Solutions: Section 2.11. Problem 1: See the Maple worksheet to get the direction field. You should see that itlooks like all solutions are approaching some curve (maybe a line?) as t → ∞. To bemore analytic, let us solve the DE using the Method of Integrating Factors.y0+ 3y = t + e−2t⇒ e3t(y0+ 3y) = e3tt + e−2t⇒e3ty(t)0= te3t+ etIntegrate b oth sides Hint: We need to use “integration by parts” to integrate te3t.Using a table as in class:+ t e3t− 1 (1/3)e3t+ 0 (1/9)e3t⇒Zte3tdt =13e3t−19e3tPutting it all together,e3ty(t) =13te3t−19e3t+ et+ Cso thaty(t) =13t −19+1e−2t+Ce3tNotice that the last two terms go to zero as t → ∞, so we see that y(t) does approacha line:13t −19as t → ∞.2. Problem 3: See Maple for the direction field. Very similar situation to Problem 1.Let’s go ahead and solve:y0+ y = te−t+ 1Multiply both sides by eRp(t) dt= et:et(y0+ y) = t + et⇒ety(t)0= t + etIntegrate both sides:ety(t) =12t2+ et+ C ⇒ y(t) =12t2e−t+ 1 + Ce−tThis could be written as:y(t) = 1 +t22et+Cetso that it is clear that, as t → ∞, y(t) → 1, which we also see in the direction field.3. Problem 11: See Maple for the direction field, where it looks like all solutions areapproaching some periodic function as t → ∞. Let’s solve it:y0+ y = 5 sin(2t)As in the last exercise, multiply both sides by et:et(y0+ y) = 5etsin(2t) ⇒ety(t)0= 5etsin(2t)To integrate the right-hand-side of this equation, we will need to use integration byparts twice. In tabular form:+ etsin(2t)− et−(1/2) cos(2t)+ et−(1/4) sin(2t)⇒Zetsin(2t) dt = −12etcos(2t)+14etsin(2t)−14Zetsin(2t) dtAdd the last integral to the left:54Zetsin(2t) dt = −12etcos(2t) +14etsin(2t)so that:Zetsin(2t) dt = −25etcos(2t) +15etsin(2t) + C1Going back to the differential equation,ety(t) = −2etcos(2t) + etsin(2t) + C2so that the general solution is :y(t) = −2 cos(2t) + sin(2t) + C2e−tWe see that, as t → ∞, y(t) does indeed go to a periodic function.In problems 13, 15, 16, solve the IVP. For these problems, I will leave the detailsout, but I will give the integrating factor. Be sure to ask in class if you’re not sure howto solve them!4. Problem 13: (You’ll need to integrate by parts!)y0− y = 2te2teRp(t) dt= e−ty(t) = e2t(2t − 2) + 3et5. Problem 15:ty0+ 2y = t2− t + 1Be sure to put in standard form before solving:y0+2ty = t − 1 +1teRp(t) dt= t2andy(t) =14t2−13t +12+112t26. Problem 16: In this problem, the integrating factor is again t2:y0+2t· y =cos(t)t2⇒ y(t) =sin(t)t27. Problem 21: See the example Maple worksheet to get the direction field. To solve theIVP (with y(0) = a):y0= −12y = 2 cos(t)The integrating factor is: e−(1/2)t:e−(1/2)ty0= 2e−(1/2)tcos(t)Use integration by parts twice:+ cos(t) e−(1/2)t− − sin(t) −2e−(1/2)t+ − cos(t) 4e−(1/2)t⇒Ze−(1/2)tcos(t) dt = −2e−(1/2)tcos(t) + 4e−(1/2)tsin(t) − 4Ze−(1/2)tcos(t) dtAdd the last integral to both sides and divide by 5:Ze−(1/2)tcos(t) dt = −25e−(1/2)tcos(t) +45e−(1/2)tsin(t) + CGoing back to get the solution (be sure to multiply the antiderivative by 2:e−(1/2)ty = −45e−(1/2)tcos(t) +85e−(1/2)tsin(t) + CSo that:y(t) = −45cos(t) +85sin(t) + Ce(1/2)tSolve for the constant in terms of the