laplace formulae

8/13/2019 LAPLACE Formulae

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CHAPTER

5

LAPLACE TRANSFORMS

5.1 Introduction and Definition

In this section we introduce the notion of the Laplace transform. We will usethis idea to solve differential equations, but the method also can be used to

sum series or compute integrals. We begin with the definition:

Laplace TransformLetf(t) be a function whose domain includes (0, ) then the Laplace trans-form off(t) is:

L(f(t))(s) =

0

f(t)est dt

Note that the Laplace transform of a function is another function whosevariable is usually denoted as s. Also note that the Laplace transform involvesan improper integral. We compute the Laplace transform of several functions:

Example 5.1 Compute the Laplace transform off(t) = 1

127


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128 CHAPTER 5. LAPLACE TRANSFORMS

Solution:

L(1)(s) =

0

1est dt= limc

1

s

estc

0

In order for this limit to exist, we must insist that s= 0 and that s >0 sothatesc has a limit (of zero). When s >0, we obtain

1

s limc

(esc 1) =1

s

So

L(1)(s) =1

s; s >0.

Example 5.2 Compute the Laplace transform off(t) = t

Solution:

L(t)(s) =

0

test dt

We integrate by Parts (letting u= t and dv = est dt) to obtain: test dt=

1

stest

1

s2est,

so

0

test dt= limc

1

stest

1

s2est

c

0In order for this limit to exist, we again must insist that s = 0 and thats >0so that esc has a limit (of zero). We obtain

1

s limc

(cesc 0) 1

s2 limc

(esc 1)

which exists for s >0 and after LHopitals rule yields

L(t)(s) = 1

s2; s >0.

The previous example can be upgraded to find the Laplace transform off(t) =tn for any positive integer n.


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5.1. INTRODUCTION AND DEFINITION 129

Example 5.3 Show that iff(t) = tn, for any positive integern then

L(tn

)(s) =n

s L(tn1

)(s)

Solution:

L(tn)(s) =

0

tnest dt

We integrate by Parts (letting u= tn and dv = est dt) to obtain:

tnest dt=

1

stnest +

1

s

ntn1est dt,

so

0

tnest dt= limc

1s

tnest c

0

+n

sL(tn1)(s).

In order for this limit to exist, we again must insist that s = 0 and thats >0so thatesc has a limit (of zero). Using LHopitals rulen times, we see that

limc

1

scnesc

= 0.

Using this result inductively, we compute:

L(t2

)(s) =

2

s L(t)(s) =

2

s

1

s2 =

2

s3

L(t3)(s) =3

sL(t2)(s) =

3

s

2

s

1

s2=

6

s4

L(t4)(s) =4

sL(t3)(s) =

4

s

3

s

2

s

1

s2 =

24

s5

Note that in all cases above, we must have s >0. In summary,

Laplace Transform of single term monic polynomials

L(tn)(s) = n!sn+1

; s >0


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Linear Combinations and Laplace Transform

Theorem 5.4 Letf(t) andg(t) be functions with Laplace transformsF(s)andG(s) respectively then for any constantsa andb

L(af(t) +bg(t))(s) =aF(s) +bG(s)

Proof:

L(af(t) +bg(t))(s) =

0

[af(t) +bg(t)]est dt

=

0

af(t)est dt+

0

bg(t)est dt= aF(s) +bG(s)

From this result we can take the Laplace transform of any arbitrary poly-nomial in the next example,

Example 5.5 FindL(4t3 + 8t2 7)(s)

Solution:

L(4t3 + 8t2 7)(s) = 4L(t3)(s) + 8L(t2)(s) 7L(1)(s)

= 4

6

s4

+ 8

2

s3

7

1

s

=24

s4 +

16

s3

7

s,

wheres >0.

