Brian Bowers (TA for Hui Sun) MATH 20D Homework Assignment 0 October 1, 2013

20D - Homework Assignment 0 1.3 #1-6 Determine the order of the given differential equation; also state whether the equation is linear or nonlinear. 2

(1) t2 ddt2y + t dy dt + 2y = sin t 2

t (2) (1 + y 2 ) ddt2y + t dy dt + y = e

(3)

d4 y dt4

(4)

dy dt

(5)

d2 y dt2

(6)

3

+

d3 y dt3

+

d2 y dt2

+

dy dt

+y =1

+ ty 2 = 0

d y dt3

+ sin(t + y) = sin t 2 3 + t dy dt + (cos t)y = t

(1) The highest derivative is a second derivative, so the order is 2. The equation matches the form of equation (11) on page 21, so it is linear. 2

(2) The highest derivative is a second derivative, so the order is 2. The term ”(1 + y 2 ) ddt2y ” means that this is a nonlinear equation since (1 + y 2 ) is not a function of t. (3) The highest derivative is a fourth derivative, so the order is 4. The equation matches the form of equation (11) on page 21, so it is linear. (4) The highest derivative is a first derivative, so the order is 1. The term ”ty 2 ” means that this is a nonlinear equation since ty is not a function of t. (5) The highest derivative is a second derivative, so the order is 2. The term ”sin(t + y)” means that this is a nonlinear equation since sin(t + y) is not a function of t. (6) The highest derivative is a second derivative, so the order is 3. The equation matches the form of equation (11) on page 21, so it is linear.

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1.3 # 7,11 Verify that each given function is a solution of the differential equation. (7) y 00 − y = 0;

y1 (t) = et , y2 (t) = cosh t

(11) 2t2 y 00 + 3ty 0 − y = 0, t > 0;

y1 (y) = t1/2 , y2 (t) = t−1

To verify that a function is a solution to a differential equation, all we need to do is plug it into the differential equation and check that the equation is true. (7) First, we plug in y1 for y: y100 − y1 = (et )00 − (et ) = (et )0 − (et ) = et − et =0

Next, we plug in y2 for y. This function involves cosh, a hyperbolic trig function. All we need to know for the purposes of this course is that (sinh)0 = cosh and (cosh)0 = sinh. y200 − y2 = (cosh t)00 − cosh t = (sinh t)0 − cosh t = cosh t − cosh t =0

Thus, y1 and y2 are solutions to the differential equation. (11) First, we plug in y1 for y: 2t2 y 00 + 3ty 0 − y = 2t2 (t1/2 )00 + 3t(t1/2 )0 − (t1/2 ) 0    1 −1/2 1 −1/2 t + 3t t − t1/2 = 2t2 2 2     −1 −3/2 1 −1/2 = 2t2 t + 3t t − t1/2 4 2 3 1 = − t1/2 + t1/2 − t1/2 2 2 =0

Next, we plug in y2 for y. 2t2 y 00 + 3ty 0 − y = 2t2 (t−1 )00 + 3t(t−1 )0 − (t−1 ) = 2t2 (−t−2 )0 + 3t(−t−2 ) − t−1 = 2t2 (2t−3 ) + 3t(−t−2 ) − t−1 = 4t−1 − 3t1 − t−1 =0

Thus, y1 and y2 are solutions to the differential equation.

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1.3 #15,17 Determine the values of r for which the given differential equation has solutions of the form y = ert . (15) y 0 + 2y = 0 (17) y 00 + y 0 − 6y = 0 (15) We plug y = ert into the differential equation and solve: y 0 + 2y = (ert )0 + 2(ert ) = rert + 2ert = (r + 2)ert

But we know from the original differential equation that this is equal to 0, i.e. (r + 2)ert = 0. We set each factor equal to zero and solve. (r + 2) = 0 =⇒ r = −2 ert = 0 =⇒ no solution

So, our only solution is r = −2. (17) We plug y = ert into the differential equation and solve: y 00 + y 0 − 6y = (ert )00 + (ert )0 − 6(ert ) = (rert )0 + (rert ) − 6(ert ) = (r2 ert ) + (rert ) − 6(ert ) = (r2 + r − 6)ert

But we know from the original differential equation that this is equal to 0, i.e. (r2 + r − 6)ert = 0. We set each factor equal to zero and solve. r2 + r − 6 = 0 =⇒ (r + 3)(r − 2) = 0 =⇒ r = −3, 2 ert = 0 =⇒ no solution

So, our only solutions are r = −3 and r = 2.

