General Solution of Second Order Linear Ordinary Differential Equation
aa r X i v : . [ m a t h . G M ] J a n GENERAL SOLUTION OF SECOND ORDER LINEARORDINARY DIFFERENTIAL EQUATION
RAJNISH KUMAR JHA
Abstract.
In this paper we present a direct formula for the so-lution of the general second order linear ordinary differential equa-tion as our main result such that the parameters required for theformula are determined using another differential equation, whichhappens to be a Riccati Equation. We also derive other resultsbased on the main result which include special cases for the con-cerned differential equation with variable coefficients, formula forsolution of concerned differential equation with constant coeffi-cients and formula for the solution of the concerned differentialequation with one complementary solution known. Introduction
The general second order linear ordinary differential equation [1] isgiven as(1) y (2) + B ( x ) y (1) + C ( x ) y = R ( x ) , where y ( k ) represents k th derivative of the dependent variable - y withrespect to the independent variable x i.e. y ( k ) = d k ydx k ,B ( x ) , C ( x ) and R ( x ) are functions of the independent variable x .When R ( x ) = 0, Eqn. 1 represents a homogeneous second orderlinear ordinary differential equation and when R ( x ) = 0, it representsa non-homogeneous second order linear ordinary differential equation.When the coefficients B ( x ) and C ( x ) are constants then Eqn. 1 isknown as second order linear ordinary differential equation with con-stant coefficients, otherwise it is known as second order linear ordinarydifferential equation with variable coefficients. (See [1])In this paper, we present a method to solve the general second orderlinear ordinary differential equation wherein we derive an expressionfor the solution having integrating factors involved and also determineanother differential equation in this regard (which is a Riccati Equation[2]) such that when this equation can be solved, the solution of thecorresponding second order differential equation can also be obtained.When the coefficients B ( x ) and C ( x ) are not constants then the dif-ferential equation given by Eqn. (1) can be difficult to solve. However,under certain cases where the coefficients satisfy some conditions, we an determine the exact solution of such differential equation. Twosuch cases are described in this paper. (See Corollary-1 and Corollary-2 )We also derive a formula for the complete solution of the differen-tial equation given by Eqn. (1) provided we know one complementarysolution of the given equation (See Corollary-4). Moreover, when thecoefficients of the differential equation given by Eqn. (1) are constantthen the complete solution can be determined exactly as it is discussedin this paper. (See Corollary-3 )The main result is stated as Theorem 1 while the other results arestated in form of its corollaries.2. Main Result
In this section we give the main result in form of a Theorem ,which gives the solution expression for the differential equation givenby Eqn. (1) in terms of R ( x ) and Integrating Factors. We presentTheorem 1 along with the proof here as follows Theorem 1.
The solution of second order linear ordinary differentialequation, as described by Eqn. (1) , is given as (2) y = 1 h ( x ) Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x ) where g ( x ) = e R P ( x ) dx , (3) h ( x ) = e R Q ( x ) dx , (4) P ( x ) + Q ( x ) = B ( x ) , (5) Q ′ ( x ) = Q ( x ) − B ( x ) Q ( x ) + C ( x ) , (6) Q ′ ( x ) represents first order derivative of Q ( x ) with respect to x and g ( x ) and h ( x ) are the first and second Integrating Factors respectively.Proof. Consider the differential equation given by Eqn. (1), y (2) + B ( x ) y (1) + C ( x ) y = R ( x ) . Let(7) B ( x ) = P ( x ) + Q ( x ) , where P ( x ) and Q ( x ) are the functions of the independent variable x .Then Eqn. (1) becomes(8) y (2) + P ( x ) y (1) + Q ( x ) y (1) + C ( x ) y = R ( x ) . We multiply Eqn. (8) by a function g ( x ), which we call here as the firstIntegrating Factor, to get(9) ( g ( x ) y (2) + g ( x ) P ( x ) y (1) )+( g ( x ) Q ( x ) y (1) + g ( x ) C ( x ) y ) = g ( x ) R ( x ) . e impose the following conditions on g ( x ), g ′ ( x ) = g ( x ) P ( x ) , (10) ( g ( x ) Q ( x )) ′ = g ( x ) C ( x ) , (11)such that Eqn. (9) becomes(12) ddx ( g ( x ) y (1) ) + ddx ( g ( x ) Q ( x ) y ) = g ( x ) R ( x ) . On integrating Eqn. (12) we get g ( x ) y (1) + g ( x ) y = Z g ( x ) R ( x ) dx + C (13) = ⇒ y (1) + Q ( x ) y = 1 g ( x ) Z g ( x ) R ( x ) dx + C g ( x )(14)Now, we multiply Eqn. (14) with another function h ( x ), which we callhere as the second Integrating Factor, to get(15) h ( x ) y (1) + h ( x ) Q ( x ) y = h ( x ) g ( x ) Z g ( x ) R ( x ) dx + C h ( x ) g ( x ) . We impose the following condition on h ( x ),(16) h ′ ( x ) = h ( x ) Q ( x ) , such that Eqn. (15) transforms into(17) ddx ( h ( x ) y ) = h ( x ) g ( x ) Z g ( x ) R ( x ) dx + C h ( x ) g ( x ) . On integrating Eqn. (17), we get h ( x ) y = Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C Z h ( x ) g ( x ) dx + C (18) = ⇒ y = 1 h ( x ) Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x ) , (19)such that Eqn. (19) gives the solution for the differential equation givenby Eqn. (1) where g ( x ) and h ( x ) are determined using conditions givenin Eqn. (10), Eqn. (11) and Eqn. (16).The conditions described in Eqn. (10), Eqn. (11) and Eqn. (16) canalso be expressed as g ′ ( x ) = g ( x ) P ( x ) = ⇒ g ( x ) = e R P ( x ) dx (20) h ′ ( x ) = h ( x ) Q ( x ) = ⇒ h ( x ) = e R Q ( x ) dx (21) ( g ( x ) Q ( x )) ′ = g ( x ) C ( x )= ⇒ g ′ ( x ) Q ( x ) + g ( x ) Q ′ ( x ) = g ( x ) C ( x ) . (22) n using Eqn. (10) in Eqn. (22), we get g ( x ) P ( x ) Q ( x ) + g ( x ) Q ′ ( x ) = g ( x ) C ( x )(23) = ⇒ Q ′ ( x ) + P ( x ) Q ( x ) = C ( x ) . (24)Also, by using Eqn. (7) in Eqn. (24), we can write an equivalent equa-tion as Q ′ ( x ) + B ( x ) Q ( x ) − Q ( x ) = C ( x )(25) = ⇒ Q ′ ( x ) = Q ( x ) − B ( x ) Q ( x ) + C ( x )(26)which is a Riccati Equation.Thus, the equations (7), (20), (21) and (26) can be used to determinethe Integrating Factors g ( x ) and h ( x ) such that they also determinethe solution expression given by Eqn. (19) for the differential equationgiven by Eqn. (1).Therefore, the solution of the general second order linear ordinarydifferential equation (1) is given by (19), i.e., y = 1 h ( x ) Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x ) , where g ( x ) = e R P ( x ) dx ,h ( x ) = e R Q ( x ) dx ,P ( x ) + Q ( x ) = B ( x ) ,Q ′ ( x ) = Q ( x ) − B ( x ) Q ( x ) + C ( x ) and g ( x ) and h ( x ) are the first and second Integrating Factors respectively. (cid:3) It is worthwhile to discuss about the complementary solution partand the particular integral part of the solution of the differential equa-tion (1) given by Eqn. (19).The complementary solution part of a general linear ordinary differ-ential equation is given by the solution of the corresponding homoge-neous equation. So, in the context of the differential equation (1), weget the corresponding homogeneous equation by putting R ( x ) = 0 , i.e.(27) y (2) + B ( x ) y (1) + C ( x ) y = 0such that by Theorem 1, its solution is given as y = 1 h ( x ) Z h ( x ) g ( x ) Z dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x )(28) = ⇒ y = ( C + C ) h ( x ) Z h ( x ) g ( x ) dx + C h ( x )= Ch ( x ) Z h ( x ) g ( x ) dx + C h ( x ) ( C = C + C ) . (29) rom the above observations, we conclude that the complementary so-lution part of the solution of the differential equation (1) given byEqn. (19) is the part containing arbitary constants while the termhaving no arbitary constant gives the particular solution (or the non-homogeneous solution) part of the solution.3. Other Results Based on the Main Result
In this section we state and prove some other important results re-lated to second order linear ordinary differential equations and basedon Theorem 1 in form of corollaries.The first two results deal with special cases of the general secondorder linear ordinary differential equation having variable coefficientswhere the coefficients satisfy some condition. The two results are de-scribed here as follows
Corollary 1.
