y sin ⎜⎟−x cos ⎜⎟ ⎬x dy ⎝⎠xx ⎝⎠ x⎩ ⎝⎠⎭ ⎩ x ⎝⎠ ⎭ dyy y⎛⎞ ⎛⎞8. yxsin ⎜⎟0 9. ydx x log ⎜⎟dy 2x dy 0x −+= +−= dxx x⎝⎠ ⎝⎠ xx x1 eydx ey 1 dy 010. y For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition: 11. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1 12. x2 dy + (xy + y2) dx = 0; y = 1 when x = 1 13. xsin2 y y dx xdy 0; y when x = 1 x4 dy y ⎛⎞y14. −+cosec ⎜⎟=0; y = 0 when x = 1dx x ⎝⎠x 22 dy15. 2xy y 2x =0+− ; y = 2 when x = 1dx dx x⎛⎞ 16. A homogeneous differential equation of the from =h⎜⎟can be solved bydy y⎝⎠making the substitution. (A) y = vx (B) v = yx (C) x = vy (D) x = v 17. Which of the following is a homogeneous differential equation? (A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (B) (xy) dx – (x3 + y3) dy = 0 (C) (x3 + 2y2) dx + 2xy dy = 0 (D) y2 dx + (x2 – xy – y2) dy = 0 9.5.3 Linear differential equations A differential equation of the from dy +Py =Qdx where, P and Q are constants or functions of x only, is known as a first order linear differential equation. Some examples of the first order linear differential equation are dy +y = sin xdx dy 1⎛⎞+⎜⎟y = ex dx x⎝⎠dy ⎛ y ⎞ 1+ =⎜⎟dx ⎝x log x ⎠x Another form of first order linear differential equation is dx +P1x =Q1dy where, P1 and Q1 are constants or functions of y only. Some examples of this type of differential equation are dx +x = cos ydy dx −2x – y+ = y2edy y To solve the first order linear differential equation of the type dy Py = Q ... (1) dxMultiply both sides of the equation by a function of x say g (x) to get dyg(x) + P.(g (x)) y =Q. g (x) ... (2)dxChoose g (x) in such a way that R.H.S. becomes a derivative of y . g (x). dy di.e. g (x) + P. g (x) y =[y . g (x)]dxdx dy dyor g (x) + P. g (x) y = g(x) + yg′ (x)dxdx ⇒ P. g (x) = g′ (x) g′()x or P = g ()x Integrating both sides with respect to x, we get ′()gx∫Pdx = ∫ dx ()gx or ∫P⋅dx =log (g (x)) or g(x) = ∫P dx e∫P dxOn multiplying the equation (1) by g(x) = e, the L.H.S. becomes the derivative ∫P dxof some function of x and y. This function g(x) = e is called Integrating Factor (I.F.) of the given differential equation. Substituting the value of g (x) in equation (2), we get P dx dy P dx P dx e P ey =Q edxd P dx dxor ye =Qe P dxIntegrating both sides with respect to x, we get yeP dx =Q e P dx dx P dx Pdx or y = e Q e dx C which is the general solution of the differential equation. Steps involved to solve first order linear differential equation: dy(i) Write the given differential equation in the form +Py =Qdx constants or functions of x only. (ii) Find the Integrating Factor (I.F) = e P dx . (iii) Write the solution of the given differential equation as y (I.F) = Q×I.F dx C In case, the first order linear differential equation is in the form where, P1 and Q1 are constants or functions of y only. Then I.F = solution of the differential equation is given by x . (I.F) = (Q× I.F )dy +C∫1dyExample 19 Find the general solution of the