Unformatted text preview: MEC702 Wk 5 Lect 1 Homogeneous Linear ODE with Constant Coecients
Summary of Cases IIII Case
I
II
III Condition
2 a − 4b > 0
a2 − 4b = 0
2 a − 4b < 0 Roots of CE
Distinct Real λ1 , λ2
Real Double λ1 = λ2 = − 12 a Basis of ODE
e
e 1
λ1 = − a + iω
2
Complex Conjugate
1
λ2 = − a − iω
2 λ1 x − ax
2 e− General Solution of ODE λ2 x y = c1 eλ1 x + c2 eλ2 x
ax
y = (c1 + c2 x)e− 2 ,e
ax
, xe− 2 ax
2 − ax
2 e cos ωx
sin ωx y = e− ax
2 (A cos ωx + B sin ωx) Example. • Case I: For the IVP y 00 + y 0 − 2y = 0, we have the characteristic equation y(0) = 4, y 0 (0) = −5 , λ2 + λ − 2 = 0. The roots are then λ1 = 1 and λ2 = −2 and the general solution is
y = c1 ex + c2 e−2x . Applying the initial conditions
y(0)
0 y (0) we have the particular solution
• Case II: For the IVP = 4 = c1 + c2 = 5 = c1 − 2c2 y = ex + 3e−2x . y 00 + y 0 + 0.25y = 0, we have the characteristic equation
We see that y(0) = 3, y 0 (0) = −3.5, λ2 + λ + 0.25 = 0. λ2 + λ + 0.25 = (λ + 0.5)2 =⇒ λ = −0.5. This gives the general solution y = (c1 + c2 x)e−0.5x . Applying the initial conditions
y(0)
0 y (0) we have the particular solution = 3.0 = c1 = c2 − 0.5c1 = 3.5 =⇒ c2 = −2,
y = (3 − 2x)e−0.5x . 1 • Case III: For the IVP
y 00 + 0.4y 0 + 9.04y = 0, we have the characteristic equation y(0) = 0, λ2 + 0.4λ + 9.04 = 0. Solving, we obtain
λ = −0.2 ± 3i and 1
ω 2 = b − a2 = 9 =⇒ ω = 3.
4 This gives the general solution y = e−0.2x (A cos 3x + B sin 3x). Applying the initial conditions
y(0)
0 y (0) we have the particular solution
Exercise. = 0=A = 3 = 3B =⇒ B = 1, y = e−0.2x sin 3x. Solve the ODEs. 1. 4y 00 − 25y = 0.
2. y 00 + 36y = 0.
3. y 00 + 6y 0 + 8.96y = 0
4. y 00 + 4y 0 + (π 2 + 4)y = 0 2 y 0 (0) = 3, EulerCauchy Equations
EulerCauchy equations are ODE'sof the form
x2 y 00 + axy 0 + by = 0 with given constants a and b and unknown function y(x). Consider a solution of the form
y = xm , we have
y0 = mxm−1 y 00 = m(m − 1)xm−2 and substituting these into the ODE, we have
x 2 m(m − 1)x m−2 x2 y 00 + ay 0 + by = 0 + amxm−1 + bxm = 0 from which we have the auxilliary equation
m2 + (a − 1)m + b = 0. (Note: a − 1, not a.) The roots then are Real Roots m1 = m2 = r
1
1
(1 − a) +
(1 − a)2 − b
2
4
r
1
1
(1 − a) −
(1 − a)2 − b.
2
4 m1 6= m2 For real roots, we have the two real solutions
y1 (x) = xm1 with the general solution
Real Double Roots and y2 (x) = xm2 y = c1 xm1 + c2 xm2 . m1 = m2 In this case,
m= 1
2 (1 − a) and we have a solution i b = 41 (a − 1)2
x y1 = x(1−a) 2 and a second linearly independent solution
y2 = y1 ln x can be obtained from the method of reduction. The general solution is
y = (c1 + c2 ln x)xm . 3 Cases
Real Distinct roots m1 6= m2
Real Double roots m1 = m2
Example. Basis General Solution y1 = xm1 , y2 = xm2
y1 = xm , y2 = xm ln x y = c1 xm1 + c2 xm2
y = (c1 + c2 ln x)xm Find the general solutions of the EulerCauchy equations. 1. x2 y 00 + 1.5xy 0 − 0.5 = 0.
2. x2 y 00 − 5xy 0 + 9y = 0.
. Solution 1. y = c1 x0.5 + c2 x−1 .
2. y = (c1 + c2 ln x)x3 . Linear Independence of Solutions. Wronskian.
Theorem. Let p(x) and q(x) in y 00 + p(x)y 0 + q(x)y = 0 be continuous on an interval I , then the two solutions y1 and y2 of the ODE on I are linearly dependent i their
'Wronskian' y
W (y1 , y2 ) = 10
y1 y2 =0
y20 at some x0 ∈ I . Furthermore, if there is an x1 in I at which y
W (y1 , y2 ) = 10
y1 y2 6= 0,
y20 then y1 and y2 are linearly independent on I .
Example. For the ODE y 00 + ω 2 y = 0 has the solutions
y1 = cos ωx y2 = sin ωx. Notice that
W (y1 , y2 ) = cos ωx −ω sin ωx sin ωx ω cos ωx = (cos ωx)(ω cos ωx) − (sin ωx)(−ω sin ωx) = ω which is a constant. The two solutions are linearly independent i ω 6= 0.
Example. Check whether the solutions y1 = ex and y2 = xex are linearly independent for the ODE
y 00 − 2y 0 + y = 0 on any interval. 4 ...
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 Winter '16
 Alveen Chand
 Linear Algebra, Linear Independence, basis, Complex number, real solutions

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