initial condition y(0) = a:a = −45+ C ⇒ C = a +45The solution to the IVP is:y(t) = −45cos(t) +85sin(t) +a +45e(1/2)tIn particular, we see that if y(0) = a = −4/5, then the solution will be the periodic part(and will not become unbounded). Otherwise (because of the exponential function),all other solutions will become unbounded as t → ∞.8. Problem 24: See the Maple sample for the direction field.To solve the IVP, first write in standard form, then find the integrating factor:y0+t + 1ty = 2e−t, t > 0, y(1) = aThe integrating factor: First compute the antiderivative-Zt + 1tdt =Z1 +1tdt = t + ln(t), t > 0And exponentiate:eRp(t) dt= et+ln(t)= eteln(t)= tetNow,tety(t)0= 2t ⇒ tety(t) = t2+ Cso that the general solution is :y(t) =t2+ CtetSolve in terms of a:y(1) =1 + Ce= a ⇒ C = ae − 1so that:y(t) =t2+ (ae − 1)tetAnalysis: If the constant ae − 1 = 0, then y(t) becomes te−t, which is zero at timet = 0. Otherwise, all other solutions are not defined at time t = 0. The value of a isthen a = 1/e ≈ 0.3679. Furthermore, as t → 0, the solution will tend to zero (as doesall solutions).9. Problem 27: Solve the IVPy0+12y = 2 cos(t), y(0) = −1Using the integrating factor of e(1/2)t,e(1/2)ty(t)0= 2e(1/2)tcos(t)To integrate the right hand side of the equation, use integration by parts twice (sincewe’ve showed this a couple of times, I leave it out here):e(1/2)ty(t) =45e(1/2)tcos(t) +85e(1/2)tsin(t) + Cso that:y(t) =45cos(t) +85sin(t) + Ce−(1/2)tSolve for C:−1 =45+ C ⇒ C = −95and the solution to the IVP is:y(t) =45cos(t) +85sin(t) −95e−(1/2)tWe now want to find the coordinates of the first local maximum, t > 0. This meansthat we want to solve for the first t for which the derivative is zero. Unfortunately, wecannot do this exactly, so we can use Maple to find a numerical approximation. Hereis the Maple code to do this:DE27:=diff(y(t),t)+(1/2)*y(t)=2*cos(t);Y27:=dsolve({DE27,y(0)=-1},y(t));dy:=diff(rhs(Y27),t);plot(dy,t=0..3);tsol:=fsolve(dy=0,t=0..2);evalf(subs(t=tsol,rhs(Y27)));so the coordinates are approximately (1.3643, 0.8201).Notes about the Maple commands:• If you look at Y27, you’ll see that:Y27 := y(t) = 4/5*cos(t)+8/5*sin(t)-9/5*exp(-1/2*t)Therefore, to plot the function, we need the right hand side of Y27. In Maple,this is rhs(Y27).• To get the FIRST value of t, I need a rough estimate for the fsolve function.That’s why we plot the derivative first. You see in the fsolve line, t=0..2, whichis the estimate I got from the graph.10. Problem 29: Solve the IVP:y0+14y = 3 + 2 cos(2t) y(0) = 0To find the solution, we see that the integrating factor is e(1/4)t. Multiply both sidesby the I.F. and integrate. Note that again we’ll need to integrate by parts twice toevaluate:Ze(1/4)tcos(2t) dt =465e(1/4)tcos(2t) +3265e(1/4)tsin(2t)Therefore,e(1/4)ty(t)= 12e(1/4)t+865e(1/4)tcos(2t) +6465e(1/4)tsin(2t) + Cso that:y(t) = 12 +865cos(2t) +6465sin(2t) + Ce−(1/4)tSolve for C:0 = 12 +865+ C ⇒ C = −78865so that the overall solution is:y(t) = 12 +865cos(2t) +6465sin(2t) −78865e−(1/4)tAs t → ∞, the last term (with the exponential) drops out, leaving the rest. Thatmeans the solution will