Thread: ODE

**kjchauhan** · January 3rd 2011, 02:10 AM

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,

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

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

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

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

Thanks ..

**Ackbeet** · January 3rd 2011, 05:05 AM

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?

**HallsofIvy** · January 3rd 2011, 12:28 PM

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.

**kjchauhan** · January 3rd 2011, 04:50 PM

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..

**kjchauhan** · January 3rd 2011, 05:07 PM

$<br /> (p^3-p^2-p+1)L(y) = 8\left(\frac{1}{(p+1)^2}\right) +p-1$

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

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

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

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

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

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

Correct???

**Ackbeet** · January 3rd 2011, 05:38 PM

Can you factor the cubic?

**kjchauhan** · January 3rd 2011, 10:29 PM

Yes

**Ackbeet** · January 4th 2011, 02:26 AM

So how does that change your expression?

**kjchauhan** · January 4th 2011, 02:13 PM

Originally Posted by Ackbeet

So how does that change your expression?

Thank you very much..

**Ackbeet** · January 6th 2011, 02:42 AM

You're welcome. Have a good one!

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