1 Some general theory for 2nd order linear nonhomogeneous

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1 Math 175 Honors ODE I Spring, 013 Notes 5 1 Some general theory for nd order linear nonhomogeneous equations 1.1 General form of the solution Suppose that p; q; and g are continuous on an interval I; and consider the equation y 00 + p (t) y 0 + q (t) y = g (t) (1) Theorem 1 Suppose that y p is a solution of (1) and y h = c 1 y 1 + c y is the general solution of the homogeneous equation y 00 + p (t) y 0 + q (t) y = 0: () Then the general solution of the non-homogeneous equation (1) is y = y p + y h : Proof. Suppose that y and y p are both solutions of (1). Let u = y y p : Then u 00 + p (t) u 0 + q (t) u = y 00 + p (t) y 0 + q (t) y yp 00 + p (t) yp 0 + q (t) y p = g g = 0; so u satis es (). Conversely, if u solves () and y p solves (1), then it is easily calculated that u+y p also satis es (1). To nd the solution of (1) such that y (0) = and y 0 (0) = ; we let u be the solution of () such that u (0) = y p (0) and u 0 (0) = yp 0 (0) : Hence any solution of (1) can be written as u + y p where u solves (), and this proves the theorem. Remark There are many possibilities for the particular solution y p ; since it can be any solution of (1). And there are many possibilities for y h ; since y 1 and y can be any pair of linearly independent solutions of (). 1

2 Some examples For an inhomogeneous equation with constant coe cients, which we will write as ay 00 + by 0 + cy = g (t) ; the main di culty is nding y p ; since we can always nd y h : For certain special functions g it is known what the general form of the solution should be. Example 3 y 00 + y = t: (3) By inspection we can see that the function y p = 1 t solves this equation. You can easily check that the general solution of y 00 + y = 0 is y h (t) = c 1 cos t + c sin t; and so the general solution of (3) is y = c 1 cos t + c sin t + 1 t: If and are real numbers, then we can nd c 1 and c such that y (0) = and y 0 (0) = ; by the method used in the proof of the theorem. Example y 00 + y = cos t: Now it is more of a challenge to nd y p : The method is to look for a solution of the form y = A sin t + B cos t: If we substitute this into the equation we get This can only be true for all t if A sin t B cos t + A sin t + B cos t = cos t: 3A = 0 3B = 1: we can set y p = 1 cos t; 3 and the general solution of (3) is y = c 1 cos t + c sin t + 1 cos t: 3 You may wonder why we included A sin t in the solution we tried. example shows that this is sometimes necessary. The next

3 Example 5 Again try y = A sin t + B cos t: We get y 00 + y 0 + y = cos t: A sin t B cos t + A cos t B sin t + A sin t + B cos t = cos t: We now get the equations 3A B = 0 3B + A = 1 so both A and B are needed: A = 13 ; B = 3 13 : Example 6 Try y = A cos t + B sin t: We get y 00 + y = cos t A cos t B sin t + A cos t + B sin t = cos t: But the left side adds to 0; so there is no choice of A and B which will work. It turns out that this is a little similar to the homogeneous case with repeated roots: we multiply our original guess by t; and try y = At cos t + Bt sin t: Now the calculations are more complicated. You should verify that we get y p = 1 t sin t: Note again that we had to include the t sin t term in our trial solution. The key to this example is that the original trial solution satis es L y = 0; and so couldn t possibly solve Ly = g if g is not zero. In this case, you multiply the trial solution by t and try again. Example 7 y 00 y = e t : We are tempted to try y p = Ae t : However, this is a solution of y 00 we must try y = Ate t : I will not give the details. y = 0: Hence To repeat, you can solve only some special types of second order linear odes (all with constant coe cients) by this trial solution method. Table summarizes what solutions you should try for what right hand sides g: Be sure to read the Notes following this table. 3