Laplace Transform of exponential functions

L(eat)(s) = 1s a

; s > a


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Proof:

L(eat)(s) =

0

eatest dt

=

0

e(as)t dt,

which can be integrated with respect to t (use u = (a s)t and du= (a s) dt). We obtain

= limb

1

a se(as)b

1

a s

The above limit exists exactly when s > a,so

L(e

at

)(s) =

1

s a , s > a

Example 5.6 Find

L(2t)(s)

Solution: Rewriting 2t =eln2t

=etln2 and taking a = ln 2,

L(e(ln2)t

)(s) =

1

s ln 2 , s >ln 2

Lastly,

Laplace Transform of sine and cosine

L(cos(t))(s) = s

s2 +2, s >0

L(sin(t))(s) = s2 +2

, s >0


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We derive the second formula and leave the derivation of the first formulaas an exercise. By definition:

L(sin(t))(s) =

0

est sin(t) dt.

Using integration by parts with u = sin(t) anddv= est dtwe obtain

0

est sin(t) dt = 1

ssin(t)est

0

0

(

s)est cos(t) dt.

Using parts again with u = cos(t) and dv = est dt,we obtain

= 1

ssin(t)est

0

+ (

s)

(

1

s )est cos(t)

0

0

(

s)est sin(t) dt

From the squeeze theorem we see that for s > 0 limc

esc cos(c) = 0 and

limc

esc sin(c) = 0, so the above reduces to

= (

s2)

2

s2

0

(est sin(t) dt

Now recall that the left-hand side of this equation (what we were solvingfor) is

0

est sin(t) dt

so we have

0

est sin(t) dt =

s2

2s2

0

est sin(t) dt

So moving the term with the integral on the right side to the left sidegives

0

est sin(t) dt+

2

s2

0

est sin(t) dt = (

s2)

Factoring, 1 +

2

s2

0

est sin(t) dt = (

s2)

So 0

est sin(t) dt = (

s2 )1 +

2

s2

= s2 +2


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NOTE: It is customary to use the independent variable s for a functionthat is an output of a Laplace transform and the independent variable t for

a function that is an output of a Laplace transform. This convention will behandy in the later sections.

The below result gives a condition that guarantees the existence of theLaplace transform.

A condition that guarantees the existence ofL[f(t)](s)Suppose that there are numbers and Mso that

|f(t)| M et

for all t >0. Then the Laplace transform off(t) exists with a domain of at

least s > .

Proof: Suppose that there are numbers and Mso that

|f(t)| M et

for all t >0. Then

L(|f(t)|)(s) = limb

b0

est|f(t)| dt

0

estM et dt= ML[et](s) = M

s

,

fors > .Therefore, since b

0

est|f(t)| dt

is increasing in b and bounded for all b,

limb

b0

est|f(t)| dt

exists. This implies that

limb

b

0

estf(t) dt


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also must exist.

Note that most all exponential functions, polynomials, and the trig func-tions sine and cosine satisfy this condition but ln x, tan x and et

2do not.

These functions do not have Laplace transforms.

Exercises

1. ComputeL[f(t)](s) for f(t) = 0.

2. ComputeL[f(t)](s) for

f(t) =

1 if t< 80 if t 8.

3. ComputeL[f(t)](s) for

f(t) =

t if t< 12 t if 1 t < 20 if t 2.

4. Compute the formula forL[cosh(t)](s), where is a constant.

5. ComputeL[sinh(t)](s),

6. Find the Laplace transform of

f(t) =

t 8 if t< 8et+6 if 8< t < 10t2 if 10< t < 110 if t> 11.

In 7-12, take the Laplace transform of the functionf(t)using the resultsin this section (do not derive using the definition)

7. f(t) =t3 7t2 + 8

8. f(t) = 4 sin t 3cos t

9. f(t) = cos(2t) e9t

10. f(t) = 1 + 7 sin(5t)


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11. f(t) = cosh(t),where is a constant. (Recall cosh(z) = ez+ez

2 ).

12. f(t) = sinh(t),where is a constant. (Recall sinh(z) = ezez

2 ).

