Chapter 9
= 2y. Therefore 1/1 log log( ) 1I.F
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−− −∫= = = = 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
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− / − = − =∫ 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.
(x - y ) dy - ( x + y) dx = 0 4. (x2 - y2) dx + 2 xy dy = 0 5. 2 2 2 2dyx x y xydx = − + 6. x dy - y dx = 2 2x y dx+ 7. cos sin sin cosy y y yx y y dx y x x dyx x x x + = − 8. sin 0dy yx y xdx x − + = 9. log 2 0yy dx x dy x dyx + − = 10. For each of the dif ferential equations in Exercises from 1 1 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. when x = 1 14. cosec 0dy y y
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− + = ; y = 0 when x = 1 15. 2 22 2 0 dyxy y x dx+ − = ; y = 2 when x = 1 16. A homogeneous differential equation of the from dx x hdy y = can be solved by making the substitution. (A) y = vx (B) v = yx (C) x = vy (D) x = v MATHEMA TICS322 17. Which of the following is a homogeneous differential equation? (A) (4x + 6y + 5) dy - (3 y + 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 9.4.3 Linear differential equations A differential equation of the from Pdy ydx + = Q 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 ydx + = sin x 1dy ydx x + = ex log dy y
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+ = 1 x Another form of first order linear differential equation is 1Pdx x dy + = Q1 where, P1 and Q1 are constants or functions of y only. Some examples of this type of differential equation are dx xdy + = cos y 2dx x dy y −+ = y2e - y To solve the first order linear differential equation of the type Pdy ydx + = Q ... (1) Multiply both sides of the equation by a function of x say g (x) to get g(x) dy dx + P. (g(x)) y = Q. g (x) ... (2) Choose g (x) in such a way that R.H.S. becomes a derivative of y . g (x). i.e. g (x) dy dx + P. g(x) y = d dx [y . g (x)] or g (x) dy dx + P. g(x) y = g (x) dy dx + y g′ (x) ⇒ P. g (x) = g′ (x) or P = ( ) ( ) / g x / g x ′ Integrating both sides with respect to x, we get Pdx∫ = ( )
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′ / ∫ or P dx⋅∫ = log (g( x)) or g (x) = P dx e∫ On multiplying the equation (1) by g(x) = P dx e∫ , the L.H.S. becomes the derivative of some function of x and y. This function g(x) = P dx e∫ is called Integrating Factor (I.F.) of the given differential equation. Substituting the value of g (x) in equation (2), we get = or d dx ye dxP∫ = Integrating both sides with respect to x, we get = Q P .e dx dx∫ ∫ or y = e e dx dx dx− ∫ ∫ +∫ P P Q C. . which is the general solution of the differential equation. MATHEMA TICS324 Steps involved to solve first order linear dif ferential equation: (i) Write the given dif ferential equation in the form P Qdy ydx + = where P, Q are constants or functions of x only. (ii) Find the Integrating Factor (I.F) = . (iii) Write the solution of the given differential equation as y (I.F) = In case, the first order linear differential equation is in the form 1 1P Qdx x dy + = , where, P 1 and Q 1 are constants or functions of y only. Then I.F = 1P dy e and the solution of the differential equation is given by x . (I.F) = ( )1Q × I.F Cdy +∫ Example 14 Find the general solution of the differential equation cosdy y xdx − = . Solution Given differential equation is of the form P Qdy ydx + = , where P = -1 and Q = cos x Therefore I. F = Multiplying both sides of equation by I.F, we get = e- x cos x or ( ) xdy yedx − = e- x cos x On integrating both sides with respect to x, we get ye- x = cos Cxe x dx− +∫ ... (1) Let I = cosxe x dx− ∫ = cos ( sin ) ( )1 x xex x e dx − − − − − − ∫ = cos sin x xx e x e dx− −− − ∫ = cos sin (- ) cos ( )x x xx e x e x e dx− − − − − − − ∫ = cos sin cosx x xx e x e x e dx− − −− + − ∫ or I = - e- x cos x + sin x e- x - I or 2I = (sin x - cos x) e- x or I = (sin cos ) xx x e −− Substituting the value of I in equation (1), we get ye- x = sin cos C2 xx x e−− + or y = sin cos C2 xx x e− + which is the general solution of the given differential equation. Example 15 Find the general solution of the differential equation 22 ( 0)dyx y x xdx + = ≠ . Solution The given differential equation is 2dyx ydx + = x 2 ... (1) Dividing both sides of equation (1) by x, we get 2dy ydx x+ = x which is a linear differential equation of the type P Qdy ydx + = , where 2P x= and Q = x. So I.F = 2 dxxe∫ = e2 log x = 2log 2xe x = log ( )[ ( )]f xas e f x = Therefore, solution of the given equation is given by y . x 2 = 2( ) ( ) Cx x dx +∫ = 3 Cx dx +∫ or y = 2C4 x x−+ which is the general solution of the given differential equation. MATHEMA TICS326 Example 16 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 dy y − = 2y This is a linear differential equation of the type 1 1P Qdx xdy + = , where 1 1P y= − and