If the coefficients of the second order linear ordinarydifferential equation (1) satisfy the condition (30) B ( x ) + 2 B ′ ( x ) − C ( x ) = 0 , then its solution is given as y = ( x + c ) e − R B ( x )2 dx Z x + c ) Z ( x + c ) e R B ( x )2 dx R ( x ) dxdx ++ C e − R B ( x )2 + C xe − R B ( x )2 dx , (31) where c is any constant.Proof. Consider the general second order linear ordinary differentialequation given by Eqn. (1),(32) y (2) + B ( x ) y (1) + C ( x ) y = R ( x ) . By Theorem 1, this equation has solution given as(33) y = 1 h ( x ) Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x ) , where g ( x ) = e R P ( x ) dx , (34) h ( x ) = e R Q ( x ) dx , (35) P ( x ) + Q ( x ) = B ( x ) , (36) Q ′ ( x ) = Q ( x ) − B ( x ) Q ( x ) + C ( x ) and(37) g ( x ) and h ( x ) are the first and second Integrating Factors respectively. et P ( x ) = B ( x )2 − f ( x )(38) = ⇒ Q ( x ) = B ( x )2 + f ( x ) . (39)Then Eqn. (37) becomes B ′ ( x )2 + f ′ ( x ) = B ( x ) f ( x ) + B ( x ) f ( x ) − B ( x ) − B ( x ) f ( x ) + C ( x )(40) = ⇒ f ′ ( x ) = f ( x ) − B ( x ) + 2 B ′ ( x ) − C ( x )4 . (41)If the coefficients B ( x ) and C ( x ) satisfy the equation(42) B ( x ) + 2 B ′ ( x ) − C ( x ) = 0 , then Eqn. (41) becomes(43) f ′ ( x ) = f ( x ) , which has solution expressed as(44) f ( x ) = − x + c ) ( c is an arbitary constant)On using f ( x ) in Eqn. (38) and Eqn. (39), we get P ( x ) = B ( x )2 + 1( x + c )(45) Q ( x ) = B ( x )2 − x + c )(46)Then on using values of P ( x ) and Q ( x ) in Eqn. (34) and Eqn. (35)respectively, we get the Integrating factors as g ( x ) = ( x + c ) e R B ( x )2 dx (47) h ( x ) = 1( x + c ) e R B ( x )2 dx , (48)such that the solution expression becomes y = ( x + c ) e − R B ( x )2 dx Z x + c ) Z ( x + c ) e R B ( x )2 dx R ( x ) dxdx ++ C ( x + c ) e − R B ( x )2 Z x + c ) dx + C ( x + c ) e − R B ( x )2 dx , (49) = ⇒ y = ( x + c ) e − R B ( x )2 dx Z x + c ) Z ( x + c ) e R B ( x )2 dx R ( x ) dxdx ++ C e − R B ( x )2 + C xe − R B ( x )2 dx ( C = C c − C ) , (50) here Eqn. (50) gives the required solution expression for the differen-tial equation (32) when the condition given by Eqn. (42) is true. (cid:3) Corollary 2.
If the coefficients of the second order linear ordinarydifferential equation (1) satisfy the condition (51) B ( x ) + 2 B ′ ( x ) − C ( x ) = k ( k is a constant, k = 0 ) , then its solution is given as y = e − R B ( x )2 dx ( e √ k ( x + c ) − e −√ k ( x + c ) ) × Z e √ k ( x + c ) − e −√ k ( x + c ) ) Z e R B ( x )2 dx ( e √ k ( x + c ) − e −√ k ( x + c ) ) R ( x ) dxdx + C e − ( R B ( x )2 dx + √ k ( x + c )) + C e ( − R B ( x )2 dx + √ k ( x + c )) (52) where c is any constant.Proof. Consider the general second order linear ordinary differentialequation given by Eqn. (1),(53) y (2) + B ( x ) y (1) + C ( x ) y = R ( x ) . By Theorem 1, this equation has solution given as(54) y = 1 h ( x ) Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x ) , where g ( x ) = e R P ( x ) dx , (55) h ( x ) = e R Q ( x ) dx , (56) P ( x ) + Q ( x ) = B ( x ) , (57) Q ′ ( x ) = Q ( x ) − B ( x ) Q ( x ) + C ( x ) and(58) g ( x ) and h ( x ) are the first and second Integrating Factors respectively.Let P ( x ) = B ( x )2 − f ( x )(59) = ⇒ Q ( x ) = B ( x )2 + f ( x ) . (60)Then Eqn. (58) becomes B ′ ( x )2 + f ′ ( x ) = B ( x ) f ( x ) + B ( x ) f ( x ) − B ( x ) − B ( x ) f ( x ) + C ( x )(61) = ⇒ f ′ ( x ) = f ( x ) − B ( x ) + 2 B ′ ( x ) − C ( x )4 . (62) f the coefficients B ( x ) and C ( x ) satisfy the equation(63) B ( x ) + 2 B ′ ( x ) − C ( x )4 = k ( k is a constant, k = 0) , then Eqn. (62) becomes(64) f ′ ( x ) = f ( x ) − k. The solution of the above equation is given as(65) f ( x ) = −√ k e √ k ( x + c ) + e −√ k ( x + c ) e √ k ( x + c ) − e −√ k ( x + c ) ! , where c is an arbitary constant.On using f ( x ) in Eqn. (59) and Eqn. (60), we obtain P ( x ) = B ( x )2 + √ k e √ k ( x + c ) + e −√ k ( x + c ) e √ k ( x + c ) − e −√ k ( x + c ) ! , (66) Q ( x ) = B ( x )2 − √ k e √ k ( x + c ) + e −√ k ( x + c ) e √ k ( x + c ) − e −√ k ( x + c ) ! . (67)Then on using P ( x ) and Q ( x ) in equations (55) and (56) respectively,we obtain the Integrating Factors g ( x ) and h ( x ) as g ( x ) = e R P ( x ) dx = e R B ( x )2 dx e R √ k (cid:18) e √ k ( x + c )+ e −√ k ( x + c ) e √ k ( x + c ) − e −√ k ( x + c ) (cid:19) dx = e R B ( x )2 dx (cid:16) e √ k ( x + c ) − e −√ k ( x + c ) (cid:17) , (68) h ( x ) = e R Q ( x ) dx = e R B ( x )2 dx e − R √ k (cid:18) e √ k ( x + c )+ e −√ k ( x + c ) e √ k ( x + c ) − e −√ k ( x + c ) (cid:19) dx = e R B ( x )2 dx (cid:16) e √ k ( x + c ) − e −√ k ( x + c ) (cid:17) . (69) n using g ( x ) and h ( x ) in Eqn. (54), the solution expression becomes y = e − R B ( x )2 dx ( e √ k ( x + c ) − e −√ k ( x + c ) ) × Z e √ k ( x + c ) − e −√ k ( x + c ) ) Z e R B ( x )2 dx ( e √ k ( x + c ) − e −√ k ( x + c ) ) R ( x ) dxdx + C e − R B ( x )2 dx (cid:16) e √ k ( x + c ) − e −√ k ( x + c ) (cid:17)Z (cid:16) e √ k ( x + c ) − e −√ k ( x + c ) (cid:17) dx + C e − R B ( x )2 dx (cid:16) e √ k ( x + c ) − e −√ k ( x + c ) (cid:17) (70) = ⇒ y = e − R B ( x )2 dx ( e √ k ( x + c ) − e −√ k ( x + c ) ) × Z e √ k ( x + c ) − e −√ k ( x + c ) ) Z e R B ( x )2 dx ( e √ k ( x + c ) − e −√ k ( x + c ) ) R ( x ) dxdx + C e − ( R B ( x )2 dx + √ k ( x + c )) + C e ( − R B ( x )2 dx + √ k ( x + c )) ( C = − C √ k − C )(71) such that Eqn. (71) gives the required solution expression for thedifferential equation (53) when the condition given by Eqn. (63) istrue. (cid:3) The third result deals with second order linear ordinary differentialequation having constant coefficients. In this case exact solution iscompletely determined. The result is stated in form of a corollary ofTheorem 1 presented here along with the proof as follows
Corollary 3.
If the coefficients of the second order linear ordinarydifferential equation (1) are constant such that B ( x ) = B and C ( x ) = C then its solution is given by (72) y = 1 h ( x ) Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x ) where g ( x ) = e (cid:18) B ± √ B − C (cid:19) x , (73) h ( x ) = e (cid:18) B ∓ √ B − C (cid:19) x . (74) Proof.
