Question 1

= 2y. Therefore 1/1 log log( ) 1I.F

−− −∫= = = = Hence, the solution of the given differential equation is 1x y = 1(2 ) Cy dy y   +  ∫ or x y = (2 ) Cdy +∫ or x y = 2y + C or x = 2y2 + C y which is a general solution of the given differential equation. Example 17 Find the particular solution of the differential equation cot +dy y xdx = 2x + x2 cot x (x ≠ 0) given that y = 0 when 2x π= . Solution The given equation is a linear differential equation of the type P Qdy ydx + = , where P = cot x and Q = 2x + x2 cot x. Therefore I.F = e e x x dx xcot log sin sin∫ = = 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 or y sin x = 2 2 22 2sin cos cos C2 2 x xx x dx x x dx   − + +       ∫ ∫ or y sin x = 2 2 2sin cos cos Cx x x x dx x x dx− + +∫ ∫ or y sin x = x2 sin x + C ... (1) Substituting y = 0 and 2x π= in equation (1), we get 0 = sin C2 2 π π    +      or C = 2/4 − π Substituting the value of C in equation (1), we get y sin x = 2 sin 4x x π− or y = 2 (sin 0)4 sinx x x π− ≠ which is the particular solution of the given differential equation. Example 18 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. Solution We know that the slope of the tangent to the curve is dy dx . Therefore, dy dx = x + xy or dy xydx − = x ... (1) This is a linear differential equation of the type P Qdy ydx + = , where P = - x and Q = x. Therefore, I. F = 2/2 x x dx e e − −∫ = MATHEMA TICS328 Hence, the solution of equation is given by 2/2 x y e − ⋅ = ( ) 2( ) C x x dxe − +∫ ... (2) Let I = 2( ) x x dxe − ∫ Let 2/2 x t− = , then - x dx = dt or x dx = - dt. Therefore, I = 2- x

− / − = − =∫ Substituting the value of I in equation (2), we get 2/2 x y e − = 2 + C − − x e or y = x e− + ... (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 y = 21 2 x e− + which is the equation of the required curve.