Question

Please Help Me!

Answer 1

My last problem!

Answer 2

ummm give me a second i might be able to help

Answer 3

Separable DE.

Answer 4

Its second order, Idk how these really work D:

Answer 5

the one you posted is first order.

Answer 6

oh ...

Answer 7

\(\color{black}{\displaystyle y''+6y'+9y=0}\) that one?

Answer 8

yea

Answer 9

So, in general, when you have an equation
\(\color{black}{\displaystyle ay''(t)+by'(t)+cy(t)=0}\)

then, you write the so called 'characteristic equation'
\(\color{black}{\displaystyle ar^2+br+c=0}\).
(where a, b, c are constants, and notice that the power of \(r\) is equivalent to the order of the derivative.)

So, suppose this equation (\(\color{black}{\displaystyle ar^2+br+c=0}\)) has real solutions in a form:
\(\color{black}{\displaystyle x=r_1}\)
\(\color{black}{\displaystyle x=r_2}\)

then, the general solution to the differential equation is:
\(\color{black}{\displaystyle y(t)=c_1e^{r_1t}+c_2e^{r_2t}}\)

Then, you are given some initial conditions.
\(y(0)=\beta\)
\(y'(0)=\delta\)

Why "initial conditions"? To solve for \(c_1\) and \(c_2\).
For the general solution you have,
\(\color{black}{\displaystyle y(t)=c_1e^{r_1t}+c_2e^{r_2t}}\)
\(\color{black}{\displaystyle y'(t)=(r_1)c_1e^{r_1t}+(r_2)c_2e^{r_2t}}\)

Then, you are plugging the initial conditions:
\(\color{black}{\displaystyle \beta=c_1e^{r_1\times 0}+c_2e^{r_2 \times 0 }}\)
\(\color{black}{\displaystyle \delta=(r_1)c_1e^{r_1\times 0}+(r_2)c_2e^{r_2\times 0}}\)

and solve for \(c_1\) and \(c_2\).

Answer 10

This is the general layout.
(And don't get frustrated. This is abstract. \(r_1\) and \(r_2\) are going to be the already known (constant) solutions to your characteristic equation.)

Answer 11

However, when the root (for the characteristic equation) is repeated.
If you have \(x=r_1\) and \(x=r_1\)

Then, your general solution should be.
\(\color{black}{\displaystyle y(t)=c_1e^{r_1t}+c_2\color{blue}{t}e^{r_1t}}\)

-------------REMARK-------------
(adding t, so that the components of the solution are linearly independent
... well .. if you just had c1e^rt+c2e^rt, then c1 and c2 are just arbitrary constants,
and then you have (c1+c2)e^rt ----> Ce^rt, and thus you only accounting for only
ONE (not two) roots)
---------------------------------

And then, the same ... plug in your initial conditions as in the general layout.
(Except that differentiation is a little harder, because you will need
the product rule for (c2)(t)e^(r_1t) ...)

Answer 12

I will do a very similar example to yours right now.

Answer 13

Okay

Answer 14

\(\color{black}{\displaystyle y''-8y'+16y=0}\)

------NOTE------
Also a perfect square trinomial, if you write the
characteristic equation for this DE, just like in your case.
(Well, the numbers are different.)
-----------------

\(\color{red}{\large [1]~{\rm Characteristic~Equation}}\rm~~(and~the~roots~of~it)\)
\(\color{black}{\displaystyle r^2-8r+16=0}\)

Notice that you can rewrite the above equation as,
\(\color{black}{\displaystyle (r-4)^2=0}\)

and thus the roots are (repeated root)
\(\color{black}{\displaystyle r_1=4}\) and \(\color{black}{\displaystyle r_2=4}\)

\(\color{red}{\large [2]~{\rm General~Solution}}\)
general the general solution is in the form
\(\color{black}{\displaystyle y(t)=c_1e^{r_1t}+c_2e^{r_2t}}\) \(\\[0.9em]\)
BUT, in order to account for both solutions WHEN ROOT IS REPEATED, you multiply one of the parts by the t (as I explained in the REMARK).

So, in this case, the root is also repeated.
We multiply times t.
\(\color{black}{\displaystyle y(t)=c_1e^{r_1t}+c_2\color{blue}{t}e^{r_1t}}\)

(I am not done, I am posting this so that you read as I go on ......)

Answer 15

In this case, the root was
\(\color{black}{\displaystyle r_1=4}\) (twice ... repeated root)

So, the general solution in this particular case is:
\(\color{black}{\displaystyle y(t)=c_1e^{4t}+c_2te^{4t}}\)
.

Answer 16

Right

Answer 17

\(\color{red}{{\large [3]~{\rm General~Solution}}~\color{blue}{\bf~~with~~initial~~conditions.}}\)

Suppose I was given
\(y(\ln [1/16])=1 \)
\(y'(\ln [1/81])=2 \)

(I am making the algebra easy for me ... sorry for laziness ... the point is the example anyway xD)

Then, I plug in my conditions.