In 13-15, use the definition to take the Laplace transform of the functionf(t) using integration by parts and formulas given in this section

13. f(t) =tet

14. f(t) =t sin(2t)

15. f(t) =t2 cos(t)

16. Recall thatL[f(t) + g(t)] = L[f(t)] + L[g(t)]. Is it true in general thatL[f(t) g(t)] = L[f(t)] L[g(t)]?

(a) Try f(t) =t and g(t) = 1 to see ifL[f(t) g(t)] = L[f(t)] L[g(t)].

(b) Is it ever true that L[f(t) g(t)] = L[f(t)] L[g(t)]?


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5.2 Laplace Transforms, The Inverse Laplace

Transform, and ODEs

In this section we will see how the Laplace transform can be used tosolve differential equations. The key result that allows us to do this isthe following:

Laplace Transform ofy (t)Suppose that L[y(t)](s) exists and that y(t) is differentiable (0, ) withderivative y (t) then

L[y(t)](s) =sL[y(t)](s) y(0)

Proof: By definition,

L[y(t)](s) =

0

esty(t) dt

Setting u = est and dv = y (t) dt and using integration by parts, weobtain

L[y(t)](s) = esty(t)0

0

(s)esty(t) dt

For s big enough, limc escy(c) = 0, since the Laplace transform of

y(t) exists (this follows since the continuous integrand of a convergent

improper integral must tend to zero we will not prove this fact here.)So we obtain

L[y(t)](s) = y(0) +sL[y(t)](s)

From this result, we derive:

Laplace Transform ofy (t)Suppose thatL[y(t)](s) exists and that y (t) is twice differentiable on (0, )with second derivative y(t) then

L[y(t)](s) = s2L[y(t)](s) sy(0) y(0)


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5.2. LAPLACE TRANSFORMS, THE INVERSE LAPLACE TRANSFORM, AND ODES137

Proof: We simply us the previous result twice

L[y

(t)](s) =sL[y

(t)](s) y

(0) =s [sL[y(t)](s) y(0)] y

(0).

Example 5.7 Solvey(t) =t using the Laplace Transform

Solution: Taking the Laplace transform of both sides we obtain

L[y(t)](s) = L[t]

L[y(t)](s) = 1

s2

so

s2L[y(t)](s) sy(0) y(0) = 1

s2

Solving for L[y(t)](s) we obtain

L(y(t))(s) = 1

s4+

y (0)

s +

y (0)

s2

Now if we could reverse engineer which function y(t) has Laplace trans-form equal to the right hand side, then we would be done.

By playing a bit, we can see that y(t) = 16t3 + y(0)2 +y

(0)t has thisLaplace transform. So voila! We have solved the differential equation.

(Note that the general solution obtained from undetermined coefficientswould be y(t) = 16t

3 +C1+ C2t which agrees with our solution sincey(0) andy(0) are unspecified constants.)

The above example illustrates a common checklist for solving DEs usingthe Laplace transform:

1. Transform the original problem to one involvingL(y(t))(s),


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2. Solve for L[y(t)](s),

3. Undo the Laplace transform to recovery(t).

Definition 5.8 A continuous function f(t) is the Inverse LaplaceTransform of a function F(s) if L[f(t)](s) = F(s). In this case, wewritef(t) =L1[F(s)]

It is customary shorthand notation to denote F(s) to be the Laplacetransform of f(t) and G(s) to be the Laplace transform of g(t). Forconstantsa, b, since

L[af(t) +bg(t)] = aL[f(t)] +bL[g(t] =aF(s) +bG(s)

we see thatL1[aF(s) +bG(s)] = af(t) +bg(t)

orL1[aF(s) +bG(s)] = aL1[F(s)] +bL1[G(s)],

so the Laplace transform of a linear combination is the linear combi-nation of the Laplace transforms.