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1.3 #19 Determine the values of r for which the given differntial equation has solutions of the form y = tr for t > 0. t2 y 00 + 4ty 0 + 2y = 0 We’ll plug y = tr into the equation and simplify: t2 y 00 + 4ty 0 + 2y = t2 (tr )00 + 4t(tr )0 + 2(tr ) = t2 (rtr−1 )0 + 4t(rtr−1 ) + 2(tr ) = t2 (r(r − 1)tr−2 ) + 4t(rtr−1 ) + 2(tr ) = t2 r2 tr−2 − t2 rtr−2 + 4trtr−1 + 2tr = r2 tr − rtr + 4rtr + 2tr = (r2 + 3r + 2)tr

But we know from the original differential equation that this is equal to 0, i.e. (r2 + 3r + 2)tr = 0. We se each factor equal to zero and solve. r2 + 3r + 2 = (r + 2)(r + 1) = 0 =⇒ r = −2, −1 tr = 0 =⇒ no solution [since t > 0]

Thus, the only solutions are r = −2 and r = −1.

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2.1 #1c,3c,4c,6c For each problem, find the general solution of the given differential equation, and use it to determine how solutions behave as t → ∞. (1c) y 0 + 3y = t + e−2t (3c) y 0 + y = te−t + 1 (4c) y 0 + (1/t)y = 3 cos 2t (6c) ty 0 + 2y = sin t Before we get into specific problems, I’ll give a brief overview of two important and useful concepts. Method of Integrating Factors: First, I will review the Method of Integrating Factors. We only use this method when we have an equation of the form y 0 + p(t)y = g(t). The goal is to multiply the entire equation by some unknown function µ and then figure out what µ should be to make the left side of the equation equal to (µy)0 . We can recall the product rule to see that (µy)0 = µy 0 + µ0 y. Multiplying our original equation by µ, we get µy 0 + p(t)yµ = µg(t). We see that the µy 0 matches up with the equation above, but we still need to take care of the µ0 y portion. Thus, we conclude that µ0 y = p(t)yµ µ0 = p(t) [separating variables] µ Z =⇒ ln µ = p(t)dt + C [integrating] Z =⇒ ln µ = p(t)dt [get rid of C! We want any old integrating factor that works.] =⇒

=⇒ µ = e

R

p(t)dt

[exponentiating]

SHORTCUT: So, if you’d prefer to skip all the above steps, you can just always use µ = e save yourself some effort.

R

p(t)dt

and

Tabular Integration: You CAN complete this course without ever using this tool, but it makes certain types of integration MUCH faster, and I hope you’ll consider learning and using it. Tabular integration is essentially a specialized shortcut for integration by parts. We use tabular integration when we want to integrate an expression with the form (polynomial)(cyclic function). What do I mean by cyclic function? I mean any function that you can integrate some number of times until you end up with a constant times the original function. For our purposes, cyclic functions are sin(at + b), cos(at + b), eat+b . Here are the steps we do every time: (1) Set up a two-column table (hence the name ”tabular”). (2) We put the polynomial on the top of the left column and the cyclic function on the top of the right column. 5

(3) For each subsequent row of the table, we differentiate the left side and integrate the right side. We stop making new rows once the left side is equal to 0. (4) Starting with the top-left entry, draw an arrow that moves right one entry and down one entry, repeating until you run out of entries on the right side of the table. Label these arrows alternating ”+,-,+,-,...”. (5) Multiply along each arrow (including multiplying by +1 for ”+” and (−1) for ” − ”. Add these products together - this is the answer. (Don’t forget to add ”+C”.) For example, let’s say we wanted to integrate t4 e2 t. With standard integration by parts techniques, this would take a VERY long time. Instead, let’s use our new method.

So, we should get the integral to be           1 2t 1 2t 1 2t 1 2t 1 2t 4 3 2 t (+1) e + 4t (−1) e + 12t (+1) e + 24t(−1) e + 24(+1) e 2 4 8 16 32 1 3 3 3 = t4 e2t − t3 e2t + t2 e2t − te2t + e2t 2 2 2 4 Now let’s do the problems. (1c) The left side of the equation is y 0 + 3y, which corresponds to y 0 + R p(t)y in our general form. Thus, p(t) = 3, which means that our integrating factor should be µ = e p(t)dt = e3t . So, we multiply both sides of the equation by this factor. e3t (y 0 + 3y) = e3t (t + e−2t ) e3t y 0 + 3e3t y = te3t + et Recall that the goal of the integrating factor is to make the left side of the equation into (µy)0 . We can check that this is the case. Thus, we have (e3t y)0 = te3t + et 1 1 =⇒ e3t y = te3t − e3t + et + C [integrating using the tabular integration below] 3 9 1 1 =⇒ y = t − + e−2t + Ce−3t [dividing by e3t ] 3 9 Below is the work for the tabular integration we did above:

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Moreover, as t gets bigger, we see that the 31 t term grows, − 19 , stays constant, and all other terms go to 0. Thus, the solutions will be asymptotic to 13 t − 91 as t → ∞. (3c) The left side of the equation is y 0 + y, which corresponds to y 0 +Rp(t)y in our general form. Thus, p(t) = 1, which means that our integrating factor should be µ = e p(t)dt = et . So, we multiply both sides of the equation by this factor. et (y 0 + y) = et (te−t + 1) et y 0 + et y = t + et

Recall that the goal of the integrating factor is to make the left side of the equation into (µy)0 . We can check that this is the case. Thus, we have (et y)0 = t + et t2 + et + C [integrating] 2 t2 e−t + 1 + Ce−t [dividing by et ] =⇒ y = 2

=⇒ et y =

Moreover, as t gets bigger, every term except 1 goes to 0. Thus, y approaches 1 as t → ∞. (4c) The left side of the equation is y 0 + (1/t)y, which corresponds to y 0 R+ p(t)y in our general form. Thus, p(t) = 1/t, which means that our integrating factor should be µ = e (1/t)dt = eln t = t. So, we multiply both sides of the equation by this factor. t(y 0 + (1/t)y) = t(3 cos 2t) ty 0 + y = 3t cos 2t

Recall that the goal of the integrating factor is to make the left side of the equation into (µy)0 . We can check that this is the case. Thus, we have (ty)0 = 3t cos 2t 3 3 =⇒ ty = t sin 2t + cos 2t + C [integrating using the tabular integration below] 2 4 3 3 C [dividing by t] =⇒ y = sin 2t + cos 2t + 2 4t t Below is the work for the tabular integration we did above:

Moreover, as t gets bigger, the second and third terms go to zero. Thus, y approaches t → ∞.

3 2

sin 2t as

(6c) First, we need to get our equation into the right form (so that it starts with y 0 , not ty 0 ). So, we divide both sides of the equation by t, giving us y 0 + (2/t)y = sint t . The left side of the equation is y 0 + (2/t)y, which corresponds to y 0 +p(t)y in our general form. Thus, p(t) = 2/t, which means that our integrating 7

factor should be µ = e factor.

R

(2/t)dt

2

= e2 ln t = eln t = t2 . So, we multiply both sides of the equation by this t2 (y 0 + (2/t)y) = t2 (sin t/t) t2 y 0 + 2ty = t sin t

Recall that the goal of the integrating factor is to make the left side of the equation into (µy)0 . We can check that this is the case. Thus, we have (t2 y)0 = t2 sin t =⇒ t2 y = −t cos t + sin t + C [integrating using the tabular integration below] − cos t sin t C + 2 + 2 [dividing by t2 ] =⇒ y = t t t

Below is the work for the tabular integration we did above:

Moreover, as t all of the terms go to zero. Thus, y approaches 0 as t → ∞.

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2.1 #14,15,20 For each problem, find the solution of the given initial value problem. (14) y 0 + 2y = te−2t ,

y(1) = 0

(15) ty 0 + 2y = t2 − t + 1, (20) ty 0 + (t + 1)y = t,

y(1) = 1/2, t > 0 y(ln 2) = 1, t > 0

For a general overview of the Method of Integrating Factors or Tabular Integration, see the introduction to 2.1 #1c. (14) The left side of the equation is y 0 + 2y, which corresponds to y 0 + R p(t)y in our general form. Thus, p(t) = 2, which means that our integrating factor should be µ = e p(t)dt = e2t . So, we multiply both sides of the equation by this factor. e2t (y 0 + 2y) = e2t (te−2t ) e2t y 0 + 2e2t y = t Recall that the goal of the integrating factor is to make the left side of the equation into (µy)0 . We can check that this is the case. Thus, we have (e2t y)0 = t t2 + C [integrating] 2 t2 C =⇒ y = 2t + 2t [dividing by e2t ] 2e e

=⇒ e2t y =

Next, we use the initial condition by plugging in t = 1 and setting equal to 0: C 1 (1)2 C 1 + 2C + 2(1) = 2 + 2 = = 0, 2e e 2e2 2e2(1) e which we can solve to see that 1 + 2C = 0, so C = −1/2. Thus, we plug C back into our above solution for y to get t2 − 1 y= . 2e2t (15) First, we need to get our equation into the right form (so that it starts with y 0 , not ty 0 ). So, we divide both sides of the equation by t, giving us y 0 + (2/t)y = t − 1 + 1t . The left side of the equation is y 0 + (2/t)y, which corresponds to y 0 + Rp(t)y in our general form. Thus, p(t) = 2/t, which means that R R 2 our integrating factor should be µ = e p(t)dt = e 2 ln tdt = e ln t = t2 . So, we multiply both sides of the equation by this factor.   1 2 0 2 t (y + (2/t)y) = t t − 1 + t y(1) =

t2 y 0 + 2ty = t3 − t2 + t Recall that the goal of the integrating factor is to make the left side of the equation into (µy)0 . We can check that this is the case. Thus, we have (t2 y)0 = t3 − t2 + t t4 t3 t2 − + + C [integrating] 4 3 2 t2 t 1 C =⇒ y = − + + 2 [dividing by t2 ] 4 3 2 t