I have:
\(\color{black}{\displaystyle y(t)=c_1e^{4t}+c_2te^{4t}}\)
\(\color{black}{\displaystyle y'(t)=4c_1e^{4t}+\underbrace{4c_2te^{4t}+c_2e^{4t}}_{\large \rm (Product~Rule)}}\)

So, let's plug in everything.
\(\color{black}{\displaystyle y(\ln 16)=c_1e^{4(\ln [1/16])}+c_2te^{4(\ln [1/16])}=1}\)
\(\color{black}{\displaystyle y'(\ln [1/81])=c_1e^{4(\ln [1/81])}+c_2te^{4(\ln [1/81])}=2}\)

\(\color{black}{\displaystyle c_1e^{-\ln [2]}+c_2te^{-\ln [2]}=1}\)
\(\color{black}{\displaystyle c_1e^{-\ln [3]}+c_2te^{-\ln [3]}=2}\)

Answer 18

Oh, sory I forgot to plug in for t into thy y' equation.
let me fix that

Answer 19

\(\color{black}{\displaystyle y'(\ln [1/81])=c_1e^{4(\ln [1/81])}+c_2(\ln [1/81])e^{4(\ln [1/81])}=2}\)

\(\color{black}{\displaystyle c_1e^{-\ln [3]}+(-4\ln [3])c_2e^{-\ln [3]}=2}\)

\(\color{black}{\displaystyle (1/3)[c_1+(-4\ln [3])c_2]=2}\)

\(\color{black}{\displaystyle c_1-4\ln [3]c_2=6}\)

Answer 20

Sorry I keep leaving things out.

Answer 21

No worries. I really appreciate the help.

Answer 22

Basically, you are plugging \(y(h)=k\) and \(y'(l)=m\)

into
So you have two equations to solve for \(c_1\) and \(c_2\).

\(\color{black}{\displaystyle y(t)=c_1e^{Rt}+c_2te^{Rt}}\)
\(\color{black}{\displaystyle y'(t)=Rc_1e^{Rt}+Rc_2te^{Rt}+c_2e^{Rt}}\)

to then solve for \(c_1\) and \(c_2\) (numerically).

Lastly, you are to plug in the solutions for \(c_1\) and \(c_2\) into the general solutions.

Answer 23

solution ***

Answer 24

Okay, then to implement it to my problem...

\[y'' + 6y'+9y=0 \rightarrow r^2+6r+9=0 \rightarrow (r+3)^2 \rightarrow r=-3\]
\[y(t) = C_1e^{-3t}+C_2 te^{-3t}\]

Plugging in the first initial value:\[y(0)=0 \rightarrow C_1 = 0\]

and if we plug in our first constant, then we get
\[y(t) = C_2te^{-3t}\]
(Since the Constant 0 makes the whole first part 0.)

Am I right so far? :D

Answer 25

after you found y(t), you went off track

Answer 26

Oh, didn't notice, you actually did find C1 correctly. Good jub.

Answer 27

Thanks :D

Answer 28

But, for to use the second initial condition \(y'(0)=1\), you will need to differentiate

\(y(t)=C_1e^{-3t}+C_2te^{-3t}\)

\(y'(t)=-3C_1e^{-3t}+-3C_2te^{-3t}+C_2e^{-3t}\)

Answer 29

Cant you just take out the first part of the equation with the constant, since you already know what it is 0.

Answer 30

You can re-write it simpler, a bit.

\(y'(t)=(-3C_1+C_2)e^{-3t}-3C_2te^{-3t}\)

you have already found \(C_1=0\), so you can plug this in already.
((Be happy you are given easier initial conditions))

\(y'(t)=C_2e^{-3t}-3C_2te^{-3t}\)

Then, since \(y'(0)=1\), therefore ....
\(1=C_2\)

Answer 31

oh, yes, I think you are able to do this in this case, very clever.
(Just don't get into this habit, because the initial conditions won't be zeros in the future problems as you move on)

Answer 32

Well, you can do it the standard way, if you lean to the standard-writers category.
Or, you can do it your way (in this case), which is very clever. Good job on that!

Answer 33

in either case, you will find \(C_2=1\) and \(C_1=0\).

Answer 34

Okie.

So then my final answer would be. \[y=te^{-3t}\]

Answer 35

So, yes, just as you impressively nailed,
\(y(t)=C_2e^{-3t}\) ....


YES, EXACTLY !

Answer 36

Um but i should replace t with x huh? for my problem?

Answer 37

Yes, \(x\).

Answer 38

I hate to use \(x\) in DE, because more often than not, it is a function of t.
(Esp, in system of linear DE)

Answer 39

Wait, we got different answers.

You got \[y = e^{-3t}\]
I got \[y= te^{-3t}\]

Answer 40

I didn't write that t, it should be there.

Answer 41

Nice catch!

Answer 42

Oh okay, cool beans :3

Answer 43

Yups :)

Answer 44

Thanks! You have really made my day. I appreciate the help. :)