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  1. #1
    Member kjchauhan's Avatar
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    ODE

    Pl help me to solve it..

    Solve: $\displaystyle y'''-y''-y'+y=8te^{-t}$, with the initial conditions y(0)=o, y'(0)=1 and y''(0)=0 using Laplace Transform.

    I've taken Laplace Transform,

    $\displaystyle L(y''')-L(y'')-L(y')+L(y)=8L(te^{-t})$

    $\displaystyle \therefore p^3L(y)-p^2y(0)-py'(0)-y''(0)-p^2L(y)+py(0)+y'(0)-pL(y)+y(0)+L(y)
    =8\left(\frac{1}{(p+1)^2}\right)$

    $\displaystyle \therefore (p^3-P^2-p+1)L(y) = 8\left(\frac{1}{(p+1)^2}\right) +p-1$

    I've simplified it but, I did not get it..

    Thanks ..
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  2. #2
    A Plied Mathematician
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    Your work so far is fine (watch the capital P on the LHS, though). On the RHS, you can get a common denominator and add. The LHS factors, and then you can solve for L(y). What do you get?
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  3. #3
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    Oh, dear, oh dear, oh, dear! I really dislike the Laplace Transform! It can be used in two diametrically opposite ways- to "look up" solutions to non-homogeneous equations or to assert a form of a solution in very abstract theorems. The first is more easily done in other ways and the second, undergraduates never see.

    I would look at that equation and immediately write down it characteristic equation: $\displaystyle r^3- r^2- r+ 1= 0$. It's easy to see, by inspection, that r= 1 is a root of that and that $\displaystyle r^3- r^2- r+ 1= (r-1)(r^2- 1)= (r- 1)^2(r+ 1)$ so that r= 1 is a double root and r= -1 is a root. That tells us that the general solution to the associated homogeneous equation is $\displaystyle y= Ae^t+ Bte^t+ Ce^{-t}$.

    The "non-homogeneos" part is $\displaystyle 8te^{-t}$. Since $\displaystyle te^{-t}$ is already a solution to the associated homogeneous equation, we try a solution of the form $\displaystyle y= (At^2+ Bt)e^{-t}$. Then $\displaystyle y'= (At+ B)e^{-t}- (At^2+ Bt)e^{-t}= (-At^2+ (A- B)t+ B)e^{-t}$, $\displaystyle y''= (-2At+ (A- B)e^{-t})e^{-t}- (-At^2+ (A- B)t+ B)e^{-t}$$\displaystyle =(At^2+ (B- 3A)t+ A- 2B)e^{-t}$, and $\displaystyle y'''= (2At+ B- 3A)e^{-t}- (At^2+ (B- 3A)t+ A- 2B)e^{-t}= (-At^2+ (5A- B)t+ 3B- 5A)e^{-t}$.

    Putting those into the differential equation will give you equations for A and B.
    Last edited by HallsofIvy; Jan 4th 2011 at 02:32 AM.
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  4. #4
    Member kjchauhan's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Oh, dear, oh dear, oh, dear! I really dislike the Laplace Transform! It can be used in two diametrically opposite ways- to "look up" solutions to non-homogeneous equations or to assert a form of a solution in very abstract theorems. The first is more easily done in other ways and the second, undergraduates never see.

    I would look at that equation and immediately write down it characteristic equation: $\displaystyle r^3- r^2- r+ 1= 0$. It's easy to see, by inspection, that r= 1 is a root of that and that $\displaystyle r^3- r^2- r+ 1= (r-1)(r^2- 1)= (r- 1)^(r+ 1)$ so that r= 1 is a double root and r= -1 is a root. That tells us that the general solution to the associated homogeneous equation is $\displaystyle y= Ae^t+ Bte^t+ Ce^{-t}$.

    The "non-homogeneos" part is $\displaystyle 8te^{-t}$. Since $\displaystyle te^{-t}$ is already a solution to the associated homogeneous equation, we try a solution of the form $\displaystyle y= (At^2+ Bt)e^{-t}$. Then $\displaystyle y'= (At+ B)e^{-t}- (At^2+ Bt)e^{-t}= (-At^2+ (A- B)t+ B)e^{-t}$, $\displaystyle y''= (-2At+ (A- B)e^{-t})e^{-t}- (-At^2+ (A- B)t+ B)e^{-t}$$\displaystyle =(At^2+ (B- 3A)t+ A- 2B)e^{-t}$, and $\displaystyle y'''= (2At+ B- 3A)e^{-t}- (At^2+ (B- 3A)t+ A- 2B)e^{-t}= (-At^2+ (5A- B)t+ 3B- 5A)e^{-t}$.

    Putting those into the differential equation will give you equations for A and B.
    You are absolutely right, and that way I can solve this equation, But it is compulsory to solve using Laplace Trasform for me..
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  5. #5
    Member kjchauhan's Avatar
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    $\displaystyle
    (p^3-p^2-p+1)L(y) = 8\left(\frac{1}{(p+1)^2}\right) +p-1$

    $\displaystyle \therefore L(y) = \frac{\frac{8}{(p+1)^2}+p-1}{(p^3-p^2-p+1)}$

    $\displaystyle \therefore L(y) = \frac{8}{(p+1)^2(p^3-p^2-p+1)}+\frac{p-1}{(p^3-p^2-p+1)}$

    $\displaystyle \therefore L(y) = \frac{8}{(p+1)^3(p-1)^2}+\frac{p-1}{(p-1)^2(p+1)}$

    $\displaystyle \therefore L(y) = \frac{1}{(p-1)^2}+ \frac{1}{2(p+1)^3}+\frac{1}{2(p-1)}-\frac{1}{2(p+1)}$

    $\displaystyle \therefore y=L^{-1} \left( \frac{1}{(p-1)^2}+ \frac{1}{2(p+1)^3}+\frac{1}{2(p-1)}-\frac{1}{2(p+1)}\right)$

    $\displaystyle \therefore y=te^t+\frac{t^2e^{-t}}{4}+\frac{e^t}{2}+\frac{e^{-t}}{2}$

    Correct???
    Last edited by kjchauhan; Jan 4th 2011 at 12:36 AM.
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  6. #6
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    Can you factor the cubic?
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  7. #7
    Member kjchauhan's Avatar
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    Yes
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  8. #8
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    So how does that change your expression?
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  9. #9
    Member kjchauhan's Avatar
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    Quote Originally Posted by Ackbeet View Post
    So how does that change your expression?
    Thank you very much..
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    You're welcome. Have a good one!
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