dt 2 roots r = 1 and r =,1, thus the solution is a linear combination of e t and e,t. conditions. We havey(0) = c 1 + c 2 =5=4 and dy (0) = c 1 + c

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1 MAE 305 Assignment #3 Solutions Problem 9, Page 8 The characteristic equation for d y,y =0isr, = 0. This has two distinct roots r = and r =,, thus the solution is a linear combination of e t and e,t. That is to say y(t) =c e t +c e,t.we nd c and c by using the boundary conditions. We havey(0) = c + c =5=4 and dy (0) = c + c =,3=4, giving y(t) = +e 4 et,t. To nd the min, we set dy = 0, solving, we nd t = ln(). The verication that this is a max can be done by looking at the second derivative or the properties of the rst derivative. The hyperbolic trig functions sinh(t) and cosh(t) are both solutions, just do the derivatives. If you didn't use a computer to the plot this time, we let you o. But that would happen again! Learn to do it; it's good for you in the long run a t Figure : Solution for Problem 9, Page 8

2 Problem 30, Page 8 To solve d y + t( dy ) =0,we follow the hint in the paragraph above the problem. After changing variables to v = dy,we arrive the following rst order equation dv + t(v) =0. This is a separable equation. Solving the equation gives,=v =,t =+c. Solving for v and remembering v = dy,we get dy = v =, c,t.we are not done yet. We still have to solve for y(t). We don't know any thing about the constant ofintegration c.aswe try to solve for y, we nd that we need to make assumptions the sign of c. If c =0, then dy ==t =) y(t) =,=t + c.ifc >0we set c = k.(this is a nice trick because now c is clearly positive.) Solving the resulting equation by partial fractions, we get y(t) = k,t ln j j+c k k+t.ifc <0, we set c =,k. The resulting integral is an inverse tangent. Thus we get y(t) = tan, ( t )+c k k. Lastly, notice that y(t) =constant is also a solution. Problem 34, Page 8 Following the hint, we substitute v = dy which gives y dv +(v) = 0. But this doesn't really help. Now wehave three variable, clearly it would be nice to get rid of one. Again following the hint we use dv = dv dy = dv v to arrive dy dy at y( dv )v + v =0. This is separable and gives v = c dy =y. Next we solve v = dy = c =y which is again separable giving a nal answer of y = c t + c. Problem 4, Page 38 First check that both are solutions. y (t) =e t =)y 0 (t)=e t,y 00 = e t =) y 00,y 0 +y = e t,e t +e t = 0 Checks!! y (t) =te t =) y 0 (t) =et +te t, y 00 =et +te t =) y 00, y 0 + y =e t +te t, (e t + te t )+e t = 0 Checks!!

3 Now check to see if it is a complete set of solutions. In other words, are they linearly independent. To do this we use the Wronskian. y y y 0 y 0 = y y 0, y 0 y W = e t We see that W is never zero for any t, hence they are linearly independent solutions. Problem 8, Page 38 Again we take inspiration from the paragraph above the question. We see that if P 00 (x), Q 0 (x) +R(x) = 0 then the equation is exact. In our case, P =,Q=x,R=. Plugging these in we nd that the equation is exact. The point is to get the equation in the form d dy [P (x) dx dx ]+ d [f(x)y] = 0 (*) dx because we can integrate this once arriving at P (x) dy dx + f(x)y = which we can solve. So, the question is what is f(x) in terms of R; Q; P.To see this, we expand equation (*). This gives lp 0 y 0 + Py 00 + f 0 y + fy 0 Py 00 +(P 0 +f)y 0 +f 0 y=0 We know that Py 00 + Qy 0 + R = 0. By equating terms in the two equations we get that Q, P 0 = f. Hence in our case f = x. After integrating equation () once, we get y 0 + xy = c. This is just a rst order ODE that we know how to solve from previous sections. We get y = c e,x = Z x x0 e x = dx + c e,x = 3

