Find general equation of x′′(t)+5x′(t)+4x(t)=0

[Arnab Patra]

Suppose $x_1(t)$ and $x_2(t)$ are two linearly independent solutions of the equations:

$x'_1(t) = 3x_1(t) + 2x_2(t)$ and $x'_2(t) = x_1(t) + 2x_2(t)$

where $x'_1(t)\text{ and }x'_2(t)$ denote the first derivative of functions $x_1(t)$ and $x_2(t)$

respectively with respect to $t$.

Find the general solution of $x''(t) + 5x'(t) + 4x(t) = 0$ in terms of $x_1(t)$ and $x_2(t)$.
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The general solution of the equation

$x''(t) + 5x'(t) + 4x(t) = 0$......(1)

is

$x(t) = c _1 e^{-4t} + c _2e^{-t}$......(2)

Now if i want to express equation (2) in term of $x_1(t)$ and $x_2(t)$ , what exactly i have to do ?

 

[LCKurtz]

Suppose $x_1(t)$ and $x_2(t)$ are two linearly independent solutions of the equations:

$x'_1(t) = 3x_1(t) + 2x_2(t)$ and $x'_2(t) = x_1(t) + 2x_2(t)$

where $x'_1(t)\text{ and }x'_2(t)$ denote the first derivative of functions $x_1(t)$ and $x_2(t)$

respectively with respect to $t$.

Find the general solution of $x''(t) + 5x'(t) + 4x(t) = 0$ in terms of $x_1(t)$ and $x_2(t)$.

I'm guessing that that equation is supposed to be $x''(t) \color{red}{\bf -} 5x'(t) + 4x(t) = 0$

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

The general solution of the equation

$x''(t) + 5x'(t) + 4x(t) = 0$......(1)

is

$x(t) = c _1 e^{-4t} + c _2e^{-t}$......(2)

Now if i want to express equation (2) in term of $x_1(t)$ and $x_2(t)$ , what exactly i have to do ?

I would first solve the 2x2 matrix system you are given for $x_1$ and $x_2$.

 

[Arnab Patra]

The question is correct . X1 = -4 and x2 = -1 . And now what ?

 

[LCKurtz]

The question is correct . X1 = -4 and x2 = -1 . And now what ?

$x_1=-4$ and $x_2=-1~$?? Where did those come from? They don't solve either your 2x2 system or your second order system.

In the last line of post #2 I suggested you solve your 2x2 matrix system for $x_1$ and $x_2$. Have you done that yet? You should get two linearly independent vectors.

After that maybe you can make some sense out of the problem.

 

[Arnab Patra]

sorry my mistake .. $$x _1(t) = 3/2 $$ and $$x _2(t) = -1/2 $$ .

 

[LCKurtz]

We are getting nowhere. You aren't showing any work and I don't know what those numbers are supposed to be. They certainly aren't solutions.

 

[Arnab Patra]

by considering $$x''(t)+5x'(t)+4x(t)=0$$ equation as $$r^2+5r+4=0$$ i get $$r=-4,-1$$ i.e. $$x'(t)=-4,-1$$ which are the two roots of the said equation. from this solution i found the general equation of the said equation i.e. $$x(t) = c_1e^{-4t} + c_2e^{-t}$$ .

And you are asking to solve $$\left(\matrix{x_1(t)\\x_2(t)}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x_1(t)\\x_2(t)}\right)$$ this ?

 

[LCKurtz]

And you are asking to solve $$\left(\matrix{x_1(t)\\x_2(t)}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x_1(t)\\x_2(t)}\right)$$ this ?

The first line in your first post refers to two linearly independent solutions of that system. So I would expect you would want to find them unless you can find a way to do the problem without finding them explicitly. And when you do that I think you will see why I thought (and still do think) that second order DE has a misprint.

 

[Mark44]

The question is correct . X1 = -4 and x2 = -1 . And now what ?

This makes no sense. $x_1(t)$ and $x_2(t)$ are functions of t, not constants.

sorry my mistake .. $$x _1(t) = 3/2 $$ and $$x _2(t) = -1/2 $$ .

This makes no sense either

by considering $$x''(t)+5x'(t)+4x(t)=0$$ equation as $$r^2+5r+4=0$$ i get $$r=-4,-1$$ i.e. $$x'(t)=-4,-1$$ which are the two roots of the said equation. from this solution i found the general equation of the said equation i.e. $$x(t) = c_1e^{-4t} + c_2e^{-t}$$ .

Isn't this given information?

And you are asking to solve $$\left(\matrix{x_1(t)\\x_2(t)}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x_1(t)\\x_2(t)}\right)$$ this ?

I believe that what LCKurtz is suggesting is that you need to solve the system of differential equations as shown in post #1, by diagonalization. To do that, you need to find the eigenvalues and eigenvectors for your matrix, and use them to find a diagonal matrix D to uncouple the system of diff. equations.

So far in your work I don't see any evidence that you are doing this.