Consider the general second order linear ordinary differentialequation given by Eqn. (1) such that B ( x ) and C ( x ) are constant i.e. ( x ) = B and C ( x ) = C ,(75) y (2) + By (1) + Cy = R ( x ) . By Theorem 1, this equation has solution given as(76) y = 1 h ( x ) Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x ) , where g ( x ) = e R P ( x ) dx , (77) h ( x ) = e R Q ( x ) dx , (78) P ( x ) + Q ( x ) = B ( x ) , (79) Q ′ ( x ) = Q ( x ) − B ( x ) Q ( x ) + C ( x ) and(80) g ( x ) and h ( x ) are the first and second Integrating Factors respectively.Since the coefficients B ( x ) and C ( x ) are constants ( B ( x ) = B, C ( x ) = C ), we can consider P ( x ) and Q ( x ) to be constants too such that P ( x ) = P , Q ( x ) = Q and(81) P ′ ( x ) = Q ′ ( x ) = 0 . Then Eqn. (79) and Eqn. (80) respectively transform into P + Q = B, (82) P Q = C. (83)On solving Eqn. (82) and Eqn. (83), we obtain P = B ± √ B − C , (84) Q = B ∓ √ B − C . (85)Then we can express the Integrating Factors g ( x ) and h ( x ) as g ( x ) = e R P ( x ) dx = e R P dx = e (cid:18) B ± √ B − C (cid:19) x , (86) h ( x ) = e R Q ( x ) dx = e R Q dx = e (cid:18) B ∓ √ B − C (cid:19) x , (87)such that on using the Integrating factors given by the above equationsin the solution expression given by Eqn. (76), the complete solution ofthe required differential equation (75) is determined. (cid:3) The fourth result deals with the general second order linear ordinarydifferential equation and gives the formula for the complete solutionwhen one of the complementary solutions is known. The result is givenin form of a corollary of Theorem 1 presented along with the proof hereas follows. orollary 4. If one of the complementary solution of the general sec-ond order linear ordinary differential equation (1) is given as f ( x ) ,then the solution of the differential equation (1) is given as y = f ( x ) Z e − R B ( x ) dx f ( x ) Z e R B ( x ) dx f ( x ) R ( x ) dxdx + C f ( x ) Z e − R B ( x ) dx f ( x ) dx + C f ( x ) . (88) Proof.
Consider the general second order linear ordinary differentialequation given by Eqn. (1),(89) y (2) + B ( x ) y (1) + C ( x ) y = R ( x ) . By Theorem 1, this equation has solution given as(90) y = 1 h ( x ) Z h ( x ) g ( x ) Z g ( x ) R ( x ) dxdx + C h ( x ) Z h ( x ) g ( x ) dx + C h ( x ) , where g ( x ) = e R P ( x ) dx , (91) h ( x ) = e R Q ( x ) dx , (92) P ( x ) + Q ( x ) = B ( x ) , (93) Q ′ ( x ) = Q ( x ) − B ( x ) Q ( x ) + C ( x ) and(94) g ( x ) and h ( x ) are the first and second Integrating Factors respectively.If we have a function f ( x ) such that it satisfies the homogeneous equa-tion,(95) y (2) + B ( x ) y (1) + C ( x ) y = 0 , corresponding to the differential equation (89) then we can put f ( x )equal to one of the terms of complementary solution in Eqn. (90), i.e.(96) f ( x ) ≡ C h ( x ) . For the ease of solving we have taken the term of complementary so-lution corresponding to arbitary constant C to be equivalent to f ( x )such that we can put C = 1 to obtain the function h ( x ) in terms of f ( x ).On putting C = 1, we get(97) f ( x ) = 1 h ( x ) = ⇒ h ( x ) = 1 f ( x ) . ow consider the following equations g ( x ) h ( x ) = e R ( P ( x )+ Q ( x )) dx = e R B ( x ) dx (98) = ⇒ g ( x ) = e R B ( x ) dx h ( x ) . (99)Using Eqn. (97), we get(100) g ( x ) = f ( x ) e R B ( x ) dx . Therefore, on using g ( x ) and h ( x ) as given in Eqn. (97) and Eqn. (100)in the solution expression in Eqn. (90), we get the solution of thesecond-order linear ordinary differential equation (89) as y = f ( x ) Z e − R B ( x ) dx f ( x ) Z e R B ( x ) dx f ( x ) R ( x ) dxdx + C f ( x ) Z e − R B ( x ) dx f ( x ) dx + C f ( x ) , (101)when one of the complementary solutions of the differential equation(89) is given as f ( x ). (cid:3) References [1] S.R.K. Iyengar R.K. Jain.
Advanced Engineering Mathematics , chapter 5, pages300–318. Alpha Science International Ltd., 2002.[2] S.R.K. Iyengar R.K. Jain.
Advanced Engineering Mathematics , chapter 4, page277. Alpha Science International Ltd., 2002.
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