Differential Equations Selected Solutions, Section 3.3 (Complex)

Selected Solutions, Section 3.3 (Complex) 7. Solving the characteristic equation, r = 1 ± i. Therefore, y = et (C1 cos(t) + C2 sin(t)) √ 8. Solving the characteristic equation, r = 1 ± 5 i. Therefore,  √  √ t y = e C1 cos( 5 t) + C2 sin( 5 t) 9. Solving the characteristic equation, r = 2, −4. Therefore, y = C1 e2t + C2 e−4t (Similarly, we solve 10-15). 25. For this one, try to go as far as you can without assistance from the computer: Let y 00 + 2y 0 + 6y = 0 with y(0) = 2 and y 0 (0) = α ≥ 0. (a) Solve it: The characteristic equation is r2 + 2r + 6 = 0. Therefore, r = −1 ± and the general solution is:  √ √  −t y(t) = e C1 cos( 5 t) + C2 sin( 5 t) √ 5i √ , so that the solution to With the initial conditions, find that C1 = 2 and C2 = α+2 5 the IVP is:   √ √ α+2 −t y(t) = e 2 cos( 5 t) + √ sin( 5 t) 5 (b) Find α so that y(1) = 0: Algebraically, we get: √ √ cos( 5) √ −2=α −2 5 sin( 5) (c) Find the smallest value of t > 0 so that y(t) = 0. We’ll write our answer as a function of α. Algebraically, solving this:   √ √ α+2 −t 0=e 2 cos( 5 t) + √ sin( 5 t) 5 will get us to: √ α+2 −2 cos( 5 t) √ = √ sin( 5 t) 5 We want to solve this for t, remembering that we want the smallest t > 0. I think it is easiest to work with the tangent rather than the cotangent, √ √ −2 5 tan( 5 t) = α+2 From here, we’re meant to use technology, so it’s OK to stop here. 1 27. Straightforward computation. Recall what we said in class- If we let r = λ + iµ, then y1 = Re(ert ) = Re(eλt+(µt)i ) = eλt cos(µt) and y2 = Im(ert ) = Im(eλt+(µt)i ) = eλt sin(µt) then W (y1 , y2 ) = µe2λt 6= 0, so y1 , y2 will form a fundamental set of solutions to our second order linear homogeneous DE with constant coefficients (in the case where we have complex roots). 35, 37 Euler Equations (We’re actually doing these after Section 3.4 this time, so I’m repeating this stuff there, too). The first thing is to convert y from a function of t to a function of x using x = ln(t). The text tell us to write dy/dt and d2 y/dt2 in terms of dy/dx and d2 y/dx2 . To do that, we use the Chain Rule. Doing this using Leibniz’ notation is easiest: dy dx dy = dt dx dt From x = ln(t), we have dx/dt = 1/t, so that 1 dy dy = dx t dx Secondly, we find a formula for d2 y/dt2 . To do that, we use our previous answer and the Product Rule:     d2 y d 1 dy 1 dy 1 d dy = =− 2 + = dt2 dt t dx t dx t dt dx In this second term, use the chain rule again. The form we are using is the same as before: d· dx d (·) = dt dx dt Where the first time around, the dot was y, the second time the dot is dy/dx. Therefore,   d dy d2 y dx 1 d2 y = 2 = dt dx dx dt t dx2 Putting it all together, with prime denoting the derivative in x, we have: d2 y y 00 − y 0 = dt2 t2 Finally, we substitute into the given equation in t to get a DE in x: t2 d2 y dy + α + βy = 0 dt2 dt (1) Substitute: 2 t  y 00 − y 0 t2   + αt 1 dy t dt  + βy = 0 2 ⇒ y 00 + (α − 1)y 0 + βy = 0 (2) Now, to solve Equation 1, we solve 2 and use the substitution x = ln(t). As a side remark, you might notice the following: If we use the ansatz y = tr for Equation 1, we get the following: t2 (r(r − 1)tr−2 ) + αt rtr−1 + βtr = 0 ⇒ r2 + (α − 1)r + β = 0 which is the same characteristic equation that is associated to Equation 2. We then have three cases (as for the case in Sections 3.1-3.4). Important Summary To solve t2 d2 y dy + αt + βy = 0, 2 dt dt we use the ansatz y = tr , from which we get the new characteristic equation: r(r − 1) + αr + β = 0 or r2 + (α − 1)r + β = 0 In solving the characteristic equation, we have three cases: • Two real solutions: r1 , r2 . We can write the solution in x, then translate to t using x = ln(t): y(x) = C1 er1 x + C2 er2 x ⇒ y(t) = C1 tr1 + C2 tr2 • One real solution, r. Using the same technique, first write in x, then translate to t: y(x) = erx (C1 + C2 x) ⇒ y(t) = tr (C1 + C2 ln(t)) • Complex solutions, r = λ + γi (we only need one of them). The solution is: y = eαx (C1 cos(βx) + C2 sin(βx)) ⇒
y = tα C1 cos(ln(tβ ) + C2 sin(ln(tβ )) 37. Solve the following, using Equation 2. 5 t2 ÿ + 3tẏ + y = 0 4 The characteristic equation is: r(r − 1) + 3r + 5 =0 4 ⇒ r2 + 2r + 5 =0 4 ⇒ 1 r = −1 ± i 2 so that, in x we would have the following, and then we translate to t: y(x) = e−x (C1 cos (x/2) + C2 sin (x/2)) y(t) = √ √  1 C1 cos(ln( t) + C2 sin( t) t 3