MA Ordinary Differential Equations

Transcription

1 MA Ordinary Differential Equations Santanu Dey Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 76 dey@math.iitb.ac.in March 21, 2014

2 Outline of the lecture Second order linear equations Method of reduction of order Second order linear equations with constant coefficients Cauchy-Euler equations Method of variation of parameters

3 Method of reduction of order We ve been looking at the second order linear homogeneous ODE y + p(x)y + q(x)y = 0. As we remarked earlier, there is no general method to find a basis of solutions. However, if we know one non-zero solution y 1 (x) then we have a method to find y 2 (x) such that y 1 (x) and y 2 (x) are linearly independent. To find such a y 2 (x), set y 2 (x) = v(x)y 1 (x) We ll choose v such that y 1 and y 2 are linearly independent. Can v be a constant? No. Now for y 2 to be a solution of the given ODE that is, y 2 + p(x)y 2 + q(x)y 2 = 0. (vy 1 ) + p(x)(vy 1 ) + q(x)(vy 1 ) = 0.

4 Second solution Thus, 0 = (v y 1 + vy 1) + p(v y 1 + vy 1) + qvy 1 = v y 1 + 2v y 1 + vy 1 + p(v y 1 + vy 1) + qvy 1 = v(y 1 + py 1 + qy 1 ) + v (2y 1 + py 1 ) + v y 1. Thus, Therefore, That is, v v = 2y 1 + py 1 y 1 = 2y 1 y 1 p. ( ) 1 ln v = ln y1 2 pdx; e R pdx v = dx. y 2 1

5 Second solution Claim: y 1 and vy 1 are linearly independent. Proof. Enough to check Wronskian! W (y 1, vy 1 ) = y 1 (v y 1 + y 1v) y 1vy 1 = y 2 1 v e R pdx = y1 2 y1 2 = e R pdx 0.

6 Example Given that y = x is a solution, find a l.i. solution of by reducing the order. y 2 = vy 1 = vx. Then, e R pdx v(x) = dx = y 2 1 (x 2 + 1)y 2xy + 2y = 0 e R 2x x 2 +1 dx x 2 dx = x x 2 dx = Hence, v(x) = x 1 ( x and y 2 = x x 1 ) = x 2 1. x Are y 1 & y 2 l.i.? What is the general solution? (1+ 1 x 2 )dx

7 Abel s formula Show that the Wronskian of any two solutions y 1 (x), y 2 (x) of y + p(x)y + q(x)y = 0 satisfies the DE W (x) = p(x)w (x) and is given by for any x 0 I. W (y 1, y 2 )(x) = W (y 1, y 2 )(x 0 )e R x x 0 p(t)dt,

8 Abel s formula Show that the Wronskian of any two solutions y 1 (x), y 2 (x) of y + p(x)y + q(x)y = 0 satisfies the DE W (x) = p(x)w (x) and is given by W (y 1, y 2 )(x) = W (y 1, y 2 )(x 0 )e R x x 0 p(t)dt, for any x 0 I. Solution : Set W (y 1, y 2 )(x) = W (x). Then, W (x) = (y 1 y 2 y 1y 2 )(x) W (x) = (y 1 y 2 y 1 y 2 )(x).

9 Contd.. Now, y 1 = p(x)y 1 q(x)y 1 y 2 = p(x)y 2 q(x)y 2. Thus, W (x) = y 1 py 2 y 1 qy 2 + y 2 py 1 + y 2 qy 1 = p(y 1 y 2 y 1y 2 ) = pw (x). Hence, W (x) = ce R x x 0 p(t)dt, for a constant c. For x = x 0, we get W (x 0 ) = c. Hence, W (y 1, y 2 )(x) = W (y 1, y 2 )(x 0 )e R x x 0 p(t)dt.

10 Second Order Linear ODE s with constant coefficients We have developed enough theory to now find all solutions of y + py + qy = 0, where p and q are in R; that is, a second order homogeneous linear ODE with constant coefficients. Suppose e mx is a solution of this equation. Then, and this implies m 2 e mx + pme mx + qe mx = 0, m 2 + pm + q = 0. This is called the characteristic equation or auxiliary equation of the linear homogeneous ODE with constant coefficients. The roots of this equation are m 1, m 2 = p ± p 2 4q. 2

11 Second Order Linear ODE s Case I: Real & unequal roots m 1, m 2 R, m 1 m 2. When p 2 4q > 0, m 1 and m 2 are distinct real numbers. Moreover, e m 1x e m 2x = e(m 1 m 2 )x is not a constant function. Hence, e m 1x and e m 2x are linearly independent. So the general solution of y + py qy = 0 is y = c 1 e m 1x + c 2 e m 2x, where c 1, c 2 R.

