Page 1 UNIT Differential Equations-I 5.1 INTRODUCTION We have studied methods of solving ordinary differential equations of first order and first degree, in chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders. Differential equations of second order arise very often in physical problems, especially in connection with mechanical vibrations and electric circuits. 5.2 LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS A differential equation of the form dy dx a dy dx a dy dx n n n n n n ++ 1 1 1 2 2 2 – – – – + ... + a n y = X ...(1) where X is a function of x and a 1 , a 2 ..., a n are constants is called a linear differential equation of n th order with constant coefficients. Since the highest order of the derivative appearing in (1) is n, it is called a differential equation of n th order and it is called linear. Using the familiar notation of differential operators: D = d dx , D 2 = d dx 2 2 , D d dx D d dx n n n 3 3 3 == ..., Then (1) can be written in the form {D n + a 1 D n–1 + a 2 D n–2 + ... a n } y = X i.e., f (D) y = X ...(2) where f (D)= D n + a 1 D n–1 + a 2 D n–2 + ... a n . Here f (D) is a polynomial of degree n in D If x = 0, the equation f (D) y =0 is called a homogeneous equation. If x ? 0 then the Eqn. (2) is called a non-homogeneous equation. 214 Page 2 UNIT Differential Equations-I 5.1 INTRODUCTION We have studied methods of solving ordinary differential equations of first order and first degree, in chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders. Differential equations of second order arise very often in physical problems, especially in connection with mechanical vibrations and electric circuits. 5.2 LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS A differential equation of the form dy dx a dy dx a dy dx n n n n n n ++ 1 1 1 2 2 2 – – – – + ... + a n y = X ...(1) where X is a function of x and a 1 , a 2 ..., a n are constants is called a linear differential equation of n th order with constant coefficients. Since the highest order of the derivative appearing in (1) is n, it is called a differential equation of n th order and it is called linear. Using the familiar notation of differential operators: D = d dx , D 2 = d dx 2 2 , D d dx D d dx n n n 3 3 3 == ..., Then (1) can be written in the form {D n + a 1 D n–1 + a 2 D n–2 + ... a n } y = X i.e., f (D) y = X ...(2) where f (D)= D n + a 1 D n–1 + a 2 D n–2 + ... a n . Here f (D) is a polynomial of degree n in D If x = 0, the equation f (D) y =0 is called a homogeneous equation. If x ? 0 then the Eqn. (2) is called a non-homogeneous equation. 214 DIFFERENTIAL EQUATIONS–I 215 5.3 SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR DIFFERENTIAL EQUATION We consider the homogeneous equation dy dx p dy dx qy 2 2 ++ = 0 ...(1) where p and q are constants (D 2 + pD + q) y = 0 ...(2) The Auxiliary equations (A.E. ) put D = m m 2 + pm + q = 0 ...(3) Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) being quadratic in m, will have two roots in general. There are three cases. Case (i): Roots are real and distinct The roots are real and distinct, say m 1 and m 2 i.e., m 1 ? m 2 Hence, the general solution of eqn. (1) is y = C 1 e m 1 x + C 2 e m 2 x where C 1 and C 2 are arbitrary constant. Case (ii): Roots are equal The roots are equal i.e., m 1 = m 2 = m. Hence, the general solution of eqn. (1) is y =(C 1 + C 2 x) e mx where C 1 and C 2 are arbitrary constant. Case (iii): Roots are complex The Roots are complex, say a ± iß Hence, the general solution is y = e ax (C 1 cos