The following table lists several inverses which come from the previoussection:

Some Basic Inverse Laplace Transform Facts

F(s) L1[F(s)]1s

1n!

sn+1 tn

1sa

eat

ss2+2 cos t

s2+2 sin t

aF(s) +bG(s) af(t) +bg(t)

Example 5.9 FindL17s3


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Solution:

L1 7

s3= 7L1

1

s3

We multiply the inner fraction by 22

to obtain the form in the previoustable.

= 7L1

2

2s3

=

7

2L1

2

s3

=7

2t2

Example 5.10 FindL12s1s2+4

Solution:

L1

2s 1

s2 + 4

= 2L1

s

s2 + 4

L1

1

s2 + 4

Again, the inner fraction of the second must be multiplied by 22 to putit in the form in the table (= 2 since 2 = 4).

= 2cos 2t L1

2

2(s2 + 4)

= 2cos 2t 1

2L1

2

s2 + 4

= 2cos 2t 1

2sin2t

Example 5.11 Solvey(t) + 9y(t) = e4t withy(0) = 1 andy(0) = 0


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Solution: We first take the Laplace transform of both sides to obtain

L[y

(t) + 9y(t)] = L[e4t

]

s2L[y(t)] sy(0) y(0) + 9L[y(t)] = 1

s 4

(s2 + 9)L[y(t)] = 1

s 4+ s

L[y(t)] = 1

(s 4)(s2 + 9)+

s

s2 + 9

We use partial fractions to decompose the first term into a sum of termsthat are recognizable inverses.

Noting that by partial fractions:

1

(s 4)(s2 + 9)=

A

s 4+

B s+C

s2 + 9 ,

where A = 125 B= 125

C= 425 .Substituting in, we obtain

L[y(t)] = 1

25

1

s 4

1

25

s

s2 + 9

4

25

1

s2 + 9

+

s

s2 + 9

L[y(t)] = 1

25

1

s 4

+

24

25

s

s2 + 9

4

25

1

s2 + 9

So

y(t) = 1

25e4t +

24

25cos3t

4

75sin 3t

(note that the 3 appeared in the demoninator since = 3 and in order

to findL1

1

s2 + 9

we multiply the inner fraction by 3

3.)

We end this section by noting the following extension


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Laplace Transform ofy (n)(t)

Suppose that L[y(t)](s) exists and thaty(t) is differentiablentimes on (0, )withnth derivative y (n)(t) then

L[y(n)(t)](s) = snL[y(t)](s)sn1y(0) sn2y(0) ... sy(n2)(0) y(n1)(0)

Exercises

In 1-7, find the Inverse Laplace transform of the function F(s).You may need partial fractions

1. F(s) = 1s+2

2. F(s) = 1s4

3. F(s) = s+5s2+16

4. F(s) = 4s+6

5. F(s) = 1(s+1)(s2+1)

6. F(s) = s(s+1)(s2+1)

7. F(s) = 3s3+s

8. Use the Laplace transform to find the general solution of

y + 6y + 5y= t


y y = et


y +y = et

11. Use the Laplace transform to find the solution of the IVP

y +y = 2, y(0) = 1, y(0) =1

12. Use the Laplace transform to find the solution of the IVP

y + 2y +y = 2, y(2) = 1, y(2) =1

(Hint: find the general solution first).


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5.3 Advanced Properties of the Laplace

Transform

In this section, we look at several theorems which can be used to solveDEs using the Laplace transform method. We start with:

Laplace Transform Exponential Shift Theorem (Forward)

L[eatf(t)](s) = L[f(t)](s a) =F(s a)

Effectively, this saysL[eatf(t)](s) is equal to L[f(t)](w), (using w for the

traditionals in the definition) and replacing w = s a.

Proof: By definition:

L[eatf(t)] =

0

eatestf(t) dt

=

0

e(sa)tf(t) dt.

Also by definition,

L[f(t)](s a) =

0

e(sa)tf(t) dt.

We can use this result to compute Laplace transforms of exponentialsmultiplied by functions whose Laplace transforms we already know.

