Solving an I.V.P. with Laplace transforms.

**bkbowser** · December 8th 2013, 11:22 AM

$y^{\prime\prime}+6y^\prime+9y=0$ and $y(0)=-1, y^\prime(0)=7$

OK so I use two properties of Laplace Transforms;

${\Lapl}{\mathcal{L}}(f^{\prime\prime})(s)=s^2 {\Lapl}{\mathcal{L}}(f)(s)-sf(0)-f^\prime(0)$ and
${\Lapl}{\mathcal{L}}{(f^\prime)}(s) = s{\Lapl}{\mathcal{L}}(f)(s) - f(0$ )

And I replace ${\Lapl}{\mathcal{L}}(f)(s)$ with $Y$

$s^2Y-sf(0)-f^\prime(0)+6(sY-f(0))+9Y=0$

using my initial conditions;

$s^2Y+s-7+6sY+6+9Y=0$

$s^2Y+6sY+9Y=-s+1$

$Y(s^2+6s+9)=-s+1$

$Y=-\frac{(s+1)}{(s^2+6s+9)}$

Carry out the division;
$Y=-\frac{s}{(s^2+6s+9)}-\frac{1}{(s^2+6s+9)}$

So I guess this is the part I'm stuck on. It seems like the rightmost term is an expression of something like $e^{at}t^n$ but I don't see any obvious way to handle the middle term. Have I made a mistake? Or do I not just see the obvious answer?

**romsek** · December 8th 2013, 01:33 PM

Originally Posted by bkbowser

$y^{\prime\prime}+6y^\prime+9y=0$ and $y(0)=-1, y^\prime(0)=7$

OK so I use two properties of Laplace Transforms;

${\Lapl}{\mathcal{L}}(f^{\prime\prime})(s)=s^2 {\Lapl}{\mathcal{L}}(f)(s)-sf(0)-f^\prime(0)$ and
${\Lapl}{\mathcal{L}}{(f^\prime)}(s) = s{\Lapl}{\mathcal{L}}(f)(s) - f(0$ )

And I replace ${\Lapl}{\mathcal{L}}(f)(s)$ with $Y$

$s^2Y-sf(0)-f^\prime(0)+6(sY-f(0))+9Y=0$

using my initial conditions;

$s^2Y+s-7+6sY+6+9Y=0$

$s^2Y+6sY+9Y=-s+1$

$Y(s^2+6s+9)=-s+1$

$Y=-\frac{(s+1)}{(s^2+6s+9)}$

I believe this should be

$Y=\frac{1-s}{s^2+6 s+9}$

**bkbowser** · December 8th 2013, 02:21 PM

$Ys^2 + s -7 +6sY+6+9Y=0$

$Ys^2+s+6sY+9Y-1=0$

$Ys^2+6sY+9Y=1-s$

$Y(s^2+6s+9)=1-s$

$Y=\frac{1-s}{(s^2+6s+9)}$

$Y=\frac{1}{(s^2+6s+9)}-\frac{s}{(s^2+6s+9)}$

$Y=\frac{1}{(s+3)^2}-\frac{s}{(s^2+6s+9)}$

Is this correct?

I still am not sure what to do with the the rightmost term. It doesn't look like I can convert it into anything at the moment.

**romsek** · December 8th 2013, 03:23 PM

Originally Posted by bkbowser

$Ys^2 + s -7 +6sY+6+9Y=0$

$Ys^2+s+6sY+9Y-1=0$

$Ys^2+6sY+9Y=1-s$

$Y(s^2+6s+9)=1-s$

$Y=\frac{1-s}{(s^2+6s+9)}$

$Y=\frac{1}{(s^2+6s+9)}-\frac{s}{(s^2+6s+9)}$

$Y=\frac{1}{(s+3)^2}-\frac{s}{(s^2+6s+9)}$

Is this correct?

I still am not sure what to do with the the rightmost term. It doesn't look like I can convert it into anything at the moment.

well it's s/(s+3)²

you can look up the inverse for 1/(s+3)², and then recall what multiplying by s in the frequency domain does in the time domain (differentiation)

You're close.

**bkbowser** · December 8th 2013, 04:56 PM

Originally Posted by romsek

well it's s/(s+3)²

and then recall what multiplying by s in the frequency domain does in the time domain (differentiation)

I don't know what this means.

**romsek** · December 8th 2013, 05:23 PM

Originally Posted by bkbowser

I don't know what this means.

you didn't get to the bit with Laplace transforms where if f(t) and F(s) are a transform pair then

d/dt f(t) <--> sF(s) - f(0) ? (usually if f(0) isn't mentioned in the problem it's just assumed to be 0)

**bkbowser** · December 8th 2013, 05:48 PM

Originally Posted by romsek

you didn't get to the bit with Laplace transforms where if f(t) and F(s) are a transform pair then

d/dt f(t) <--> sF(s) - f(0) ? (usually if f(0) isn't mentioned in the problem it's just assumed to be 0)

No we have that. I just don't see how that applies here since the problem gives that f(0)=-1 I'd need a -1 hanging around wouldn't I?