4 Problem 7, Page 44 Problem Find the Wronskian of two solutions of the given dierential equation without solving the equation, x y 00 + xy 0 +(x, )y=0: Bessel's equation () Solution Here we will use the Abel's theorem. In this problem, P (x) = =x and Q(x) =(x, )=x. The Wronskian is given by W (y ;y )(x) =cexp[, Z p(x)dx] () i.e., W (x) =cexp[,z dx] =cexp(, ln x): (3) x Thus, we have W(x)= c x : (4) Problem 4, Page 50 Problem Consider the initial value problem 5u 00 +u 0 +7u=0; u(0)=; u 0 (0)=: (5) (a) Find the solution u(t) of this problem. Solution The characteristic equation for this problem is 5r +r+7=0: (6) Solving this equation, we get r =(,+ p 34i)=5 and r =(,, p 34i)=5. Thus, u = c exp[(,+ p 34i)t=5] + c exp[(,, p 34i)t=5]: (7) 4

5 Re-organizing equation (7), we have p 34it=5 u=(c e + c e,p 34it=5 )e,t=5 : (8) Plugging the initial conditions into equation (8) yields u(t) =e,t=5 [cos( p 34t=5)+7= p 34sin( p 34t=5)]: (9) (b) Find the smallest T such that ju(t)j 0: for all t>t. Solution For all t, ju(t)j MAX(cos( p 34t=5)+7= p 34sin( p 34t=5))e,t=5 : (0) where MAX(cos( p 34t=5)+7= p 34sin( p 34t=5))=:33: () Thus, the problem is to nd the smallest T, such that j:33e,t=5 j0:. Solving this equation, we have T= 5:74. Notice: this estimated T is larger than the smallest T that could be obtained from the plot or the calculator. Solution : Matlab see Fig. Problem 39, Page 53 Problem Use the hint for solving the Euler Equations to solve the following given equation t y 00 + ty 0 + y =0 () Solution First, we know that y 0 = t of x =lntinto equation () yields dy dx and y00 = ( dy t d x, dy dx ). Substitution y 00 + y =0: (3) 5

6 .5 Solution for Problem 4, Page 50.5 u t Figure : Solution for Problem 4, Page 50 6

7 The characteristic equation for the above homogeneous equation is where r = i. Thus, r +=0; (4) and t>0. y = c cosx + c sinx = c cos(ln t)+c sin(ln t); (5) Problem 3, Page 6 Problem Use the method of reduction of order to nd a second solution of the given dierential equation t y 00, 4ty 0 +6y=0; y (t)=t (6) Solution Assume y (t) =v(t)t, then y 0 = v 0 t +vt, and y 00 = v 00 t + 4v 0 t +v. Substitute these relations into equation (6), we have Then, v 00 =0: (7) Thus, the second solution is v = c t + c : (8) y (t) =t 3 : (9) 7

8 Problem, Page 77 Problem Use the method of variation of parameters to nd a particular solution of the given dierential equation. Then check your answer by using the method of undetermined coecients. y 00, 5y 0 +6y=e t (0) Solution : Variation of Parameters First, let's take a look of the homogeneous equation The general solution for this equation is y 00, 5y 0 +6y=0: () y(t) =c e t +c e 3t : () Then a particular solution for the nonhomogeneous equation will have the form Dierentiating equation (3) yields y = u (t)e t + u (t)e 3t : (3) y 0 = u 0 et +u (t)e t +u 0 e3t +3u (t)e 3t : (4) We set the term involving u 0 and u 0 equal to zero, i.e., u 0 et + u 0 e3t =0; (5) and y 0 =u (t)e t +3u (t)e 3t : (6) Then, dierentiating again, we get 8

9 y 00 =u 0 +4u et (t)e t +3u 0 +9u e3t (t)e 3t : (7) Plug equation (3) (6) and (7) into equation (0), we have u 0 +3u 0 et =e t : (8) e3t Solving equation (5) and (8), wehave Thus, the general solution is u =,e,t +c ; u =e,t +c : (9) and the particular solution is y(t)=c e t +c e 3t +e t (30) Y (t) =e t : (3) Solution :Undetermined coecients has the form e t,we assume Since the nonhomogeneous term Substitute Y(t) into equation (0, we get Y (t) =Ae t : (3) Then, A =. (A, 5A +6A)e t =e t : (33) 9