Laplace Transform Solution to Second Order ODE IVP

[tetrakis]

Homework Statement

y''+6y=f(t), y(0)=0, y'(0)=-2

f(t)= t for 0≤t<1 and 0 for t≥1

Homework Equations

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

Thank you for your time

 

[vela]

Homework Statement

y''+6y=f(t), y(0)=0, y'(0)=-2

f(t)= t for 0≤t<1 and 0 for t≥1

Homework Equations

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

I think you're fine up to here.

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

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

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)

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.

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

Thank you for your time

 

[ChemistryNat]

I think you're fine up to here.


What happened to the -2 from the LHS?


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.

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)

 

[tetrakis]

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.

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?

 

[vela]

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

 

[tetrakis]

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

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) ?

 

[vela]

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

 

[tetrakis]

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

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?

 

[vela]

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
$$e^{-s}\frac{1}{s^2+6} \rightarrow \frac{1}{6}(t-1)u(t-1) - \frac{1}{6\sqrt{6}} \sin \sqrt{6}(t-1)\,u(t-1).$$

 

[tetrakis]

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
$$e^{-s}\frac{1}{s^2+6} \rightarrow \frac{1}{6}(t-1)u(t-1) - \frac{1}{6\sqrt{6}} \sin \sqrt{6}(t-1)\,u(t-1).$$

okay, so for

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

 

[tetrakis]

making the entire thing

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

 

[vela]

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.

 

[tetrakis]

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.

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?

 

[vela]

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