# Homework Help: Laplace Transform Solution to Second Order ODE IVP

1. Nov 29, 2013

### tetrakis

1. The problem statement, all variables and given/known data
y''+6y=f(t), y(0)=0, y'(0)=-2

f(t)= t for 0≤t<1 and 0 for t≥1

2. Relevant equations

3. The attempt at a solution

L{y''}+6L{y}=L{t}-L{tμ(t-1)} where μ(t-1) is Unit Step

Y(s)=L{y}
sY(s)-y(0)=L{y'} and y(0)=0
s2Y(s)-sy(0)-y'(0)+6Y(s) where y(0)=0 and y'(0)=-2

LHS:
s2Y(s)+2+6Y(s)
Y(s)(s2+6)-2

RHS:
L{t}-L{tμ(t-1)}
L{t}=1/s2
L{tμ(t-1)}=L{(t-1)μ(t-1)+L{μ(t-1)}=e-s/s2+e-s/s
L{t}-L{tμ(t-1)}=1/s2-e-s/s2-e-s/s

Y(s)= -1/(s2+6) -e-s/(s2(s2+6))-e-s/s(s2+6)

taking the inverse laplace of the first term
-L-1{1/(s2+6)}=(-1/6)L-1{1/s}+(1/6)L-1{1/(s+6)}=(-1/6)+1/6e-6t)

how could I take the laplace of the other 2 terms?
I was thinking L-1{e-s/(s2(s2+6))}=y(t-1)μ(t-1)?

or I could just break it up into (1/6)L-1{1/s2}-(1/6)L-1{1/(s2+6)} which I know the inverse laplace transforms for, however when I do so I get
y(t)= (1/6)t-(1/6)+(1/6)e-6t
which seems a bit bizarre to me, I would have assumed that the answer would have a unit step function

2. Nov 29, 2013

### vela

Staff Emeritus
I think you're fine up to here.

You didn't do the algebra correctly. The first term is wrong.

You didn't factor the denominator correctly. $s^2 + 6 \ne s(s+6)$. Check your table for a Laplace transform with a quadratic denominator.

Last edited: Nov 29, 2013
3. Nov 29, 2013

### ChemistryNat

definitely an algebra error present,

Y(s)(s2+6)+2=1/s2-e-s/s2-e-s/s

then bring the 2 over
Y(s)(s2+6)=1/s2-e-s/s2-e-s/s-2

divided out the (s2+6) term

Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6)

4. Nov 29, 2013

### tetrakis

So, with the algebra corrected I have
Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6)

the inverse of the first term
L-1{1/(s2(s2+6))} I can't find anything like this in my table, most of the entries have s2+k2 which I can't break 6 down into?

5. Nov 29, 2013

### vela

Staff Emeritus
You have to use partial fractions on that term to break it up. If $k^2=6$, then $k = \sqrt{6}$, no?

6. Nov 29, 2013

### tetrakis

right! so I have 1/(s2(s2+6))=(1/6)(1/s2) - (1/6)(1/(s2+6)

which the inverse laplace transform is

(1/6)t-(√6/6)sin(√6t) ?

7. Nov 29, 2013

### vela

Staff Emeritus
Almost. You're off by a factor of $\sqrt{6}$ in the second term:
$$\frac{1}{6}\frac{1}{s^2+6} = \frac{1}{6\sqrt{6}} \frac{\sqrt{6}}{s^2+6} \rightarrow \frac{1}{6\sqrt{6}}\sin\sqrt6 t$$

8. Nov 29, 2013

### tetrakis

right, okay, and I can also apply this to the last term?

L-1{2/(s2+6}=(2/√6)sin(6t)?

how could I approach the exponential terms?

9. Nov 29, 2013

### vela

Staff Emeritus
The exponential indicates a time shift. You found, for example, that
$$\frac{1}{s^2+6} \rightarrow \frac{1}{6}t - \frac{1}{6\sqrt{6}} \sin \sqrt{6}t,$$ so
$$e^{-s}\frac{1}{s^2+6} \rightarrow \frac{1}{6}(t-1)u(t-1) - \frac{1}{6\sqrt{6}} \sin \sqrt{6}(t-1)\,u(t-1).$$

10. Nov 29, 2013

### tetrakis

okay, so for

$$e^{-s}\frac{1}{s(s^2+6)} \rightarrow \frac{1}{6}e^{-s}\frac{1}{s} -\frac{1}{6}e^{-s}\frac{s}{s^{2}+6} \rightarrow \frac{1}{6}u(t-1) - \frac{1}{6}cosh(\sqrt{6})u(t-1)$$?

11. Nov 29, 2013

### tetrakis

making the entire thing

$$\frac{1}{6}t - \frac{1}{6\sqrt{6}}sin(\sqrt{6}t) - \frac{1}{6}(t-1)u(t-1) + \frac{1}{6\sqrt{6}}sin(\sqrt{6}(t-1))u(t-1) - \frac{1}{6}u(t-1) + \frac{1}{6}cosh(\sqrt{6}t)u(t-1) - \frac{2}{\sqrt{6}}sin(\sqrt{6}t)$$

12. Nov 29, 2013

### vela

Staff Emeritus
Where did you get cosh from? It should be cos. The argument of that function is slightly incorrect. The rest of the solution looks good.

Last edited: Nov 29, 2013
13. Nov 29, 2013

### tetrakis

okay, fixed.

$$\frac{1}{6}t - \frac{1}{6\sqrt{6}}sin(\sqrt{6}t) - \frac{1}{6}(t-1)u(t-1) + \frac{1}{6\sqrt{6}}sin(\sqrt{6}(t-1))u(t-1) - \frac{1}{6}u(t-1) + \frac{1}{6}cos(\sqrt{6}t)u(t-1) - \frac{2}{\sqrt{6}}sin(\sqrt{6}t)$$

I'm sorry but I'm not sure what you mean by the argument? should it be t-1 instead of t?

14. Nov 29, 2013

### vela

Staff Emeritus
Yes, t-1 instead of t. Everywhere t appears in the unshifted function, you have to replace it by t-1.