Solving a second order diff. equation

[Alphaboy2001]

Homework Statement

Hi

I need to solve the following:

[itex]\frac{d^{2} x}{dt} + 2 b \frac{dx}{dt} + (b^2 - \frac{db}{dt}) x = 0[/tex]

2. The attempt at a solution

Isn't this equivalent to??::

$x^{''} + 2bx^{'}+ (b^2 - b^{'})x = 0 \iff x^2 + 2x +1 = 0$ If I let b = 1.

which gives me the solution $$\lamda = -1$$

Thus giving me the particular solution

$$x(t) = c \cdot e^{-t}$$


Have I understood correctly? There is no precondition regarding either t or b.

Best Regards

Alphaboy

 

[tiny-tim]

Welcome to PF!

Hi Alphaboy! Welcome to PF!

(type tex instead of itex, and it comes out bigger )

Homework Statement

Hi

I need to solve the following:

$$\frac{d^{2} x}{dt} + 2 b \frac{dx}{dt} + (b^2 - \frac{db}{dt}) x = 0$$

2. The attempt at a solution

Isn't this equivalent to??::

[tex]x^{''} + 2bx^{'}+ (b^2 - b^{'})x = 0 \iff x^2 + 2x +1 = 0[/itex] If I let b = 1.

which gives me the solution $$\lamda = -1$$

Thus giving me the particular solution

$$x(t) = c \cdot e^{-t}$$

Have I understood correctly? There is no precondition regarding either t or b.

Yes … technically , that's correct …

for any constant b, x = C e^-bt is a solution.

But where does that get you?

Is this a part of some bigger problem?

 

[Alphaboy2001]

Hello tiny-tim,

I was referred to this forum by a friend and then I got stuck with the bigger problem which will follow below, I decided to check this forum out :)

by the way there is typoo is my original equation:

$\frac{d^{2}x}{dt^2} - 2b \frac{dx}{dt} + (b^2 - \frac{db}{dt})x = 0$
(the right one)

Here is the greater problem:

Let $\phi \subset \mathbb{R}$ be an open interval, and $b \in C^{1}(\phi; \mathbb{C})$ where $C^{1}$ is the area on the solutions for

$\frac{d^{2}x}{dt^2} - 2b \frac{dx}{dt} + (b^2 - \frac{db}{dt})x = 0$


(1) Show that if $(\phi, x)$ solves the equation $\frac{dx}{dt} - bx = 0$ then $x \in C^{2}(\phi;\mathbb{C})$ thus $(\phi,x)$ is the solution of the original equation.

I invision the following way of showing this:

In for the solution x to belong to the greater Disk $C^2$ and show that the equation $\frac{dx}{dt} - bx = 0$ have a commen solution Disk then:

$\frac{dx}{dt} - bx = 0 \iff \frac{dx}{dt} = bx$ and by substituting this into

$(bx)^{''} - 2b (bx)^{'} + (b^2 - \frac{db}{dt})x = 0$

then I re-write the above to become

$(bx)^2-2b^2 \cdot x -1 = 0$ which be assuming b = 1

$x^2-2 \cdot x -1 = 0$ which gives the solution

$\lambda = \pm (\sqrt{2} \pm 1)$

Then the two equation mentioned originally must have their solution within the same Disk and thusly $(\phi,x)$ defines the solution for both
$(bx)^{''} - 2b (bx)^{'} + (b^2 - \frac{db}{dt})x = 0$

and

$\frac{dx}{dt} - bx = 0$

(2)

Finding the particular solution for
$\frac{d^{2}x}{dt^2} - 2b \frac{dx}{dt} + (b^2 - \frac{db}{dt})x = 0$

being

$\lambda = \pm (\sqrt{2} \pm 1)$

thus

$$x(t) = (C_{1}+C_{2}) \cdot e^{\pm (\sqrt{2} \pm 1) \cdot (t)}$$

Mister Tim have caught the essentials now?

There is one more question in the entire problem, but I leave that out for now.

Then $C_{1} = \frac{1}{e^{\sqrt{2} +1}}$ and $C_{2} = \frac{1}{e^{-\sqrt{2} - 1}}$

if and only if t = 1

$$x(t) = (\frac{1}{e^{\sqrt{2} +1}} + \frac{1}{e^{-\sqrt{2} - 1}}) \cdot e^{\pm (\sqrt{2} \pm 1) \cdot (t)} = 0$$

How does this look, this the complete solution for the first equation isn't it?

best regards
Alphaboy

 

[tiny-tim]

Hello Alphaboy!