differential equation dx Solution Given differential equation is of the form dy +Py =Q, where P = –1and Q = cos xdx 1 dx Therefore I.F = e ex Multiplying both sides of equation by I.F, we get dy−x −xe −ey = e –x cos xdx dy ye−xor ()= e –x cos x dx On integrating both sides with respect to x, we get ye – x = e−x cos x dx +C∫ Let I = ∫e−x cos x dx ⎛−x ⎞e −x = cos x⎜⎟ ( sin x)( −e )−− ⎠∫⎝−1 where P, Q are dx +P x =Q1 1,dy P1 dy and thee−=y cos x . ... (1) dx −x − x = −cos xe −∫sin xe dx −x −x − x = −cos xe −⎡sin x(–e ) − cos x (−e ) dx ⎤⎣∫ ⎦ −x − x −x = −cos xe + sin xe −∫cos xe dx or I =– e –x cos x + sin xe –x – I or 2I =(sin x – cos x) e –x (sin x − cos xe−x)or I = 2 Substituting the value of I in equation (1), we get ⎛ sin x − cos x ⎞−xye – x = ⎜⎟e+ C ⎝ 2 ⎠ ⎛ sin x − cos x ⎞ xor y = ⎜ ⎟+ Ce ⎝ 2 ⎠ which is the general solution of the given differential equation. dy 2Example 20 Find the general solution of the differential equation x + 2y = x (x ≠0) .dx Solution The given differential equation is dyx +2y = x2 ... (1)dx Dividing both sides of equation (1) by x, we get dy 2+y = xdx x dy 2which is a linear differential equation of the type + Py = Q , where P = and Q = x.dx x dx = e2 log xlog x 2 log fx() So I.F = e ∫ 2 x = e 2 = x [as e = f ( )] x Therefore, solution of the given equation is given by y . x2 = xx ) dx + C = xdx +()( 2 3C∫∫ 2x or y = + C x−2 4 which is the general solution of the given differential equation. Example 21 Find the general solution of the differential equation y dx – (x + 2y2) dy = 0. Solution The given differential equation can be written as dx x− =2ydy y dx 1This is a linear differential equation of the type +P1x =Q1 , where P1 =− and dy y 1∫−dy −log log( ) −11Q1 = 2y. Therefore I.F =ey =ey =ey = y Hence, the solution of the given differential equation is ⎛⎞11(2 ) dy Cyx = ∫ ⎜⎟+ yy⎝⎠ x or = ∫(2dy) +C yx or =2y + C yor x =2y2 + Cy which is a general solution of the given differential equation. Example 22 Find the particular solution of the differential equation dy +y cot x =2x + x2 cot x (x ≠0)dxπgiven that y = 0 when x =. 2 dySolution The given equation is a linear differential equation of the type +Py =Q,dx where P = cot x and Q = 2x + x2 cot x. Therefore cot xdx log sin xI.F = ee sin x Hence, the solution of the differential equation is given by y . sin x =∫(2x + x2 cot x) sin x dx + C or y sin x =∫2x sin x dx + ∫x2 cos x dx + C ⎛2x2 ⎞⎛2x2 ⎞ or y sin x = sin x⎜ ⎟−cos x⎜⎟dx +x2 cos x dx +C∫∫⎝2 ⎠⎝2 ⎠ 22 2 or y sin x = x sin x −x cos x dx +x cos x dx +C∫∫ or y sin x = x2 sin x + C ... (1) πSubstituting y = 0 and x = in equation (1), we get2 ⎛⎞π2 π⎛⎞0 = sin ⎜⎟+C⎜⎟⎝⎠22⎝⎠ −π2 or C = 4 Substituting the value of C in equation (1), we get π2 y sin x = x2 sin x − 4 2 π2 or y = x − (sin x ≠0) 4sin x which is the particular solution of the given differential equation. Example 23 Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point. dySolution We know