ß x + C 2 sin ß x) where C 1 and C 2 are arbitrary constants. Note. Complementary Function (C.F.) which itself is the general solution of the D.E. 1. Solve dy dx –5 dy dx 6y 0 2 2 += . Solution. Given equation is (D 2 – 5D + 6) y = 0 A.E. is m 2 – 5m + 6 = 0 i.e., (m–2) (m– 3)= 0 i.e., m = 2, 3 ? m 1 = 2, m 2 = 3 ? The roots are real and distinct. Page 3 UNIT Differential Equations-I 5.1 INTRODUCTION We have studied methods of solving ordinary differential equations of first order and first degree, in chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders. Differential equations of second order arise very often in physical problems, especially in connection with mechanical vibrations and electric circuits. 5.2 LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS A differential equation of the form dy dx a dy dx a dy dx n n n n n n ++ 1 1 1 2 2 2 – – – – + ... + a n y = X ...(1) where X is a function of x and a 1 , a 2 ..., a n are constants is called a linear differential equation of n th order with constant coefficients. Since the highest order of the derivative appearing in (1) is n, it is called a differential equation of n th order and it is called linear. Using the familiar notation of differential operators: D = d dx , D 2 = d dx 2 2 , D d dx D d dx n n n 3 3 3 == ..., Then (1) can be written in the form {D n + a 1 D n–1 + a 2 D n–2 + ... a n } y = X i.e., f (D) y = X ...(2) where f (D)= D n + a 1 D n–1 + a 2 D n–2 + ... a n . Here f (D) is a polynomial of degree n in D If x = 0, the equation f (D) y =0 is called a homogeneous equation. If x ? 0 then the Eqn. (2) is called a non-homogeneous equation. 214 DIFFERENTIAL EQUATIONS–I 215 5.3 SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR DIFFERENTIAL EQUATION We consider the homogeneous equation dy dx p dy dx qy 2 2 ++ = 0 ...(1) where p and q are constants (D 2 + pD + q) y = 0 ...(2) The Auxiliary equations (A.E. ) put D = m m 2 + pm + q = 0 ...(3) Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) being quadratic in m, will have two roots in general. There are three cases. Case (i): Roots are real and distinct The roots are real and distinct, say m 1 and m 2 i.e., m 1 ? m 2 Hence, the general solution of eqn. (1) is y = C 1 e m 1 x + C 2 e m 2 x where C 1 and C 2 are arbitrary constant. Case (ii): Roots are equal The roots are equal i.e., m 1 = m 2 = m. Hence, the general solution of eqn. (1) is y =(C 1 + C 2 x) e mx where C 1 and C 2 are arbitrary constant. Case (iii): Roots are complex The Roots are complex, say a ± iß Hence, the general solution is y = e ax (C 1 cos ß x + C 2 sin ß x) where C 1 and C 2 are arbitrary constants. Note. Complementary Function (C.F.) which itself is the general solution of the D.E. 1. Solve dy dx –5 dy dx 6y 0 2 2 += . Solution. Given equation is (D 2 – 5D + 6) y = 0 A.E. is m 2 – 5m + 6 = 0 i.e., (m–2) (m– 3)= 0 i.e., m = 2, 3 ? m 1 = 2, m 2 = 3 ? The roots are real and distinct. 216 ENGINEERING MATHEMATICS—II ? The general solution of the equation is y = C 1 e 2x + C 2 e 3x . 2. Solve dy dx – dy dx –4 dy dx 4y 0 3 3 2 2 += . Solution. Given equation is (D 3 – D 2 – 4D + 4) y = 0 A.E. is m 3 – m 2 – 4m + 4 = 0 m 2 (m – 1) – 4 (m – 1) = 0 (m – 1) (m 2 – 4) = 0 m = 1, m = ± 2 m 1 = 1, m 2 = 2, m 3 = – 2 ? The general solution of the given equation is y = C 1 e x + C 2 e 2x + C 3 e –2x . 3. Solve dy dx – dy dx –6y 0 2 2 = . Solution. The D.E. can be written as (D 2 – D – 6) y =0 A.E. is m 2 – m – 6 = 0 ? (m – 3) (m + 2) = 0 ? m = 3, – 2 ? The general solution is y = C 1 e 3x + C 2 e –2x . 