Thread: anyone want some DE pratice?

1. anyone want some DE pratice?

University won't give me mark schemes for past papers. Gits.

x'' + 2x' + x = t with x(0) = x'(0) = 1.

My answer is x = t^2(1-t) + t - 1 + e^-t(t+1) using variation of parameters.

x'' + 2mx' + (w^2 + m^2)x = 0 with x(0) = 0, x'(0) = A.

Much appreciated (although I am providing good practice for other people)

2. Originally Posted by poirot
University won't give me mark schemes for past papers. Gits.
Is this assignment, then, for marks? It is forum policy not knowingly to help with such problems.

x'' + 2x' + x = t with x(0) = x'(0) = 1.

My answer is x = t^2(1-t) + t - 1 + e^-t(t+1) using variation of parameters.

x'' + 2mx' + (w^2 + m^2)x = 0 with x(0) = 0, x'(0) = A.

Much appreciated (although I am providing good practice for other people)
You can always double-check solutions to DE's yourself by plugging back the solution into the original DE and see if you get equality. It is also straight-forward to check the initial conditions.

I'm not sure what there is for helpers to do here.

3. no I am practicing for my exam in may. Ok yes I could check them, thanks for reminding me. Weird font , who are you to judge lol. One person's liked font is another person's hated.

4. Originally Posted by poirot
no I am practicing for my exam in may. Ok yes I could check them, thanks for reminding me. Weird font , who are you to judge lol.
Contrary to many peoples' opinions (at least in the USA; less so in, say, Germany), there are such things as authorities. On the MHF, my rank of MHF Helper is a rank of authority. In addition to that, I am a moderator for the Differential Equations forum in which you posted. So, while I can't hand out infractions (only mods and admins can do that), I can do a fair bit. And I will do a fair bit to keep the forums I moderate useful to users. The "fair bit" that I will do definitely involves judging. So yes, I will judge lots of things, and I should. Such judging is for the benefit of legitimate users of MHF such as yourself. I try to judge according to the rules (posted as a sticky in almost every forum) and according to common sense.

One person's liked font is another person's hated.
The reason I changed the font is that if there are a lot of weird fonts in a post, it becomes much harder to do the "Reply With Quote", because there will be a whole raft of markup tags in the text. I'm not the only person who helps people out on MHF who prefers the straight-forward, simple font. Naturally, it's perfectly fine to use italics, bold, whatever. However, putting lots of non-default fonts in a particular post makes quoting much more difficult.

As I said, the whole reason I do things on MHF is to make the site more useful to users, including you. Therefore, if you believe the preceding sentence, then the course of action that will turn out the best for you, that will make MHF more useful to you in your study of math, would be to follow the rules and cooperate with staff. And you shouldn't take that as an accusation, because generally I think you do follow the rules.

Cheers.

5. Originally Posted by Ackbeet
Is this assignment, then, for marks? It is forum policy not knowingly to help with such problems.

You can always double-check solutions to DE's yourself by plugging back the solution into the original DE and see if you get equality. It is also straight-forward to check the initial conditions.

I'm not sure what there is for helpers to do here.
Ok I checked them but i couldn't simplify to 0. Where have I gone wrong. I wasn't being serious (I added the lol to emphasise the tone of my sentence)

6. Originally Posted by poirot
Ok I checked them but i couldn't simplify to 0. Where have I gone wrong.
Impossible to say without you showing your work. Show how you took the derivatives, and show how you plugged those into the original DE.

7. Originally Posted by Ackbeet
Impossible to say without you showing your work. Show how you took the derivatives, and show how you plugged those into the original DE.
x'=2t(1-t)-t^2-(2t+1)e^-t(t+1)
x''= 2(1-t) -2t -2e^-t(t+1)-(2t+1)^2e^-t(t+1)
plugging in: 4t(1-t) -2t^2 +2(1-t) -2t +t^2(1-t)+t-1+e^-t(t+1)*((2t+1)^2-2(2t+1) -1)

8. Yeah, I think you might have made one or two small errors in taking your derivatives. I get the following:

And right away, you can see there's a problem, because the cubed term in x(t) can't be canceled by anything in the other derivatives, and hence will remain. But there's no cubed term on the RHS of the original DE.