that the slope of the tangent to the curve is dx . dyTherefore, = x + xydxdyor −xy = x ... (1)dx dyThis is a linear differential equation of the type +Py =Q , where P = – x and Q = x.dx −x2 −xdx ∫ 2Therefore, I . F = e =e Hence, the solution of equation is given by −x22−x ye = x 2 dx +C⋅2 ∫() (e ) ... (2) −x2 () dxLet I = xe 2∫ −x2 Let =t , then – x dx = dt or x dx = – dt.2 −x2 Therefore, I = −∫t =et – eedt −= 2 Substituting the value of I in equation (2), we get −x22x ye 2 = e 2 +C 2x −+eor y = 1C 2 ... (3) Now (3) represents the equation of family of curves. But we are interested in finding a particular member of the family passing through (0, 1). Substituting x = 0 and y = 1 in equation (3) we get 1 = – 1 + C . e0 or C = 2 Substituting the value of C in equation (3), we get 2x y =12 2−+e which is the equation of the required curve. For each of the differential equations given in Exercises 1 to 12, find the general solution: dy dy dyy−2x 21. 2y =sin x 2. +3ye 3. +=x+= dx dx dxx dy ⎛ π⎞ 2 dy ⎛π4. +(sec xy) tan x⎜0 ≤<⎟ 5. cos xy tan 0 x ⎟= x += x ⎜≤<⎞dx ⎝ 2 ⎠ dx ⎝ 2 ⎠ dy 2 dy 26. +2yx log x 7. xlog x +=y log x = xdx dxx 8. (1 + x2) dy + 2xy dx = cot x dx (x ≠ 0) dy dy+−+ 10. (xy) =9. x y xxy cot x =0( x ≠0) + 1 dx dx 2 dy11. y dx + (x – y2) dy = 0 12. (x +3y ) =y (y >0) .dx For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition: dy π13. +2tan yx =sin x; y =0 when x = dx 3 2 dy 1 14. (1+x ) +2xy = 2; y =0 when x =1dx 1+x dy π15. −3cot x = xy =2 when xy sin2 ; = dx 2 16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point. 17. Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5. dy 2−=18. The Integrating Factor of the differential equation xy 2x isdx 1 –y(A) e –x (B) e (C) (D) x x 19. The Integrating Factor of the differential equation 2 dx(1−y ) +yx = ay (1 y1) isdy 1 11 1 (A)2 (B)2 (C) 2 (D) 2y 1 y −11−y 1−y Miscellaneous Examples Example 24 Verify that the function y = c1 eax cos bx + c2 eax sin bx, where c1, c2 are arbitrary constants is a solution of the differential equation 2dy dy 22−2a +(a +b )y =0 dx2 dx Solution The given function is y = eax [c1 cosbx + c2 sinbx] ... (1) Differentiating both sides of equation (1) with respect to x, we get dy ax ax= e – bc sin bx bc cos bx c cos bx c sin bxe a12 12dxdyor = eax [(bc 2 + ac 1)cos bx + (ac 2 − bc 1)sin bx ] ... (2)dxDifferentiating both sides of equation (2) with respect to x, we get 2dy ax= e [(bc ac )( bsin bx )(ac 2 bc )( bcos bx )]221 1dx+ [(bc + ac ) cos bx + (ac − bc ) sin bx ] eax .a21 21 ax 222 2= e [( ac − 2abc − bc ) sin bx + (ac + 2abc − bc ) cos bx]212 121 2dy dy Substituting the values of 2 , and y in the given differential equation, we getdxdx ax 2222L.H.S. = e [ac −2abc −bc )sin bx + (ac + 2abc −bc )cos bx ]212 121 2ae ax [( bc ac )cos bx (ac bc )sin bx]21 21 22 ax(ab )e [c1 cos bx c2 sin bx ] 2 22 22 = eax ⎡(ac2 −2abc 1 −bc 2 −2ac2 + 2abc 1 + ac2 +bc 2 )sin bx ⎥⎤ ⎢ ⎢ 2 2 222 ⎥⎣+(ac 1 + 