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

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

OK so I use two properties of Laplace Transforms;

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

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

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

using my initial conditions;

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

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

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

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

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

So I guess this is the part I'm stuck on. It seems like the rightmost term is an expression of something like $\displaystyle e^{at}t^n$ 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?

Originally Posted by bkbowser

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

OK so I use two properties of Laplace Transforms;

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

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

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

using my initial conditions;

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

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

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

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

I believe this should be

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

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

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

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

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

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

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

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

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.

Originally Posted by bkbowser

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

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

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

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

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

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

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

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.

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.

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)

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?

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

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.

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.

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

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

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 $\displaystyle r^2+ 6r+ 9= (r+ 3)^2= 0$ and so has r= -3 as a double root. That tells us that the general solution can be written as $\displaystyle y(t)= C_1e^{-3t}+ C_2te^{-3t}$. From that, $\displaystyle y'(t)= -3C_1e^{-3t}+ C_2e^{-3t}- 3C_2te^{-3t}$.

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

I have always disliked the Laplace transform method!

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 $\displaystyle r^2+ 6r+ 9= (r+ 3)^2= 0$ and so has r= -3 as a double root. That tells us that the general solution can be written as $\displaystyle y(t)= C_1e^{-3t}+ C_2te^{-3t}$. From that, $\displaystyle y'(t)= -3C_1e^{-3t}+ C_2e^{-3t}- 3C_2te^{-3t}$.

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

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 $\displaystyle C_2te^{-3t}$

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

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

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

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

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

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

Originally Posted by bkbowser

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

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

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

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

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

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.

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

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

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

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