 

[epenguin]

And you are asking to solve $$\left(\matrix{x_1(t)\\x_2(t)}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x_1(t)\\x_2(t)}\right)$$ this ?

LOL it's not transcendentally difficult after all, looks like the only way that might just throw some light on the question.

You do you see why people are saying it could be a misprint? With a change of sign there is a fairly evident relation between the two parts of the question.

If at the end it turns out your Prof says he made a mistake in transcription, or that there was a misprint in the book, it wouldn't be the first time that happened here.

 

[Arnab Patra]

sorry guys . i was wrong about the question .. i found the question and i am attaching the question's image

 

[LCKurtz]

So why did you tell me the equation was correct, without checking it, when I asked about it? And when are you going to show us how you solve the 2x2 system?

 

[Mark44]

sorry guys . i was wrong about the question .. i found the question and i am attaching the question's image

If we don't see some effort toward solving the 2x2 system real soon, I'm going to close this thread.

 

[Arnab Patra]

calculating eigenvalues are 4 and 1 . Eigen vectors are (2,1) and (-1,1)

and

$$\left(\matrix{x_1(t) \\ x_2(t)}\right)=c_1e^{4t} \left(\matrix{2 \\ 1 }\right) +c_2e^{t} \left(\matrix{-1 \\ 1 } \right)$$
or

$$\left(\matrix{x_1(t) \\ x_2(t)}\right)=\left(\matrix{2 & -1\\ 1 & 1 }\right) \left(\matrix{c_1e^{4t} \\ c_2e^{t} } \right)$$

so

$$ x_1(t) = 2c_1e^{4t} - c_2e^{t} $$ and

$$ x_2(t) = c_1e^{4t} + c_2e^{t} $$

these are the values of x1 and x2 i found so far .. am i on right direction ?

 

[LCKurtz]

calculating eigenvalues are 4 and 1 . Eigen vectors are (2,1) and (-1,1)

and

$$\left(\matrix{x_1(t) \\ x_2(t)}\right)=c_1e^{4t} \left(\matrix{2 \\ 1 }\right) +c_2e^{t} \left(\matrix{-1 \\ 1 } \right)$$
or

$$\left(\matrix{x_1(t) \\ x_2(t)}\right)=\left(\matrix{2 & -1\\ 1 & 1 }\right) \left(\matrix{c_1e^{4t} \\ c_2e^{t} } \right)$$

so

$$ x_1(t) = 2c_1e^{4t} - c_2e^{t} $$ and

$$ x_2(t) = c_1e^{4t} + c_2e^{t} $$

these are the values of x1 and x2 i found so far .. am i on right direction ?

This looks correct, and this is where your problem becomes confusing. Your 2x2 system is a vector system of the form$$
\vec x' = A\vec x$$where$$
\vec x'=\left(\matrix{x_1(t)\\x_2(t)}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x_1(t)\\x_2(t)}\right)$$and you have a general solution$$
\vec x = c_1e^{4t} \left(\matrix{2 \\ 1 }\right) +c_2e^{t} \left(\matrix{-1 \\ 1 } \right)$$You could write this as$$
\vec x = c_1\vec x_1+c_2\vec x_2$$where$$
\vec x_1 = \left(\matrix{2e^{4t} \\ e^{4t} }\right),~~\vec x_2= \left(\matrix{-e^{t} \\ e^{t} } \right)$$Writing it this way, you have two linearly independent vectors forming a basis for your vector solution. Don't confuse the vectors $\vec x_1,~\vec x_2$ with their components which you are calling $x_1(t)$ and $x_2(t)$.

Now the confusion arises in your original post where you state

Suppose $x_1(t)$ and $x_2(t)$ are two linearly independent solutions of the equations:

$x'_1(t) = 3x_1(t) + 2x_2(t)$ and $x'_2(t) = x_1(t) + 2x_2(t)$.

Find the general solution of $x''(t) + 5x'(t) + 4x(t) = 0$ in terms of $x_1(t)$ and $x_2(t)$.

When you talk about the two linearly independent solutions of those two equations, you would be talking about the independent vectors we have above. But the general solution of your second order equation would be a scalar, not a vector. So at this point, it isn't clear to me what the question is really asking for. You might have to ask your teacher to clarify this.

 

[epenguin]

My understanding is that you are not meant to solve any equation explicitly. (Even though it is not hard to solve any of them.)

My suggestion is to add the first two equations, and you will then see that you have a simple differential equation in a new variable.

Maybe you can then see that the third equation is really the same differential equation again but in yet another variable.

 

[vela]

When you talk about the two linearly independent solutions of those two equations, you would be talking about the independent vectors we have above. But the general solution of your second order equation would be a scalar, not a vector. So at this point, it isn't clear to me what the question is really asking for. You might have to ask your teacher to clarify this.

The problem statement seems strangely worded. I'd guess the intent is to solve for $e^{4t}$ and $e^t$ in terms of $x_1(t)$ and $x_2(t)$.