=⇒ t2 y =

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Next, we use the initial condition by plugging in t = 1 and setting equal to 1/2: y(1) =

3 − 4 + 6 + 12C 12 1 1 C 1 − + + 2 = = , 4 3 2 1 12 2

which we can solve to see that C = 1/12. Thus, we plug C back into our above solution for y to get y=

t2 t 1 1 . − + + 4 3 2 12t2

(20) First, we need to get our equation into the right form (so
that it starts with y 0 , not ty 0 ). So, we divide both
sides of the equation by t, giving us y 0 + 1 + 1t y = 1. The left side of the equation is y 0 + 1 + 1t y, which corresponds to y 0 + p(t)y in our general form. Thus, p(t) = 1 + 1t , which means R R 1 that our integrating factor should be µ = e p(t)dt = e (1+ t )dt = et+ln t = et eln t = tet . So, we multiply both sides of the equation by this factor.   1 te (y + 1 + y) = tet (1) t t

0

tet y 0 + tet y + et y = tet

Recall that the goal of the integrating factor is to make the left side of the equation into (µy)0 . We can check that this is the case. Thus, we have (tet y)0 = tet =⇒ tet y = tet − et + C [integrating using the tabular integration below] C 1 =⇒ y = 1 − + t [dividing by tet ] t te

Below is the work for the tabular integration we did above:

Next, we use the initial condition by plugging in t = ln 2 and setting equal to 1: y(ln 2) = 1 −

1 C C 2 ln 2 − 2 + C 1 + + = = 1, =1− ln 2 ln 2 ln 2e ln 2 2 ln 2 2 ln 2

which we can solve to see that C = 2. Thus, we plug C back into our above solution for y to get y =1−

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1 2 + t. t te

2.1 #31 Consider the initial value problem 3 y 0 − y = 3t + 2et , 2

y(0) = y0 .

Find the value of y0 that separates solutions that grow positively as t → ∞ from those that grow negatively. How does the solution that corresponds to this criticial value of y0 behave as t → ∞? Let’s solve the differential equation in general and then see what we can do about the initial condition question. The left side of the equation is y 0 − 32 y, which corresponds to y 0 + p(t)y in our general form. Thus, R p(t) = − 23 , which means that our integrating factor should be µ = e p(t)dt = e−3t/2 . So, we multiply both sides of the equation by this factor.   3 e−3t/2 y 0 − y = e−3t/2 (3t + 2et ) 2 e−3t/2 y 0 −

3e−3t/2 y = 3te−3t/2 + 2e−t/2 2

Recall that the goal of the integrating factor is to make the left side of the equation into (µy)0 . We can check that this is the case. Thus, we have (e−3/2t y)0 = 3te−3t/2 + 2e−t/2 =⇒ e−3t/2 y = −2te−3t/2 − =⇒ y = −2t −

4e−3t/2 − 4e−t/2 + C [integrating using the tabular integration below] 3

4 − 4et + Ce3t/2 [dividing by e−3t/2 ] 3

Below is the work for the tabular integration we did above:

Next, we use the initial condition by plugging in t = 0 and setting equal to y0 : y(0) = −2(0) −

4 4 − 4e(0) + Ce3(0)/2 = 0 − − 4 + C = y0 , 3 3

which we can solve to see that C = y0 + 16/3. Thus, we plug C back into our above solution for y to get y = −2t −

4 − 4e−t/2 + (y0 + 16/3)e3t/2 . 3

The problem asks us to find the value of y0 that splits the solutions between growing positively and growing negatively. Looking at our solution, we see that (y0 + 16/3)e3t/2 is the biggest term. This is positive when (y0 +16/3) is positive and negative when (y0 +16/3) is negative. Thus, the critical value is when y0 +16/3 = 0, or y0 = −16/3. Next, we examine the case when y0 = −4/3. In this case, we see that y = −2t − 34 − 4e−t/2 . In this case, we see that the last term goes to 0 as t gets bigger, so we determine that y becomes asymptotic to −2t − 34 (and goes to −∞) as t → ∞.

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