4 3 Using complex valued functions to solve nonhomogeneous linear equations with constant coe cients. I will illustrate this method with the example To solve this, we make use of Euler s formula, to write y 00 3y 0 y = 8e t sin t: () e (+i)t = e t (cos t + i sin t) : the right hand side of equation () is the imaginary part of 8e (1+i)t : Instead of (), therefore, we consider the equation z 00 3z 0 z = 8e t (5) where = 1 + i. We will nd a solution z (t) = u (t) + iv (t) : Note that z 00 3z 0 z = u 00 3u 0 u+i (v 00 3v 0 v) = 8e t = 8e t cos t+i 8e t sin t : is a particular solution of (): To solve (5) we try a solution Then Substituting this into (5) gives v (t) = Im (z (t)) z (t) = ce t : z 0 = ce t ; z 00 = c e t : ce t 3 = 8e t ;

5 from which we get that c = 8 3 : (When would this method fail?) Since = 1 + i; we nd that 3 = 10 i; and so z (t) = e t c = 8 10 i = (5 i) 5 + i 5 i = 0 i 6 = i i (cos t + i sin t) = e t 13 cos t + 10 sin t +ie t sin t cos t : 13 a particular solution to () is 10 y p (t) = e t 13 sin t cos t 13 You can also look at Example 3, pg 178, and compare the answer there with what we would get above if the right side of () were 8e t cos t. Some applications to forced oscillations.1 resonance We now apply the same method to the equation This time the complex equation is Because we will nd y from the formula Substituting z = ce it into (7) gives y 00 + y = cos t: (6) z 00 + z = e it : (7) cos t = Re e it ; y = Re z: ce it ( + ) = e it : (8) 5

6 Obviously there is no solution for c. This happened because cos t is a solution of the homogeneous equation y 00 + y = 0; and so there is no way that a constant multiple of cos t can solve the inhomogeneous equation (3) ; and correspondingly, a constant multiple of e it cannot solve (7). This turns out to be rather like the case of repeated roots for a homogeneous equation. Instead of using e it ; we try z = cte it : The derivative is now more complicated, and we get Substituting in the ode for z gives z 0 = c e it + ite it z 00 = c ie it te it : ce it (i t + t) = e it c = 1 i = 1 i; and z = 1 i (cos t + i sin t) = 1 t sin t 1 it cos t: Taking the real part, we nd that a particular solution to (6) is y p (t) = 1 t sin t: A tedious calculation can verify that this actually is a solution of (). The graph of this solution will be discussed in class, and see page 1 of the text for a similar graph.). Beats Now we change the equation just a little: y 00 + y = cos! t 6

7 where we assume that j!j 6= : This time, following the same procedure, we try z = e i!t ; and instead of equation (8) ; we get ce i!t! + = e i!t : c = 1! : Since we assumed that! 6=, we haven t divided by zero. Continuing on, we nd that y p = 1 cos!t:! The general solution is then y = c 1 cos t + c sin t + 1 cos!t: (9)! An interesting solution is found using what are called homogeneous initial conditions, namely y (0) = 0; y 0 (0) = 0: (10) To substitute the boundary conditions, we need y 0 : y 0 (t) = c 1 sin t + c cos t and setting t = 0 in (9) and (11), equations (10) imply that y = c 1 + 1! = 0 c = 0: 1 1 cos t + cos!t:!!! sin!t; (11)! This is the di erence of two oscillations of two di erent periods (or frequencies). We use trigonometry to rewrite this. The key trig identity used is probably less familiar to you than others used before: sin A sin B = 1 (cos (A B) cos (A + B)) 7

8 Setting A = t and B =!t, y = 1! =! sin (cos!t cos t) ( +!) t sin (!) t : Now we assume that! is quite close to : Suppose, for example, that! = 1:9: Then,! = :039, and y = (sin 1:95t) (sin :05t) : :039 The graph of this will also be discussed in class. 13. A similar graph appears on page 5 Homework: Due Feb 6. Don t forget to sketch graphs of the solutions when asked (1), # , #5 (156, # 39) (163) #. 17 (171), #1, but with di erent initial conditions: y (0) = ; y 0 (0) = 6: 5. pg. 185 (18), # 3 8