4. Solve dy dx 8 dy dx 16y 0 2 2 ++ = . Solution. The D.E. can be written as (D 2 + 8D + 16) y =0 A.E. is m 2 + 8m + 16 = 0 ? (m + 4) 2 =0 (m + 4) (m + 4) = 0 m = – 4, – 4 ? The general solution is y =(C 1 + C 2 x) e –4x . 5. Solve dy dx wy 0 2 2 2 += . Solution. Equation can be written as (D 2 + w 2 ) y =0 A.E. is m 2 + w 2 =0 Page 4 UNIT Differential Equations-I 5.1 INTRODUCTION We have studied methods of solving ordinary differential equations of first order and first degree, in chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders. Differential equations of second order arise very often in physical problems, especially in connection with mechanical vibrations and electric circuits. 5.2 LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS A differential equation of the form dy dx a dy dx a dy dx n n n n n n ++ 1 1 1 2 2 2 – – – – + ... + a n y = X ...(1) where X is a function of x and a 1 , a 2 ..., a n are constants is called a linear differential equation of n th order with constant coefficients. Since the highest order of the derivative appearing in (1) is n, it is called a differential equation of n th order and it is called linear. Using the familiar notation of differential operators: D = d dx , D 2 = d dx 2 2 , D d dx D d dx n n n 3 3 3 == ..., Then (1) can be written in the form {D n + a 1 D n–1 + a 2 D n–2 + ... a n } y = X i.e., f (D) y = X ...(2) where f (D)= D n + a 1 D n–1 + a 2 D n–2 + ... a n . Here f (D) is a polynomial of degree n in D If x = 0, the equation f (D) y =0 is called a homogeneous equation. If x ? 0 then the Eqn. (2) is called a non-homogeneous equation. 214 DIFFERENTIAL EQUATIONS–I 215 5.3 SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR DIFFERENTIAL EQUATION We consider the homogeneous equation dy dx p dy dx qy 2 2 ++ = 0 ...(1) where p and q are constants (D 2 + pD + q) y = 0 ...(2) The Auxiliary equations (A.E. ) put D = m m 2 + pm + q = 0 ...(3) Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) being quadratic in m, will have two roots in general. There are three cases. Case (i): Roots are real and distinct The roots are real and distinct, say m 1 and m 2 i.e., m 1 ? m 2 Hence, the general solution of eqn. (1) is y = C 1 e m 1 x + C 2 e m 2 x where C 1 and C 2 are arbitrary constant. Case (ii): Roots are equal The roots are equal i.e., m 1 = m 2 = m. Hence, the general solution of eqn. (1) is y =(C 1 + C 2 x) e mx where C 1 and C 2 are arbitrary constant. Case (iii): Roots are complex The Roots are complex, say a ± iß Hence, the general solution is y = e ax (C 1 cos ß x + C 2 sin ß x) where C 1 and C 2 are arbitrary constants. Note. Complementary Function (C.F.) which itself is the general solution of the D.E. 1. Solve dy dx –5 dy dx 6y 0 2 2 += . Solution. Given equation is (D 2 – 5D + 6) y = 0 A.E. is m 2 – 5m + 6 = 0 i.e., (m–2) (m– 3)= 0 i.e., m = 2, 3 ? m 1 = 2, m 2 = 3 ? The roots are real and distinct. 216 ENGINEERING MATHEMATICS—II ? The general solution of the equation is y = C 1 e 2x + C 2 e 3x . 2. Solve dy dx – dy dx –4 dy dx 4y 0 3 3 2 2 += . Solution. Given equation is (D 3 – D 2 – 4D + 4) y = 0 A.E. is m 3 – m 2 – 4m + 4 = 0 m 2 (m – 1) – 4 (m – 1) = 0 (m – 1) (m 2 – 4) = 0 m = 1, m = ± 2 m 1 = 1, m 2 = 2, m 3 = – 2 ? The general solution of the given equation is y = C 1 e x + C 2 e 2x + C 3 e –2x . 3. Solve dy dx – dy dx –6y 0 2 2 = . Solution. The D.E. can be written as (D 2 – D – 6) y =0 A.E. is m 2 – m – 6 = 0 ? (m – 3) (m + 2) = 0 ? m = 3, – 2 ? The general solution is y = C 1 e 3x + C 2 e –2x . 