# Find Laplace Transform of (e^t + 2t^4 - cos (4t) + 10)

1. Apr 22, 2004

### Lucy77

Find Laplace Transform of...

(e^t + 2t^4 - cos (4t) + 10)

Thank you

2. Apr 22, 2004

Remember that the Laplace Transform is a linear operator. That means that you can treat each one of the four terms as a separate little problem and then add them all back up at the end.

Other than that, just apply the definition of the Laplace Transform or look at a table of transforms.

3. Apr 22, 2004

### Lucy77

But I don't know how to apply what you said to this problem. Can you please show me how?

4. Apr 22, 2004

### Tom Mattson

Staff Emeritus
He meant that when calculating:

L{et+2t4-cos(4t)+10},

you can use the fact that the Laplace operator is linear to get:

L{et}+2L{t4}-L{cos(4t)}+L{10}.

That is, you can use linearity to turn your big problem into 4 small ones.

edit: fixed a superscript bracket

5. Apr 22, 2004

### Lucy77

Is this right

e

1/(s-1) + 2* 4!/s^5- s/(s^2+16) + 10/s

am I done?

Thanks

6. Apr 22, 2004

### Tom Mattson

Staff Emeritus
I get the same thing.

7. Apr 23, 2004

### Lucy77

Hi I just did two more Laplace problems can anyone please tell me if they're correct.

1.Solve the following intital value problem using the method of Laplace transforms.

y'' - 10y' + 25y = e^5t y(0)=0 y'(0)=0

y"- 10y'+ 25y= e5t becomes -y'(0)- sy(0)+ L(y)- 10f(0)+ 10L(y)+ 25L(y)= L(e5x).

Since y'(0)= y(0)= 0, this is just 34L(y)= L(e5x) so that
L(y)= L(e5x)/34 and, since L(cf)= cL(f) for any constant c,
y(x)= e5x/34.

2.Find Laplace transform inverse

L^-1 (4/s + 1/(s-1)^2 + (3s-16)/(s^2+64)

4 +te^t + 3cos8t + -2sin8t

Thank you

8. Apr 23, 2004

### faust9

Number 2 is right.

Number 1 isn't.

y'' - 10y' + 25y = e^5t y(0)=0 y'(0)=0

$$L \{y^{\prime \prime}\} -10L \{y^\prime\} + 25L \{y\} = L \{e^{5t}\}$$

$$s^2Y(s) -sy^\prime(0) - y(0) -10(sY(s) -y(0)) +25Y(s)= \frac{1}{s-5}$$

$$s^2Y(s) -10sY(s) +25Y(s)= \frac{1}{s-5}$$

$$Y(s)(s^2 -10s +25)= \frac{1}{s-5}$$

$$Y(s)(s-5)^2= \frac{1}{s-5}$$

$$Y(s)= \frac{1}{(s-5)^3}$$

$$L^{-1}\{Y(s)\}=L^{-1} \{\frac{1}{(s-5)^3}\}$$

$$y(t)=\frac{1}{2}t^2e^{5t}$$

The above should be correct.
Well anyway, good Luck and I hope it helps.

9. Apr 24, 2004

### Lucy77

Hi,

I have a couple of more that I did. Could someone please check and see if these are right.
1. FIND THE LAPLACE Transform of the unknown solution function for the following initial value problem:

y" + 4y' - 5y = te^(t), y(0) = 1, y'(0)=0

(Do Not actually find the function, only its transform. Then, without carrying out the steps, inidicate briefly how you would proceed to find the solution function.)

s^2L = s + 4sL - 4 - 5L = Lte^t = 1/(s-1)^2

(s^2 + 4s - 5)l = 1/(s-1)^2 + s +4 = (s^3+ 2s^2-7s+4)/(s-1)^2

simplify
= (s-1)^2(s+4)/(s-1)^3(s+5) = s=4/(s-1)(s+5)

at this point I would do partial fractions.

2.Find Laplace Inverse

L^-1 { (s+10)/(s^2+8s+20)}

s+10)/(s^2+8s+20) = (s+10)/(s^2+8s+16+4) = (s+10)/((s+4)^2+4)

(s+10)/((s+4)^2+4) = (s-(-4))+ 3*2)/(s-(-4))^2+2^2) = (s-(-4)/((s-(-4))^2+2^2) + 3*2/(s-(-4)+2^2)

The Inverse Laplace Transform of that is

e^-4t cos 2t + 3 e^-4t sin 2t.

3. Find Laplance Transform
Find L (f(t)} where f(t) = e^-t (0 <= t < 5)
-1 (t >= 5)

For t in [0,5) the solution would be Int((e^{-t})(e^{-st},0,infinity,dt)=Int(e^{-t(s+1)},0,infinity,dt}=-e^{-t(s+1)}/(s+1)|0,infinity) = 1/(s+1).

For t in [5,infinity) Int(-1*e^{-st},0,infinity,dt} =(1/s)e^{-st}|0,infinity)=-1/s.

In piecewise notation you write L(f(t))={1/s, 0 <= t < 5 and -1/s, t >= 5}

4. Find Laplace Inverse of

L^-1 { (3s-4)/s(s-4) }

so

3s-4)/s(s-4) = A/s + B/(s-4)

s*(3s-4)/(s(s-4)) = A + B*s/(s-4)

Taking limit ( s to 0)

-4/-4 = A

A =1

And

(s-4)*(3s-4)/(s(s-4)) = A*(s-4)/s + B

Taking limit ( s to 4)

8/4 = B = 2

so

(3s-4)/s(s-4) = 1/s + 2/(s-4)

Thank you

Last edited: Apr 24, 2004