So, I would say there's an error in your implementation of the variation of parameters approach. Can you please show that work?

9. sorry that differentiation was all wrong because I was reading e^t(t-1) on the computer as e^(t(t-1))

complementary function; x=Ate^-t+Be^-t

Particular solution; w= -e^-2t, w1= t^2e^-t -e^-t +te^-t , w2= -te^-t

u1' = -t^2+e^t-te^t, so u1=(t-t^2)e^t

u2'=te^t so u2 =(t-1)e^t

solution follows

10. Hmm. Well, the linearly independent solutions of the homogeneous equation are

but I think your other computations are off. I get

Try carrying those corrections through. Perhaps it is good to mention here that W_1 and W_2 are not Wronskians, although W is. They are determinants used in the context of Kramer's Rule to solve a system of equations.

11. I had the function in w1, w2 on the other side so how do you decide which is x1 and which is x2

I got x=(2-t)e^-t+3t-t^2-2 but it all hinged on choosing x1(t) and x2(t) the other way.

12. It shouldn't matter, as long as you're consistent. That is, if you look here, you'll see that as long as you pop the (0,f(x)) vector into the correct column, you're good to go.

I'm having some difficulties myself. The particular "solution" I end up with doesn't actually solve the DE! I simply have to end up with a first-order polynomial as the particular solution, because anything higher order is going to have powers of t that are too high, and won't be canceled by anything in the derivative terms on the LHS of the DE.

13. Originally Posted by poirot
University won't give me mark schemes for past papers. Gits.

x'' + 2x' + x = t with x(0) = x'(0) = 1.

My answer is x = t^2(1-t) + t - 1 + e^-t(t+1) using variation of parameters.
Why are we using variation of parameters for this? Clearly, the solution to the homogeneous equation is

$x = c_1 e^{-t}+c_2te^{-t}$

and if we assume the particular solution is of the form

$x_p = A+Bt$

then we observe that

$x^{\prime\prime}+2x^{\prime}+x = t \implies 2B+A+Bt = t$

Thus 2B+A = 0 and B = 1, implying that A=-2.

Thus, the general solution is

$x = c_1e^{-t}+c_2te^{-t}-2+t$.

Applying the initial conditions x(0)= x'(0) = 1, we observe that

\begin{aligned} 1 &= c_1-2 \\ 1 &= -c_1-c_2+1\end{aligned}

Thus, c_1 = 3 and c_2 = -3.

Therefore, our particular solution should be

$x = 3e^{-t}-3te^{-t}-2+t$

-----------------------

If you really want to use variation of parameters, you need to compute the Wronskians:

$W_1 = \begin{vmatrix} 0 & te^{-t}\\ t & (1-t)e^{-t}\end{vmatrix} = -t^2e^{-t}$

$W_2 = \begin{vmatrix} e^{-t} & 0 \\ -e^{-t} & t\end{vmatrix} = te^{-t}$

$W = \begin{vmatrix} e^{-t} & te^{-t}\\ -e^{-t} & (1-t)e^{-t}\end{vmatrix} = (1-t)e^{-2t}+te^{-2t} = e^{-2t}$

So now, it follows that

$u_1^{\prime} = \frac{W_1}{W} = \frac{-t^2e^{-t}}{e^{-2t}} = -t^2e^t$

and

$u_2^{\prime} = \frac{W_2}{W} = \frac{te^{-t}}{e^{-2t}} = te^t$

I leave it for you to verify that

$u_1 = -(t^2-2t+2)e^t$

and

$u_2 = (t-1)e^t$

So it follows that the general solution is

$x = c_1e^{-t}+c_2te^{-t}-(t^2-2t+2)+t(t-1)=c_1e^{-t}+c_2te^{-t}-2+t$

Applying the initial conditions should give you the equation

$x = 3e^{-t}-3te^{-t}-2+t$

which is the same as above.

I hope this clarifies things.

14. Ah, thanks, Chris. I see now I had some sign errors in there.

15. Your gray cells function well, unlike mine yesterday - spotted an error in my determinant treating a '0' as a '1'. Quite disconcerting