4. Solve dy dx 8 dy dx 16y 0 2 2 ++ = . Solution. The D.E. can be written as (D 2 + 8D + 16) y =0 A.E. is m 2 + 8m + 16 = 0 ? (m + 4) 2 =0 (m + 4) (m + 4) = 0 m = – 4, – 4 ? The general solution is y =(C 1 + C 2 x) e –4x . 5. Solve dy dx wy 0 2 2 2 += . Solution. Equation can be written as (D 2 + w 2 ) y =0 A.E. is m 2 + w 2 =0 DIFFERENTIAL EQUATIONS–I 217 m 2 =– w 2 = w 2 i 2 (i 2 = – 1) m = ± wi This is the form a ± iß where a = 0, ß = w. ? The general solution is y = e 0t (C 1 cos wt + C 2 sin wt) ? y = C 1 cos wt + C 2 sin wt. 6. Solve dy dx 4 dy dx 13y 0 2 2 ++ = . Solution. The equation can be written as (D 2 + 4D + 13) y =0 A.E. is m 2 + 4m + 13 = 0 m = –– 41652 2 ± =–2 ± 3i (of the form a ± iß) ? The general solution is y = e –2x (C 1 cos 3x + C 2 sin 3x). 7. Solve dy dx – dy dx 0 2 2 = given y'(0) = 0, y(0) = 1. Solution. Equation is (D 2 – D) y = 0 A.E. is m 2 – m = 0 m (m – 1) = 0 ? m = 0, 1 ? The general solution is y = C 1 e 0x + C 2 e x i.e., y = C 1 + C 2 e x when x = 0, y = 0 (Given) y (0) = 0 ? 0= C 1 + C 2 ...(1) y'(x)= C 2 e x Given, when x = 0, y' = 1 y (0) = 1 ? 1= C 2 e 0 ? C 2 = 1 ...(2) From (1) and (2) ? C 1 = –1. ? The general solution is y = e x – 1. Page 5 UNIT Differential Equations-I 5.1 INTRODUCTION We have studied methods of solving ordinary differential equations of first order and first degree, in chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders. Differential equations of second order arise very often in physical problems, especially in connection with mechanical vibrations and electric circuits. 5.2 LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS A differential equation of the form dy dx a dy dx a dy dx n n n n n n ++ 1 1 1 2 2 2 – – – – + ... + a n y = X ...(1) where X is a function of x and a 1 , a 2 ..., a n are constants is called a linear differential equation of n th order with constant coefficients. Since the highest order of the derivative appearing in (1) is n, it is called a differential equation of n th order and it is called linear. Using the familiar notation of differential operators: D = d dx , D 2 = d dx 2 2 , D d dx D d dx n n n 3 3 3 == ..., Then (1) can be written in the form {D n + a 1 D n–1 + a 2 D n–2 + ... a n } y = X i.e., f (D) y = X ...(2) where f (D)= D n + a 1 D n–1 + a 2 D n–2 + ... a n . Here f (D) is a polynomial of degree n in D If x = 0, the equation f (D) y =0 is called a homogeneous equation. If x ? 0 then the Eqn. (2) is called a non-homogeneous equation. 214 DIFFERENTIAL EQUATIONS–I 215 5.3 SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR DIFFERENTIAL EQUATION We consider the homogeneous equation dy dx p dy dx qy 2 2 ++ = 0 ...(1) where p and q are constants (D 2 + pD + q) y = 0 ...(2) The Auxiliary equations (A.E. ) put D = m m 2 + pm + q = 0 ...(3) Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) being quadratic in m, will have two roots in general. There are three cases. Case (i): Roots are real and distinct The roots are real and distinct, say m 1 and m 2 i.e., m 1 ? m 2 Hence, the general solution of eqn. (1) is y = C 1 e m 1 x + C 2 e m 2 x where C 1 and C 2 are arbitrary constant. Case (ii): Roots are equal The roots are equal i.e., m 1 = m 2 = m. Hence, the general solution of eqn. (1) is y =(C 1 + C 2 x) e mx where C 1 and C 2 are arbitrary constant. Case (iii): Roots are complex The Roots are complex, say a ± iß Hence, the general solution is y = e ax (C 1 cos ß x + C 2 sin ß x) where C 1 and C 2 are arbitrary constants. Note. Complementary Function (C.F.) which itself is the general solution of the D.E. 1. Solve dy dx –5 dy dx 6y 0 2 2 += . Solution. Given equation is (D 2 – 5D + 6) y = 0 A.E. is m 2 – 5m + 6 = 0 i.e., (m–2) (m– 3)= 0 i.e., m = 2, 3 ? m 1 = 2, m 2 = 3 ? The roots are real and distinct. 