= (aa + bb)e ax cos bx + ( ba + ab)e ax sin bx. + b2 One can similarly prove (ii).

Transcription

1 Example Prove the following by means of solving differential equations y (x) = e ax cos bx and u (x) = e ax sin bx: (i) e ax cos bx dx = eax (a cos bx + b sin bx) a 2 + b 2 + C, and (ii) e ax sin bx dx = eax (a sin bx b cos bx) a 2 + b 2 + C, where a 2 + b 2 = 0 Solution (i) The characteristic equation of homogeneous ODE y (x) = 0 is λ = 0, so y c (x) = Ce 0 x = C, for some constant C As a 2 + b 2 = 0, then (a, b) are non-zero vector in R 2, so it follows from the method of undetermined coefficients without modification, a particular solution is y p (x) = e ax (A cos bx + B sin bx) for some constants A and B 1e ax cos bx = y p(x) = ae ax (A cos bx + B sin bx) + e ax ( Ab sin bx + Bb cos bx) = (aa + bb)e ax cos bx + ( ba + ab)e ax sin bx { aa + bb = 1; Hence, ba + ab = 0, which gives A = a a 2 + b 2, and B = So, e ax a cos bx dx = y g (x) = a 2 cos bx + b + b2 One can similarly prove (ii) b a 2 + b 2 a 2 sin bx + C + b2

2 Ẹxample Find the general solution of y (x) y (x) y (x) + y(x) = 0 Solution Write down the characteristic equation λ 3 λ 2 λ + 1 = 0 by turning the order of derivative into power of λ, so 0 = λ 2 (λ 1) (λ 1) = (λ 2 1)(λ 1) = (λ 1) 2 (λ + 1), ie λ = 1, 1, 1 As λ = 1 is repeated, so the general solution of the original homogeneous ODE is y g (x) = Ae x + Be x + Cxe x = Ae x + (Cx + B)e x Remark If the real roots λ = α are repeated n times, then the general solution contains (A n 1 x n 1 + A n 2 x n 2 + A 1 x + A 0 )e αx, where A i are constants (0 i n 1) Ẹxample Find the general solution of y (4) (x) + 20y (x) + 64y(x) = 0 Solution The associated characteristic equation is 0 = λ λ = (λ )(λ 2 + 4), so the roots are λ = 2i, 2i and 4i, 4i Then the general solution of the original homogeneous ODE is y g (x) = A sin 2x + B cos 2x + C sin 4x + D cos 4x Remark The complex conjugate roots λ = α + iβ of char equation give e αx (A cos βx + B sin βx) in the general solution y g (x)

3 Example Find the solution of the equation 4y (x) + 16y (x) + 7y(x) = 0, with y(0) = 1 and y(π) = 0 Solution The associated characteristic equation is 0 = 4λ λ + 17 = (2λ + 4) 2 + 1, ie λ = 4± 1 2 = 2 ± 2 i It follows that the general solution is given by y(x) = e 2x ( A cos(x/2) + B sin(x/2) ) Then by plugging y(0) = 1 and y(π) = 0, we have 1 = y(0) = e 2 0 (A cos 0 + B sin 0) = A; 0 = y(π) = e 2 π (A cos(π/2) + B sin(π/2) = Be 2π ; (A, B) = (1, 0), ie the solution of the boundary value problem is y(x) = e 2t cos(x/2)

4 Ẹxample Find the general solution of x 2 y (x) 3xy (x) + 3y (x) = 0 Discussion First note that the given equation is homogeneous linear with variable coefficients, so the simple method of solving characteristic equation could not apply Though it is not in the Euler form yet, however one can rewrite the given ODE in the form of Euler s equation x n y (n) (x) + A n 1 x n y (n) (x) + A 1 xy (x) + A 0 y(x) = 0 by multiplying by x Solution After multiplying by x, we have x 3 y (x) 3x 2 y (x) + 3xy (x) = 0, and its associated modified characteristic equation 0 = m(m 1)(m 2) 3m(m 1) + 3m = m(m 2 3m + 2 3m + 3) = m(m 2 6m + 5) = m(m 1)(m 5) It follows that m = 0, 1, 5, so the general solution y g (x) of the original homogeneous ODE is y g (x) = Ax 0 + Bx 1 + Cx 5 = A + Bx + Cx 5

5 Ẹxample Find the general solution of y (x) 2y (x) y (x) + 2y(x) = 8x 3 Solution First note that the given equation is non-homogeneous linear Write down the characteristic equation 0 = λ 3 2λ 2 λ + 2 = λ 2 (λ 2) (λ 2) = (λ 2 1)(λ 2), so λ = 1, 1, 2 The general solution y c (x) of the associated homogeneous equation y (x) 2y (x) y (x) + 2y(x) = 0 is y c (x) = Ae 1 x + Be 1 x + Ce 2x = Ae x + Be x + Ce 2x This solution is called the complementary solution of original equation Next is to determine a particular solution y p (x), it follows from the method of undetermined coefficients (without modification) that it is given by y p (x) = Ex 3 + Fx 2 + Gx + H a polynomial of degree at most 3 Putting y p (x) into the original equation, we have 8x 3 = y p (x) 2y p (x) y p(x) + 2y p (x) = (6E) + 3(6Ex + 2F) (3Ex 2 + 2Fx + G) + 2(Ex 3 + Fx 2 + Gx + H) = 2Ex 3 + (2F 3E)x 2 + (2G 2F + 18E)x + (2H G + 6F + 6E), so E = 8 2 = 4, F = 3E 2 = 3 4 2F 18E = 6, G = = F 9E = 30, and 2 2 6E 6F + G H = = 3E 3F + G = = 45 (Next page) 2 2