12 Second Order Linear ODE s Case II: Equal roots m 1 = m 2 R. m 1 = m 2 p 2 4q = 0, and in this case m = p 2. Hence e px 2 other solution, let Then, g(x) = v(x)e px 2. e R pdx v(x) = dx e px = ax + b, is one solution. To find the for some a, b R. Choose v(x) = x. Then, g(x) = xe px 2. Hence the general solution is with c 1, c 2 R. y = c 1 e px 2 + c2 xe px 2,

13 Second Order Linear ODE s Case III : Complex roots m 1 m 2 C\R. m 2 + px + m = 0 has distinct complex roots if and only if p 2 4q < 0. In this case, let Thus, and m 1 = a + ıb, m 2 = a ıb. e m 1x = e (a+ıb)x = e ax (cos bx + ı sin bx), e m 2x = e (a ıb)x = e ax (cos bx ı sin bx). As we are only interested in real valued functions, we take and f (x) = em 1x + e m 2x 2 g(x) = em 1x e m 2x 2ı = e ax cos bx, = e ax sin bx.

14 Second Order Linear ODE s Now, g(x) = tan bx is not a constant function. Thus the general f (x) solution is of the form y = e ax (c 1 cos bx + c 2 sin bx), with c 1, c 2 R.

15 Example 1 Solve 4y 8y + 3y = 0, y(0) = 2, y (0) = 1 2. The characteristic equation is 4m 2 8m + 3 = 0 = m = 3 2, 1 2. The general solution is y = c 1 e 3 2 x + c 2 e 1 2 x. Now, y = 3 2 c 1e 3 2 x c 2e 1 2 x y(0) = 2 = c 1 + c 2 = 2 y (0) = 1 2 = 3 2 c c 2 = 1 2 Solving, c 1 = 1 2, c 2 = 5 2. Therefore, y = 1 2 e 3 2 x e 1 2 x.

16 Example 2 Solve y 4y + 4y = 0, y(0) = 3, y (0) = 1. The characteristic equation is (m 2) 2 = 0 = m = 2 The general solution is y = c 1 e 2x + c 2 xe 2x = (c 1 + c 2 x)e 2x. Now, Hence, c 2 = 5. Therefore, y = 2(c 1 + c 2 x)e 2x + c 2 e 2x y(0) = 3 = c 1 = 3, y (0) = 1 = 2c 1 + c 2 = 1. y = (3 5x)e 2x.

17 Example 3 Solve y 6y + 25y = 0, y(0) = 3, y (0) = 1. The characteristic equation is m 2 6m + 25 = 0 The general solution is Now, = m 1 = 3 + 4ı, m 2 = 3 4ı. y = e 3x (c 1 cos 4x + c 2 sin 4x). y = 3e 3x (c 1 cos 4x + c 2 sin 4x) + e 3x ( 4c 1 sin 4x + 4c 2 cos 4x) Therefore, y(0) = 3 = c 1 = 3 y (0) = 1 = 3c 1 + 4c 2 = c 2 = 2. y = e 3x ( 3 cos 4x + 2 sin 4x).

18 Cauchy-Euler Equations The equation x 2 y + axy + by = 0 where a, b R is called a Cauchy-Euler equation. Assume x > 0. Suppose y = x m is a solution to this DE. Then, We get: that is, x 2 m(m 1)x m 2 + axmx m 1 + bx m = 0. m(m 1) + am + b = 0. m 2 + (a 1)m + b = 0. This is called the auxiliary equation of the given Cauchy-Euler equation. The roots are m 1, m 2 = (1 a) ± (a 1) 2 4b. 2

19 Cauchy-Euler Equations Case I: Distinct real roots. Are x m 1 and x m 2 linearly independent? Yes. Hence the general solution is given by for c 1, c 2 R. y = c 1 x m 1 + c 2 x m 2,

20 Cauchy-Euler Equations Case II: Equal real roots. that is, Hence m 1 = m 2 = 1 a 2. y = f (x) = x 1 a 2 is a solution. To get a solution g(x) linearly independent from f (x), set g(x) = v(x)f (x). Hence, v(x) = e R a x dx x 1 a dx = g(x) = (ln x)x 1 a 2. Thus the general solution is given by c 1, c 2 R. dx x = ln x y = c 1 x 1 a 2 + c 2 x 1 a 2 ln x,

21 Examples Solve: 1 2x 2 y + 3xy y = 0, x > 0. 2 x 2 y + 5xy + 4y = 0, x > 0. 3 x 2 y + xy + y = 0, x > 0. Solutions : 1 y = c 1 x + c2 /x 2 y = x 2 (c 1 + c 2 ln x). 3 y = c 1 cos(ln x) + c 2 sin(ln x).