Example 5.12 FindL[e2t cos(3t)](s)

Solution: By the shifting theorem

L[e2t cos(3t)](s) = L[cos(3t)](s (2)) =L[cos(3t)](s+ 2) =


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5.3. ADVANCED PROPERTIES OF THE LAPLACE TRANSFORM 143

The Laplace transform of cos(3t) with (different) variable w is

L[cos(3t)](w) = w

w2 + 32 ,

so, replacing w with s + 2,

=L[cos(3t)](s+ 2) = s+ 2

(s+ 2)2 + 9.

= s+ 2

s2 + 4s+ 13.

Further note that in order for L[cos(3t)](w) to exist, w > 0 so the

domain is s + 2> 0 or s >2.

Example 5.13 FindL[t3et](s)

Solution: By the shifting theorem

L[t3et](s) =L[t3](w) = 6

w4,

wherew = s 1,so

L[t3et](s) = 6

(s 1)4

Similarly, we obtain the following result:

Laplace Transform Exponential Shift Theorem (Backward)

L1[F(s a)]) =eatf(t),

whereF(w) is the Laplace transform off(t) (with variable w).


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To apply this result, we look for familiar outputs of Laplace transformswith the variable s replaced by a shift.

Example 5.14 FindL1

24(s8)5

Solution: This is just a shift of

24

w5

which is the Laplace transform of t4 (with variable w0. So 24(s8)5

=

F(s 8) where Fis the Laplace transform off(t) =t4. Therefore, bythe Exponential Shifting Theorem

L1[F(s 8)] =e8tt4.

Example 5.15 FindL1

ss2+2s+5

Solution: The trick here is to complete the square in the denominatorand then use the backward shift theorem.

s

s2 + 2s+ 5 =

s

s2 + 2s+ 1 + 4 =

s

(s+ 1)2 + 4

We also need an s +1 in the numerator to use the backwards shift withcosine, so we write the numerator as:

= s+ 1 1

(s+ 1)2 + 4=

s+ 1

(s+ 1)2 + 4

1

(s+ 1)2 + 4

Now we can take the inverse of each term using the backward shifttheorem to get

L1

s

s2 + 2s+ 5

= L1

s+ 1

(s+ 1)2 + 4

L1

1

(s+ 1)2 + 4


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=et cos2t 1

2

et sin2t

The next result equates differentiation of the Laplace transform Fwithrespect to s to the Laplace transform of the product tf(t).

Differentiation TheoremLet F(s) be the Laplace transform off(t). Then

d

ds [F(s)] = L[tf(t)](s)

or

d

ds[F(s)] = L[tf(t)](s)

Proof: By definition:

F(s) = lims0

F(s+ s) F(s)

s

Note that

F(s+s)F(s) =

0

e(s+s)tf(t) dt

0

estf(t) dt=

0

[e(s+s)test]f(t) dt

SoF(s+ s) F(s)

s =

0

e(s+s)t est

s f(t) dt

Therefore

lims0

F(s+ s) F(s)s

=

0

lims0

e(s+s)t est

s f(t) dt


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Noting that lims0e(s+s)test

s is the definition of d

ds(est),we know

this limit is test.

Substituting, we obtain

F(s) =

0

(test)f(t) dt.

By definition, the right hand side isL[tf(t)](s)

We provide several examples which illustrate how to use this result.

Example 5.16 FindL [te2t]

By the Theorem,L[tf(t)](s) = F(s).

Here f(t) = e2t, so F(s) = 1s2 .

NowF(s) = (s 2)2

so

L[te2t](s) =((s 2)2) = 1

(s 2)2

Note that we could have also obtained this answer by using the ForwardExponential Shift Theorem.

Example 5.17 FindL [t cos(3t)]

By the Theorem,L[tf(t)](s) = F(s).

Here f(t) = cos(3t), so F(s) = ss2+9 .