6 Example Find a general solution of (D 4 1)y(x) = 30e 2x x 2, where D = dx d Solution Just like the previous example, the characteristic equation 0 = λ 4 1 = (λ 2 + 1)(λ 2 1) = (λ 2 + 1)(λ 1)(λ + 1), so (λ = 1, 1, i, and i So the complementary solution y c (x) = Ae x + Be x + E cos x + F sin x } {{ } λ=i, i It follows from the method of undetermined coefficients that a particular solution y p (x) = Gx 2 + Hx + J + Ke 2x, where G, H, J, K are some constants to be determined by the equation (D 4 1)y p (x) = 30e 2x x 2, ie x e 2x = (D 4 1)(Gx 2 + Hx + J + Ke 2x ) = (Gx 2 + Hx + J) + K(D 4 1)e 2x = Gx 2 Hx J + K(2 4 1)e 2x, so G = 1, H = 0, J = 0 and K = 30/15 = 2 Hence the general solution y g (x) = y p (x) + y c (x) = x 2 + 2e 2x + Ae x + Be x + E cos x + F sin x, where B, E and F are constants

7 Example Use the reduction of order to find the general solution of 2t 2 y (t) + ty (t) 3y(t) = 0 for t > 0, if y 1 (t) = t 1 is a solution Solution In order to apply the reduction of order, we first rewrite the given equation in the form y (t) + p(t)y (t) + q(t)y(t) = 0, ie y (t) + 1 2t y (t) 3 2t 2 y(t) = 0 Set y 2 (t) = v(t)y 1 (t), where v (t) = U(x) = exp( p(t) dt) y 1 (t) 2 = t 2 exp( dt 2t ) = t2 exp( ln t 2 ) = t2 1 2 = t 3/2 Then integrating, we have v(t) = v (t) dt = t 3/2 dt = 2t5/2 3, so y 2 (t) = v(t)y 1 (t) = t 1 2t5/2 3 = 2 3 t3/2 It follows that the general solution of original linear equation is At 1 + Bt 3/2 Solution II As the original equation is of Euler-Cauchy equation, its characteristic equation is 0 = 2m(m 1) + m 3 = 2m 2 m 3 = (2m 3)(m + 1) so m = 1 or 3/2 Hence, the general solution y g (t) = At 1 + Bt 3/2, where A and B are arbitrarily constant

8 Example Solve the initial value problem: x 2 y (x) + 3xy (x) + y(x) = 0, y(1) = 4, y (1) = 2 Solution This is Euler-Cauchy equation, and its characteristic equation 0 = m(m 1) + 3m + 1 = m 2 + 2m + 1 = (m + 1) 2, so m = 1, 1 (repeated root) Then the general solution of the homogeneous equation is y g (x) = x 1 (A ln x + B) (If you don t know what happen, read the lecture note) As the initial value conditions is at x = 1, so we only consider the interval x > 0, ie ln x = ln x we have 4 = y(1) = A + B, and 2 = y (1) = [ x 1 (A/x) x 2 (A ln x + B) ] = A B So A = 1 and x=1 B = 3 It follows that the solution of the IVP is y(x) = x 1 (3 + ln x), which is defined on (0, + )

9 Ẹxample Find the general solution of y 2y 9y + 18y = e 2x Solution This is non-homogeneous constant coefficient linear third order ODE, and its characteristic equation is 0 = λ 3 2λ 2 9λ + 18 = (λ 2 9)(λ 2) = (λ 3)(λ 2)(λ + 3), so λ = 3, 2, 3, so the complementary solution y c (x) = Ae 3x + Be 2x + Ce 3x In order to determine the particular solution y p (x), we observe that ( dx d 3)( dx d + 3)( dx d 2)(xe2x ) = ( dx d 3)( dx d + 3)(e2x ) = 5e 2x ( ), so y p (x) = xe 2x /5 Hence, the general solution is y c (x) + y p (x) = Ae 3x + Be 2x + Ce 3x xe 2x /5 Remark Though we are not using method of undetermined coefficients with modification, we choose the test function xe 2x in (*), which simplifies a lot of calculation

10 Ẹxample Find the general solution of y + 4y + 4y = e 2x 2 cos 2x Solution The characteristic equation of the non-homogeneous constant coefficient linear ODE is given by 0 = λ 2 + 4λ + 4 = (λ + 2) 2, so λ = 2, 2 with multiplicity = 2 So the complementary solution y c (x) = (Ax + B)e 2x As the ODE is non-homogeneous, one can apply the method of undetermined coefficients with modification to write down a particular solution y p (x) = Cx 2 e 2x + (D cos 2x + E sin 2x) Next we are going to determine the coefficients C, D and E in y p (Not finished yet!) It remains to determine the coefficients by subsitituing the y p in the original ODE So e 2x 2 cos 2x = y p + 4y p + 4y p = ( dx d + 2)2 (Cx 2 e 2x ) + ( dx d d dx )(D cos 2x + E sin 2x) = 2Ce 2x + 4 dx d (D cos 2x + E sin 2x) = 2Ce 2x + (2E cos 2x 2D sin 2x), ie (C, E, D) = (1/2, 1, 0) And y g (x) = (Ax + B)e 2x + x 2 e 2x /2 sin 2x Remark ( d dx λ)(xn e λx ) = nx n 1 e λx + λx n e λx λx n e λx = nx n 1 e λx, and so ( d dx λ)2 (x n e λx ) = ( d dx λ)(nxn 1 e λx ) = n(n 1)e λx for n 2