Exercise 9.2

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. 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. 9.4.1 Differential equations with variables separable A first order-first degree differential equation is of the form dy dx = F(x, y) ... (1) If 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 dx = h (y) . g(x) ... (2) If h(y) ≠ 0, separating the variables, (2) can be rewritten as ( )h y dy = g (x) dx ... (3) Integrating both sides of (3), we get ( ) dyh y∫ = ( )g x dx∫ ... (4) Thus, (4) provides the solutions of given differential equation in the form H (y) = G (x) + C Here, H (y) and G (x) are the anti derivatives of 1 ( )h y and g (x) respectively and C is the arbitrary constant. Example 4 Find the general solution of the differential equation 1 dy x dx y += − , (y ≠ 2) Solution We have dy dx = 1 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 ) y dy−∫ = ( 1)x dx+∫ or 2 2 yy − = 1C2 x x+ + or x2 + y2 + 2x - 4 y + 2 C 1 = 0 or x2 + y2 + 2x - 4 y + C = 0, where C = 2C 1 which is the general solution of equation (1). MATHEMA TICS308 Example 5 Find the general solution of the differential equation 2/2 1/1 dy y dx x += + . Solution Since 1 + y2 ≠ 0, therefore separating the variables, the given differential equation can be written as dy y+ = 21 dx x+ ... (1) Integrating both sides of equation (1), we get dy y+∫ = 21 dx x+∫ or tan-1 y = tan-1 x + C which is the general solution of equation (1). Example 6 Find the particular solution of the differential equation 24dy xydx = − given that y = 1, when x = 0. Solution If y ≠ 0, the given differential equation can be written as dy y = - 4 x dx ... (1) Integrating both sides of equation (1), we get dy y∫ = 4 x dx− ∫ or 1 y − = - 2 x2 + C or y = 2 2 Cx − ... (2) 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 given differential equation as 2 2 1 y x = + . Example 7 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 dy* = or dy = 12x dxx   +   ... (1) Integrating both sides of equation (1), we get dy∫ = 12x dxx   +  ∫ 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 8 Find the equation of a curve passing through the point (-2, 3), given that the slope of the tangent to the curve at any point (x, y) is 2 2x y . Solution We know that the slope of the tangent to a curve is given by dy dx . so, dy dx = 2 2x y ... (1) Separating the variables, equation (1) can be written as y2 dy = 2x dx ... (2) Integrating both sides of equation (2), we get 2y dy∫ = 2x dx∫ or 3/3 y = x2 + C ... (3) dy dx due to Leibnitz is extremely flexible and useful in many calculation and formal transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers. By treating 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 - V erlog New York. MATHEMA TICS310 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 2 53 y x= + or 2 3(3 15)y x= + Example 9 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 dt = 5 P100   ×   or dp dt = P 20 ... (1) separating the variables in equation (1), we get P dp = 20 dt ... (2) Integrating both sides of equation (2), we get log P = 1C20 t + or P = 1C20 t e e ⋅ or P = 20C t e (where 1C Ce = ) ... (3) Now P = 1000, when t = 0 Substituting the values of P and t in (3), we get C = 1000. Therefore, equation (3), gives P = 1000 20 t e Let t years be the time required to double the principal. Then 2000 = 1000 20 t e ⇒ t = 20 loge2

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