1. ODE

Pl help me to solve it..

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

I've taken Laplace Transform,

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

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

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

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

Thanks ..

2. 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?

3. 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: $r^3- r^2- r+ 1= 0$. It's easy to see, by inspection, that r= 1 is a root of that and that $r^3- r^2- r+ 1= (r-1)(r^2- 1)= (r- 1)^2(r+ 1)$ 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 $y= Ae^t+ Bte^t+ Ce^{-t}$.

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

Putting those into the differential equation will give you equations for A and B.

4. Originally Posted by HallsofIvy
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: $r^3- r^2- r+ 1= 0$. It's easy to see, by inspection, that r= 1 is a root of that and that $r^3- r^2- r+ 1= (r-1)(r^2- 1)= (r- 1)^(r+ 1)$ 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 $y= Ae^t+ Bte^t+ Ce^{-t}$.

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

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..

5. $
(p^3-p^2-p+1)L(y) = 8\left(\frac{1}{(p+1)^2}\right) +p-1$

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

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

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

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

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

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

Correct???

6. Can you factor the cubic?

7. Yes

8. So how does that change your expression?

9. Originally Posted by Ackbeet
So how does that change your expression?
Thank you very much..

10. You're welcome. Have a good one!