**romsek** · December 8th 2013, 05:59 PM

Originally Posted by bkbowser

No we have that. I just don't see how that applies here since the problem gives that f(0)=-1 I'd need a -1 hanging around wouldn't I?

ok.. you took care of initial conditions at the start.. sorry a bit rusty on this

forget about the f(0) term. Just deal with the fact that F(s) = s * (1/(s+3)²) <--> d/dt g(t)

where g(t) is the Inverse Laplace Transform of 1/(s+3)²

**bkbowser** · December 9th 2013, 07:07 AM

Originally Posted by romsek

ok.. you took care of initial conditions at the start.. sorry a bit rusty on this

forget about the f(0) term. Just deal with the fact that F(s) = s * (1/(s+3)²) <--> d/dt g(t)

where g(t) is the Inverse Laplace Transform of 1/(s+3)²

I'm still not following you.

**romsek** · December 9th 2013, 07:26 AM

Originally Posted by bkbowser

I'm still not following you.

you've got F(s) = s/(s+3)²

If you look at the transform table you'll see an entry for a pair where G(s) = 1/(s-a)ⁿ⁺¹ which is what you've got here with a=-3, n=1

that transforms back to some g(t) that you'll see in the table.

Now, your F(s) is actually sG(s). Multiplying by s in the s domain corresponds with differentiation in the time domain.

so sG(s) <==> d/dt g(t)

so look up g(t) using the G(s) above and then take it's time derivative.

**bkbowser** · December 9th 2013, 08:15 AM

the only thing I'm seeing on these tables that looks close is;

${\Lapl}{\mathcal{L}}(tf(t))=-\frac{d}{ds}{\Lapl}{\mathcal{L}}(f(t))$

**HallsofIvy** · December 9th 2013, 12:08 PM

Are there any examples of differential equations problems where "Laplace transform method" is worth doing?

Here, the first thing I would do is see that y''+ 6y'+ 9y= 0 has characteristic equation $r^2+ 6r+ 9= (r+ 3)^2= 0$ and so has r= -3 as a double root. That tells us that the general solution can be written as $y(t)= C_1e^{-3t}+ C_2te^{-3t}$ . From that, $y'(t)= -3C_1e^{-3t}+ C_2e^{-3t}- 3C_2te^{-3t}$ .

All that remains is to find $C_1$ and $C_2$ so that $y(0)= C_1= -1$ and $y'(0)= -3C_1+ C_2= 7$ .

I have always disliked the Laplace transform method!

**bkbowser** · December 9th 2013, 12:30 PM

Originally Posted by HallsofIvy

Are there any examples of differential equations problems where "Laplace transform method" is worth doing?

Here, the first thing I would do is see that y''+ 6y'+ 9y= 0 has characteristic equation $r^2+ 6r+ 9= (r+ 3)^2= 0$ and so has r= -3 as a double root. That tells us that the general solution can be written as $y(t)= C_1e^{-3t}+ C_2te^{-3t}$ . From that, $y'(t)= -3C_1e^{-3t}+ C_2e^{-3t}- 3C_2te^{-3t}$ .

All that remains is to find $C_1$ and $C_2$ so that $y(0)= C_1= -1$ and $y'(0)= -3C_1+ C_2= 7$ .

I have always disliked the Laplace transform method!

lol ya I hadn't looked at it from that point of view. But I'm pretty sure we'll be required to solve some IVP's using only Laplace transforms.

Your explanation makes this clearer too though. I just need to figure out what rule will let me correctly reference the fact that my last term is $C_2te^{-3t}$

**bkbowser** · December 9th 2013, 12:44 PM

${\Lapl}{\mathcal{L}}(t^{n}f(t))(s)=(-1^n)\frac{d^n}{ds^n}{\Lapl}{\mathcal{L}}(f(s))$

So starting from $C_2te^{-3t}$

${\Lapl}{\mathcal{L}}(C_2t^1e^{-3t})=-1^1\frac{d^1}{ds^1}{\Lapl}{\mathcal{L}}(f(s))$

So basically I'd ignore the $t$ and take the 1st derivative of the Laplace transform of $e^{-3t}$

$-1\frac{d}{ds}(\frac{1}{s+3})=-1(\frac{(s+3)^\prime}{(s+3)^2})=\frac{-1}{(s+3)^2}$

Did I make an error or is this just the wrong property that I'm using?

**romsek** · December 9th 2013, 12:59 PM

Originally Posted by bkbowser

${\Lapl}{\mathcal{L}}(t^{n}f(t))(s)=(-1^n)\frac{d^n}{ds^n}{\Lapl}{\mathcal{L}}(f(s))$

So starting from $C_2te^{-3t}$

${\Lapl}{\mathcal{L}}(C_2t^1e^{-3t})=-1^1\frac{d^1}{ds^1}{\Lapl}{\mathcal{L}}(f(s))$

So basically I'd ignore the $t$ and take the 1st derivative of the Laplace transform of $e^{-3t}$

$-1\frac{d}{ds}(\frac{1}{s+3})=-1(\frac{(s+3)^\prime}{(s+3)^2})=\frac{-1}{(s+3)^2}$

Did I make an error or is this just the wrong property that I'm using?

I'm just going to show you how it's done.

$\frac{1}{(s+3)^2}\Longleftrightarrow \left(t e^{-3 t} u(t)\right)$

$s F(s)\Longleftrightarrow \left(\frac{d}{dt}f(t)\right)$

$\frac{d}{dt}t e^{-3 t} u(t)=\delta (t) t e^{-3 t}+\left(e^{-3 t}-3 t e^{-3 t}\right) u(t) = \left(e^{-3 t}-3 t e^{-3 t}\right) u(t)$

I'll let you mess with getting the constants right.

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