2abc2 −bc 1 −2abc2 −2ac 1 + ac 1 +bc 1 )cos bx ⎦ = eax [0 × sin bx + 0cos bx ]= eax × 0 = 0 = R.H.S. Hence, the given function is a solution of the given differential equation. Example 25 Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes. Solution Let C denote the family of circles in the second quadrant and touching the coordinate axes. Let (–a, a) be the coordinate of the centre of any member of this family (see Fig 9.6). Equation representing the family C is = a2 Y(x + a)2 + (y – a)2... (1) or x2 + y2 + 2ax – 2ay + a2 =0 ... (2) Differentiating equation (2) with respect to x, we get dy dy X’ 2x + 2y + 2a −2a =0dx dx dy ⎛ dyor x +y = a⎜−1⎟⎞ dx ⎝ dx ⎠ Y’ x + yy′ Fig 9.6or a = y′−1 Substituting the value of a in equation (1), we get 2 22⎡ x + yy′⎤ ⎡ x + yy ′⎤ ⎡ x + yy′⎤ x ++ y − =⎢ ⎥⎢ ⎥⎢⎥⎣ y′−1 ⎦⎣ y′−1 ⎦⎣ y′−1 ⎦ or [xy′ – x + x + y y′]2 + [y y′ – y – x – y y′]2 = [x + y y′]2 or (x + y)2 y′2 + [x + y]2 = [x + y y′]2 or (x + y)2 [(y′)2 + 1] = [x + y y′]2 which is the differential equation representing the given family of circles. ⎛ dy ⎞Example 26 Find the particular solution of the differential equation log ⎜ ⎟= 3x + 4 y⎝ dx ⎠ given that y = 0 when x = 0. Solution The given differential equation can be written as dy = e(3x + 4y) dxdyor = e3x . e4y ... (1)dxSeparating the variables, we get dy = e3x dx4 ye 4 y 3xTherefore ∫e− dy = ∫edx −4 y 3xeor = e +C−43 or 4 e3x + 3 e– 4y + 12 C =0 ... (2) Substituting x = 0 and y = 0 in (2), we get −74 + 3 + 12 C = 0 or C = 12 Substituting the value of C in equation (2), we get 4e3x + 3 e– 4y – 7 = 0, which is a particular solution of the given differential equation. Example 27 Solve the differential equation ⎛⎞yy⎛⎞(x dy – y dx) y sin ⎜⎟= (y dx + x dy) x cos ⎜⎟.⎝⎠xx⎝⎠Solution The given differential equation can be written as ⎡ yy ⎤⎡ yy ⎤⎛⎞ 2 ⎛⎞ ⎛⎞ 2 ⎛⎞ ⎢xysin ⎜⎟−x cos ⎜⎟⎥dy =⎢xy cos ⎜⎟+y sin ⎜⎟⎥dx ⎣ ⎝⎠xx ⎣ ⎝⎠ x⎝⎠⎦ x ⎝⎠⎦ y⎛⎞y 2 ⎛⎞ xy cos⎜⎟+y sin ⎜⎟dy ⎝⎠xx⎝⎠ or = yydx⎛⎞ 2 ⎛⎞ xy sin ⎜⎟−x cos ⎜⎟⎝⎠xx⎝⎠ Dividing numerator and denominator on RHS by x2, we get ⎛⎞⎛ ⎞ y yy2 ⎛⎞ycos ⎜⎟+ 2 sin ⎜⎟⎜⎟ dyx xx ⎝⎠x⎝⎠⎝ ⎠ = ... (1)dxyy ⎛⎞y⎛⎞sin ⎜⎟−cos ⎜⎟ xx ⎝⎠x ⎝⎠ dy y⎛⎞Clearly, equation (1) is a homogeneous differential equation of the form =g ⎜⎟.dx x⎝⎠To solve it, we make the substitution y = vx ... (2) dy dv or = vx+dxdx dv v cosv +v2 sin v or vx = (using (1) and (2))+v sin v −cos vdx dv 2cos vv or x = dxv sin v −cos v ⎛vsin v −cos v ⎞ 2 dx or ⎜⎟dv = ⎝ v cos v ⎠x ⎛vsin v −cos v ⎞ 1Therefore ∫⎜ ⎟dv =2∫dx⎝ v cos v ⎠ x 11 or tan vdv − dv =2 dx∫∫∫vx −log | v | = 2log| x | +log|C | or log sec v 1 sec vlog or = log |C1|2vxsecv or 2 =±C1 ... (3)vxyReplacing v by in equation (3), we getxy⎛⎞sec⎜⎟⎝⎠x = C where, C = ± C⎛⎞x2y 1 ⎜⎟()x⎝⎠ y⎛⎞or sec ⎜⎟=C xyx⎝⎠which is the general solution of the given differential equation. Example 28 Solve the differential equation (tan–1y – x) dy =(1 + y2) dx. Solution The given differential equation can be written as dx x tan −1y+ = 2 ... (1)dy 1+y21+y dxNow (1) is a linear differential equation of the form +P1 x = Q1,dy 1 tan −1y .where, P1 = 2 and Q1 = 21+y 1+y Therefore, I.F = ∫1+ 1 y2 dy tan −1 ye =e Thus, the solution of the given differential equation is ⎛ −1tan y ⎞tan −1y−xetan 1 y = ∫⎜ 2 ⎟e dy +C ... (2)⎝1+y ⎠ ⎛ −1tan y ⎞tan −1 yLet I = ∫⎜ 2 ⎟e dy ⎝1+y ⎠ 1Substituting tan–1 y = t so that ⎛ 1 y ⎞dy =dt , we get⎜ 2 ⎟⎝+ ⎠ tI = ∫te dt = t et – ∫1 . et dt = t et – et = et (t – 1) or I = tan −1 y (tan–1y –1)e Substituting the value of I in equation (2), we get −1 y tan −1tan y −1 . =e (tan y −+Cxe 1) −1 tan −1 yor x = (tan y −+1) C e− which is the general solution of the given differential equation. Miscellaneous Exercise on Chapter 9 1. For each of the differential equations given below, indicate its order and degree (if defined). 2 32dydy dy dy 2 ⎛⎞ ⎛⎞−⎛⎞(i) 2 +5x⎜ ⎟−6y =log x (ii) ⎜⎟ 4⎜ ⎟ +7 y =sin x dxdx dx ⎝⎠ dx ⎝⎠ ⎝⎠ 43dy ⎛dy ⎞ (iii) 4 −sin ⎜ 3 ⎟=0 dx ⎝dx ⎠ 2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. 2dy dy 2(i) y = a ex + b e–x: x +2 −+ −= x 2+ x2 xy 0 dx2 dx 2dy dy (ii) y = ex (a cos x + b sin x): 2 −2 +2y =0 dx dx 2dy 9 x =0 dx2(iii) y = x sin 3x : +− y 6cos3 2 2 dy(iv) x2 = 2y2 log y :(x +y ) −xy =0 dx 3. Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2, where a is an arbitrary constant. 4. Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter. 5. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes. dy 1−y2 6. Find the general solution of the differential equation + 2 =0.dx 1−x dy y2 y 17. Show that the general solution of the differential equation + ++=0 isdx 2 x 1x ++given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter. ⎛ π⎞8. Find the equation of the curve passing through the point ⎜0, ⎟whose differential⎝ 4 ⎠equation is sin x cos y dx + cos x sin y dy = 0. 9. Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0. 10. Solve the differential equation ye dx =xe +y dy ( y ≠0) .x ⎛ x ⎞ y ⎜ y 2 ⎟⎝⎠ 11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t) −2 x⎡e y ⎤ dx 12. Solve the differential equation ⎢ −⎥=1(x ≠ 0) . ⎣ xx ⎦ dy dy13. Find a particular solution of the differential equation +y cot x = 4x cosec xdx π .(x ≠ 0), given that y = 0 when x = 2 dy14. Find a particular solution of the differential equation (x + 1) = 2 e –y – 1, givendxthat y = 0 when x = 0. 15. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009? ydx − x dy 16. The general solution of the differential equation =0 is y (A) xy = C (B) x = Cy2 (C) y = Cx (D) y = Cx2 dx17. The general solution of a differential equation of the type + P1x =Q1 isdy ∫P1 dy ∫P1 dy (A) ye =∫(Q1e ) dy + C ∫P1 dx ∫P1 dx(B) . =∫(Q1e dx + Cye ) ∫P1 dy ∫P1 dy(C) xe =∫(Q1e ) dy + C P1 dx P1 dx ∫∫(D) xe =∫(Q1e ) dx + C 18. The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is (A) x ey + x2 = C (B) x ey + y2 = C (C) y ex + x2 = C (D) y ey + x2 = C

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