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Homogeneous Linear Differential Equations with Constant Coefficients

Homogeneous Linear Differential Equations with Constant Coefficients. A Lecture in ENGIANA. Auxillary Equation. Consider a second order equation ay’’ + by’ + cy = 0 where a, b, and c are constants. If we try to find a solution of the form y = e mx , then after substitution of

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Homogeneous Linear Differential Equations with Constant Coefficients

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  1. Homogeneous Linear Differential Equationswith Constant Coefficients A Lecture in ENGIANA

  2. Auxillary Equation • Consider a second order equation ay’’ + by’ + cy = 0 where a, b, and c are constants. • If we try to find a solution of the form y = emx, then after substitution of y’ = memx and y’’ = m2emx, the equation becomes am2emx + bmemx + cemx = 0

  3. Auxillary Equation • Solving am2emx + bmemx + cemx = 0, emx(am2 + bm + c) = 0 • The quantity in parenthesis, a quadratic equation, is called the auxiliary equation. • This means that to find the solution y (see previous slide), we must solve for m.

  4. Auxillary Equation There are three possible cases: • m1 m2; distinct real roots • m1 = m2; repeated real roots • m1 m2; conjugate complex roots

  5. Case 1: Distinct Real Roots For this case, we have And hence, Or

  6. Example Find the general solution of (D2 + D – 6) y = 0

  7. Case 2: Real Repeated Roots • Having two real, repeated roots means • Now, one solution is

  8. Case 2: Real Repeated Roots Recall that a2(x)y’’ + a1(x)y’ + a0(x)y = 0 can be written as y” + P(x)y’ + Q(x)y = 0 where P(x) = a1(x)/a2(x) Q(x) = a0(x)/a2(x)

  9. Case 2: Real Repeated Roots In our case, the coefficients are constants: ay’’ + by’ + cy = 0 Thus, y” + Py’ + Qy = 0 where P = b/a Q = c/a

  10. Case 2: Real Repeated Roots Recall also that another solution y2 is

  11. Case 2: Real Repeated Roots Hence,

  12. Case 2: Real Repeated Roots The general solution is then

  13. Example Find the general solution of y’’ + 8y’ + 16y = 0

  14. Case 3: Conjugate Complex Roots • If m1 and m2 are complex, then we have m1 =  + i m2 =  - i where  and  are real and positive • Hence, we can write y = C1e( + i)x + C2e( - i)x

  15. Case 3: Conjugate Complex Roots • However, in practice we prefer to work with real functions instead of complex exponentials. • To this end, we use Euler’s formula: ei = cos + isin where  is any real number

  16. Case 3: Conjugate Complex Roots • Thus, we have e ix = cosx + isinx e- ix =cosx - isinx • Note that e ix + e- ix = 2cosx & e ix – e- ix = 2isinx

  17. Case 3: Conjugate Complex Roots • Our solution is then y = C1e (+i)x + C2e(-i)x • If we let C1 = 1 and C2 = 1: y1 = e (+i)x + e(-i)x y1 = e x(eix + e-ix) y1 = e x(2cosx) y1 = 2e xcosx

  18. Case 3: Conjugate Complex Roots • If we let C1 = 1 and C2 = -1: y2 = e (+i)x - e(-i)x y2 = e x(eix - e-ix) y2 = e x(2isinx) y2 = 2ie xsinx

  19. Case 3: Conjugate Complex Roots Thus, the solution to y = C1e( + i)x + C2e( - i)x is y = c1y1 + c2y2 y = c1(e xcosx) + c2(e xsinx) or y = e x(c1cosx + c2sinx)

  20. Example Find the general solution of (D2 – 4D + 7) y = 0

  21. Higher-Order (n>2) Equations: Distinct Roots Consider the case where the auxiliary equation has distinct roots. Say we are given f(D)y = 0. Then one possible solution is emx, f(D)emx = 0, if the auxiliary equation is f(m) = 0

  22. Higher-Order (n>2) Equations: Distinct Real Roots In other words, if the distinct roots of the auxiliary equation are m1, m2, …, mn, then the corresponding solutions are exp(m1x), exp(m2x), …, exp(mnx). The general solution is

  23. Example Find the general solution of (D3 + 6D2 + 11D + 6) y = 0

  24. Higher-Order (n>2) Equations: Repeated Real Roots Consider the case where the auxiliary equation has repeated roots. Say we are given f(D)y = 0. If there are several identically repeated roots m1 = m2 = … = mn = b, then this means (D - b)n y = 0

  25. Higher-Order (n>2) Equations: Repeated Roots If we let y = xkebx [k = 0, 1, 2, …, (n-1)] Then, (D – b)n y = (D – b)n [xkebx] But (D – b)n [xkebx] = ebxDn[xk] = ebx (0) Thus, (D – b)n y = (D – b)n [xkebx] = 0

  26. Higher-Order (n>2) Equations: Repeated Roots The functions yk = xkebx [e.g., e7x, xe7x, x2e7x, etc.], where k = 0, 1, 2, …, (n – 1) are linearly independent because, aside from the common factor ebx, they contain only the respective powers x0, x1, x2, …, xn-1. The general solution is thus y = c1ebx + c2xebx + … + cnxn-1ebx

  27. Example Find the general solution of (D4 + 6D3 + 9D2) y = 0

  28. Higher-Order (n>2) Equations: Repeated Imaginary Roots • Repeated imaginary roots lead to solutions analogous to those brought in by repeated real roots. • For instance, if the conjugate pair m = a  bi occur three times, the corresponding general solution is y = (c1 + c2x + c3x2) eaxcosbx + (c4 + c5x + c6x2) eaxsinbx

  29. Example Find the general solution of (D4 + 18D2 + 81) y = 0

  30. Exercises Find the solution required: • (D2 – 2D – 3)y = 0 y(0)=0; y’(0)=-4 • (D3 – 4D)y = 0 y(0)=0; y’(0)=0; y’’(0)=2 • (D4 + 2D3 + 10D2)y = 0 • (D6 + 9D4 + 24D2 + 16)y = 0 • (D3 + 7D2 + 19D + 13)y = 0 y(0)=0; y’(0)=2; y’’(0)=-12 • (4D4 + 4D3 – 3D2 – 2D + 1)y = 0 • (D4 – 5D2 – 6D – 2)y = 0

  31. Exercises Find the solution required: • (D3 + D2 – D – 1)y = 0 y(0)=1; y(2)=0; • Find for x = 2 the y value for the particular solution required: (D3 + 2D2)y = 0 y(0)=-3; y’(0)=0; y’’(0)=12

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