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Math Help - Solving an I.V.P. with Laplace transforms.

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

    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?
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by bkbowser View Post
    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}
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    Re: Solving an I.V.P. with Laplace transforms.

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

    Quote Originally Posted by bkbowser View Post
    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.
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by romsek View Post
    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.
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by bkbowser View Post
    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)
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by romsek View Post
    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?
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by bkbowser View Post
    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
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by romsek View Post
    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.
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by bkbowser View Post
    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.
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    Re: Solving an I.V.P. with Laplace transforms.

    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))
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    Re: Solving an I.V.P. with Laplace transforms.

    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!
    Last edited by HallsofIvy; December 9th 2013 at 12:22 PM.
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by HallsofIvy View Post
    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}
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    Re: Solving an I.V.P. with Laplace transforms.

    {\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?
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    Re: Solving an I.V.P. with Laplace transforms.

    Quote Originally Posted by bkbowser View Post
    {\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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