216 ENGINEERING MATHEMATICS—II ? The general solution of the equation is y = C 1 e 2x + C 2 e 3x . 2. Solve dy dx – dy dx –4 dy dx 4y 0 3 3 2 2 += . Solution. Given equation is (D 3 – D 2 – 4D + 4) y = 0 A.E. is m 3 – m 2 – 4m + 4 = 0 m 2 (m – 1) – 4 (m – 1) = 0 (m – 1) (m 2 – 4) = 0 m = 1, m = ± 2 m 1 = 1, m 2 = 2, m 3 = – 2 ? The general solution of the given equation is y = C 1 e x + C 2 e 2x + C 3 e –2x . 3. Solve dy dx – dy dx –6y 0 2 2 = . Solution. The D.E. can be written as (D 2 – D – 6) y =0 A.E. is m 2 – m – 6 = 0 ? (m – 3) (m + 2) = 0 ? m = 3, – 2 ? The general solution is y = C 1 e 3x + C 2 e –2x . 4. Solve dy dx 8 dy dx 16y 0 2 2 ++ = . Solution. The D.E. can be written as (D 2 + 8D + 16) y =0 A.E. is m 2 + 8m + 16 = 0 ? (m + 4) 2 =0 (m + 4) (m + 4) = 0 m = – 4, – 4 ? The general solution is y =(C 1 + C 2 x) e –4x . 5. Solve dy dx wy 0 2 2 2 += . Solution. Equation can be written as (D 2 + w 2 ) y =0 A.E. is m 2 + w 2 =0 DIFFERENTIAL EQUATIONS–I 217 m 2 =– w 2 = w 2 i 2 (i 2 = – 1) m = ± wi This is the form a ± iß where a = 0, ß = w. ? The general solution is y = e 0t (C 1 cos wt + C 2 sin wt) ? y = C 1 cos wt + C 2 sin wt. 6. Solve dy dx 4 dy dx 13y 0 2 2 ++ = . Solution. The equation can be written as (D 2 + 4D + 13) y =0 A.E. is m 2 + 4m + 13 = 0 m = –– 41652 2 ± =–2 ± 3i (of the form a ± iß) ? The general solution is y = e –2x (C 1 cos 3x + C 2 sin 3x). 7. Solve dy dx – dy dx 0 2 2 = given y'(0) = 0, y(0) = 1. Solution. Equation is (D 2 – D) y = 0 A.E. is m 2 – m = 0 m (m – 1) = 0 ? m = 0, 1 ? The general solution is y = C 1 e 0x + C 2 e x i.e., y = C 1 + C 2 e x when x = 0, y = 0 (Given) y (0) = 0 ? 0= C 1 + C 2 ...(1) y'(x)= C 2 e x Given, when x = 0, y' = 1 y (0) = 1 ? 1= C 2 e 0 ? C 2 = 1 ...(2) From (1) and (2) ? C 1 = –1. ? The general solution is y = e x – 1. 218 ENGINEERING MATHEMATICS—II 8. Solve (4D 4 – 4D 3 – 23D 2 + 12D + 36) y = 0. Solution. A.E. is 4m 4 – 4m 3 – 23m 2 + 12m + 36 = 0 If m = 2, 64 – 32 – 92 + 24 + 36 = 0 ? m = 2 is a root of inspection. By synthetic division, 2 4423 1236 8 8 30 36 4415 18 0 –– –– –– i.e.,4m 3 + 4m 2 – 15m – 18 = 0 If m =2 32 + 16 – 30 – 18 = 0 Again m = 2 is a root. By synthetic division 2 44 824 18 412 9 0 –15 –18 4m 2 + 12m + 9 = 0 (2m + 3) 2 =0 m = – , – 3 2 3 2 ? The roots of the A.E. are 2, 2, – , – . 3 2 3 2 Thus, y = (C 1 + C 2 x) e 2x + (C 3 + C 4 x) e –3x/2 . 9. Solve (D 5 – D 4 – D + 1) y = 0. Solution. A.E. is m 5 – m 4 – m + 1 = 0 i.e., m 4 (m – 1) – 1 (m – 1) = 0 (m – 1) (m 4 – 1) = 0 (m – 1) (m 2 – 1) (m 2 + 1) = 0 (m – 1) (m – 1) (m + 1) (m 2 + 1) = 0 ? The roots of the A.E. are 1, 1, –1, ± i. Thus y = (C 1 + C 2 x) e x + C 3 e –x + (C 4 cos x + C 5 sin x). 10. Solve y? + 4y' + 4y = 0 given that y = 0, y' = –1 at x = 1. Solution. We have (D 2 + 4D + 4) y =0 A.E. is m 2 + 4m + 4 = 0Read More

- Computer Science And Engineering Past Year Paper - III July 2016
Test | 75 questions | 150 min

- Computer Science And Engineering Model Test Paper -II
Test | 50 questions | 75 min

- Computer Science And Engineering Past Year Paper - III Jan 2017
Test | 75 questions | 150 min

- Computer Science And Engineering Past Year Paper - III Dec 2014
Test | 75 questions | 150 min

- Test: Introduction To EduRev
Test | 10 questions | 10 min