Review. To solve ay + by + cy = 0 we consider the characteristic equation. aλ 2 + bλ + c = 0.

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1 Review To solve ay + by + cy = 0 we consider the characteristic equation aλ 2 + bλ + c = 0. If λ is a solution of the characteristic equation then e λt is a solution of the differential equation. if there are two distinct roots then we have the general solution, i.e. a fundamental set of solutions. We need complex exponentials to solve ay + by + cy = 0 when the roots of the characteristic equation are not real. Complex exponentials work like this: e it = cos t + i sin t e s+it = e s cos t + i sin t

2 The case of equal roots What if the two roots of the characteristic equation are equal? Then they are both real (since we assume a, b, and c are real), so they must be complex conjugates, but a number equal to its complex conjugate is real.

3 The solution Suppose the characteristic equation has a double root λ. Then e λt is one solution, as we know. The two roots of aλ 2 + bλ + c are λ = b 2a ± 2a b b 2 4ac If they are equal then λ = b/2a. We will show that another solution is te λt.

4 te λt is a solution y = te λt y = e λt + tλe λt = (tλ + 1)e λt y = λe λt + (tλ + 1)λe λt = (2λ + tλ 2 )e λt ay + by + cy = a(2λ + tλ 2 )e λt + b(tλ + 1)e λt + cte λt = e λt (a(2λ + tλ 2 ) + b(tλ + 1) + ct) = te λt (aλ 2 + bλ + c) + (2aλ + b)e λt ) = te λt 0 + (2aλ + b)e λt = (2aλ + b)e λt = 0 since λ = b/2a Compare this with the different (and longer) calculation on page 168 of the textbook.

5 e λ t and te λt form a fundamental set All we have to do is show that the Wronskian does not vanish. u = e λt v = te λt v = tλe λt + e λt = (tλ + 1)e λt W (u, v) = uv vu = e λt (tλ + 1)e λt te λt λe λt = e 2λt (tλ + 1 tλ) = e 2λt 0

6 Summary To solve ay + by + cy = 0 (with real coefficients): Solve the characteristic equation aλ 2 + bλ + c = 0. If the roots λ and µ are distinct then the general complex-valued solution is Ae λt + Be µt. If the roots are equal then the general solution is Ae λt + Bte λt. If the roots are not real, they are λ = p + iq, and we can write the general real-valued solution as Ae pt cos(qt) + Be pt sin qt. (This formula does not involve complex numbers).

7 Reduction of Order Above we just verified that te λt is a solution. But how was that discovered? We knew one solution y 1 = e λt, and we wanted another solution y. We could just try y = v(t)y 1. Plug it in to the equation and see what equation v should satisfy to make y a solution. v will need to satisfy a first-order equation, which is simpler than second-order. (Hence reduction of order. ) If we do this for ay + by + cy = 0 we find out v = t works. This procedure can be tried even if the original equation does not have constant coefficients.

8 Reduction of order with constant coefficients Here is how we could have discovered the solution y = te λt of ay + by + cy = 0. y = vu where u is a known solution e λt y = v u + vu y = v u + v u + v u + vu = v u + 2v u + vu ay + by + cy = a(v u + 2v u + vu ) +b(v u + vu ) + cuv With constant coefficients we have u = e λt and u = λu and with equal roots we have λ = b/a. So ay + by + cy = a(v u + 2v λu + vλ 2 u) + bu(v + λv) + cuv = (aλ 2 + bλ + c)uv + auv + uv (2aλ + b) = auv since aλ 2 + bλ + c = 0 and λ = b/2a Since au 0, we have v = 0, so v is a linear function of t. Done!

9 Reduction of order in general We got v = 0. This can be regarded as a first-order equation for v. We can try the same trick for a linear equation with non-constant coefficients, if we know one solution. For example, if we knew J 0 but did not know Y 0.

10 Reduction of order example Consider the equation 2t 2 y + 3ty y = 0 It has one solution u = 1/t. Find a fundamental set of solutions. We look for a solution y = vu = v/t. y = tv v t 2 = v t v t 2 y = tv v t 2 = v t 2 v t 2 + 2v t 3 t2 v 2vt t 4

11 Reduction of order example Consider the equation 2t 2 y + 3ty y = 0 It has one solution u = 1/t. Find a fundamental set of solutions. We look for a solution y = vu = v/t. y = tv v t 2 = v t v t 2 y = tv v t 2 t2 v 2vt t 4 = v t 2 v t 2 + 2v t 3 2t 2 y = 2tv 4v + 4v/t 3ty = 3v 3v/t 2t 2 y + 3ty y = 2tv 4v + 4v/t + 3v 3v/t v/t = 2tv v

12 We re working on 2t 2 y + 3ty y = 0 We looked for a solution y = v/t and found that it works if 0 = 2tv v. That is, if v satisfies the first-order equation 0 = 2tw w.

13 So v should satisfy That we can solve: 0 = 2tw w. 2dw = dt w t 2 ln w = ln t + C = ln At where A = e c > 0 ln w 2 = ln At w 2 = At w = E t where E is a positive constant, E = A Hence v = E t. Then v is a constant time t 3/2. Hence our fundamental set of solutions for is {1/t, t 3/2 }. 2t 2 y + 3ty y = 0

14 Reduction of Order Was it a miracle (or accident) that the coefficients of v canceled out and left us a first-order equation for v? We will see that it was not. It had to happen!

15 Reduction of Order Suppose u + pu + qu = 0 where p and q are not necessarily constant. We look for a solution y = vu. Then y = v u + vu y = uv + 2v u + vu y + py + qy = uv + 2v u + vu + p(v u + vu ) + qvu = uv + v (2u + pu) + v(u + pu + qu) = uv + v (2u + pu)

16 Reduction of Order Suppose u + pu + qu = 0 where p and q are not necessarily constant. We look for a solution y = vu. Then y = v u + vu y = uv + 2v u + vu y + py + qy = uv + 2v u + vu + p(v u + vu ) + qvu = uv + v (2u + pu) + v(u + pu + qu) = uv + v (2u + pu) The coefficient of v vanished because u is a solution. That leaves a first-order equation for v

17 Summary For the linear equation with constant coefficients, ay + by + cy = 0, when the roots of the characteristic equation are equal, the general solution is y = Ae λt + Bte λt. For the linear homogeneous equation with not-necessarily-constant coefficients, if we have one solution u, we can find the general solution Au + Bvu using the method of reduction of order. We look for a solution y = vu, and when we plug this into the equation, the coefficients of v will cancel out and leave a first-order equation for v. Solve that equation, and then integrate to find v. Of course, this method doesn t tell us how to find the first solution u.