Now

F(s) =(s2 + 9) s(2s)

(s2 + 9)2 =

9 s2

(s2 + 9)2


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We can invert this using the shift theorem, so the inverse of the firstterm is:

y(0)e3t cos t 3y(0)e3t sin t

Similarly, the second term in ??can be realized as a shift, It is

y(0) + 6y(0)

(s+ 3)2 + 1

which has inverse transform

(y(0) + 6y(0))e3t sin t.

Lastly, we use partial fractions on the final term to get

2

(s2 + 4)(s2 + 6s+ 10)=

As+B

s2 + 4 +

Cs +D

s2 + 6s+ 10

whereA = 115

, B= 115 , C= 115

, D= 13 .(The reader should be familiarwith the method of Partial Fractions, normally covered in a secondsemester calculus class).

We again can easily invert the first term

As+B

s2 + 4 =A

s

s2 + 4+

B

2

2

s2 + 4

which inverts to

A cos2t+B

2sin2t.

The second term, after again setting up for the shifting theorem bycompleting the square, becomes

Cs +D

s2 + 6s+ 10=

C(s+ 3 3)

(s+ 3)2 + 1 +

D

(s+ 3)2 + 1

=C (s+ 3)

(s+ 3)2

+ 1

+ (D 3C) 1

(s+ 3)2

+ 1Taking the inverse, we obtain:


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=C e3t cos t+ (D 3C)e3t sin t

Putting it altogether,

y(t) =y(0)e3t cos t 3y(0)e3t sin t+ (y(0) + 6y(0))e3t sin t

+Ce3t cos t+ (D 3C)e3t sin t

1

15cos 2t+

1

30sin2t,

whereC=

1

15 , D=

1

3 .Note that this simplifies to

y(t) = (y(0) +C)e3t cos t+ ((D 3C) +y(0) + 3y(0))e3t sin t

1

15cos 2t+

1

30sin2t.

If we had used the method of undetermined coefficients, we would haveobtained (possibly in a lot less time)

y(t) = C1e3t cos t+C2e

3t sin t 1

15cos2t+

1

30sin 2t.

The previous example could (and probably should) have been solvedusing other methods. The next example is one where these other meth-ods would not apply.

Example 5.19 Find the solution to

y +ty 2y= 2, y(0) = 0, y(0) = 0


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We take the Laplace transform of both sides of the differential equation.

L[y] + L[ty] 2L[y] =2s

,

s2L[y] d

ds(L[y]) 2L[y] =

2

s,

s2L[y] d

ds(sL[y]) 2L[y] =

2

s,

Using the product rule, note dds

(sL[y]) = sL[y] + L[y] (recall: L[y] isalso a function ofs). So,

s2L[y] sL[y] L[y] 2L[y] =2

s,

or

(s2 3)L[y] sL[y] =2

s,

This is a first order linear differential equation in L with variable s!

L[y] + s+3

sL[y] =

2

s2

,

This has integrating factor es2

2+3ln s which simplifies to s3e

s2

2 , so

L[y] =

s3e

s2

2( 2s2

) ds+C

s3es2

2

This resolves to

2es2

2 +C

s3es2

2

We takeC= 0 and obtain L[y] = 2s3 (all other choices do not give validoutputs of Laplace transforms).


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Which implies that y(t) = t2 solves the DE. (One may easily checkthat, indeed y(t) =t2 does solve the DE/IVP.

Exercises

In 1-8, solve the ODE/IVP using Laplace Transform

1. y + 4y + 3y = 0, y(0) = 1, y(0) = 0

2. y + 4y + 3y = t2, y(0) = 1, y(0) = 0

3. y 3y + 2y= sin t, y(0) = 0, y(0) = 0

4. y 3y + 2y= et, y(0) = 1, y(0) = 0

5. y 2y = t2, y(0) = 1, y(0) = 0

6. y

4y = e2t, y(0) = 0, y

(0) =17. y + 3ty y = 6t, y(0) = 0, y(0) = 0

8. y+ty3y= 2t, y(0) = 0, y(0) = 1 (You will need integrationby parts or use technology)

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