(please don't do fractions in itex … they're too small! )

$\phi \subset \mathbb{R}$ be an open interval, and $b \in C^{1}(\phi; \mathbb{C})$ where $C^{1}$ is the area on the solutions for

$$\frac{d^{2}x}{dt^2} - 2b \frac{dx}{dt} + (b^2 - \frac{db}{dt})x = 0$$

(1) Show that if $(\phi, x)$ solves the equation $$\frac{dx}{dt} - bx = 0$$ then $x \in C^{2}(\phi;\mathbb{C})$ thus $(\phi,x)$ is the solution of the original equation.

I don't follow what you've done …

you can't put b = 1.

All you need is, if x' = bx,

then x'' = b'x + bx' = b'x + b²x,

so x'' - 2bx' + (b² - b')x = … ?

 

[Alphaboy2001]

Hello Alphaboy!

(please don't do fractions in itex … they're too small! )


I don't follow what you've done …

you can't put b = 1.

All you need is, if x' = bx,

then x'' = b'x + bx' = b'x + b²x,

so x'' - 2bx' + (b² - b')x = … ?

Sorry Mr. Tiny Tim I made such a stupid error in the post above that I should be send to Guantanamo :sad:

Anyway using Your calculations

x'' - 2bx' + (b² - b')x = bx' + b^2x - 2bx' + b^2x - bx' = 0

Putting this into order.

2bx^2 - 2bx' = 0

and since

x' = bx

Then

2bx^2 -2b(bx) = 0

Therefore 2bx^2 - 2bx^2 = 0 <-> 0 = 0

Thusly using terminology put forward in my task.

Since describes the solution (phi, x) for x' - bx = 0 and then x' = bx as proved above is also a solution forx''(t) - 2bx(t) + (b^2 - b'(t))x = 0 then $$x \in C^2(\phi; \mathbb{C})$$

connecting this with part(2)

Find the general solution for for

x''(t) - 2bx(t) + (b^2 - b'(t)) = 0

Which must imply that any for any belonging to C^1

then

x''(t) - 2bx(t) + (b^2 - b'(t))x = 0 <-> x^2 - 2bx - (b^2-b) = 0

thus the solution being

$$\lambda = \frac{2bx \pm \sqrt{(-2bx)^2 - 4 \cdot - (b^2 -b)}}{2}$$

which gives the the solutions $$\lambda = \pm 2bx$$

Then the general solution must be

$$x(t) = C_1 \cdot e^{2bx} + C_2 \cdot e^{-2bx}$$

This must be it? Isn't it?

Final question

Given the equation unhomogenous equation

$$x''(t) - 2bx(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} (b(s) ds)$$

solve it then x(t_0) = x'(t_0) = 0

I know that $$x^2 - 2bx + (b^2-b) = e^{t - t_0}$$

Which gives us

$$x(t) = C_1 \cdot e^{2bx} + C_2 \cdot e^{-2bx} + \ldots$$

Don't I substitute (t-t0) for X in the original differential equation in order to obtain for the inhomogenous version?

Alphaboy

Best Regards
Alphaboy

 

[tiny-tim]

Hi Alphaboy!

Sorry Mr. Tiny Tim I made such a stupid error in the post above that I should be send to Guantanamo :sad:

(Guantanamo Bay, actually … Guantanamo itself is a rather nice Cuban fishing village … the song "Guantanamera" means "girl from Guantanamo" )

But it's ok … I reported you to the Department of Homeland Security, and they say it didn't actually endanger the free world …

but they are keeping an eye on you …

connecting this with part(2)

Find the general solution for for

x''(t) - 2bx(t) + (b^2 - b'(t)) = 0
…
x''(t) - 2bx(t) + (b^2 - b'(t))x = 0 <-> x^2 - 2bx - (b^2-b) = 0

Why are you going back to the original equation?

The question itself tells you exactly what to do … solve x' = bx.

Assuming that b(t) and x(t) are independent (you didn't say), that means:

x'(t)/x(t) = b(t), so … ?

(though I must confess I don't yet see how that gives you all the solutions )

 

[Alphaboy2001]

Hi Alphaboy!


(Guantanamo Bay, actually … Guantanamo itself is a rather nice Cuban fishing village … the song "Guantanamera" means "girl from Guantanamo" )

But it's ok … I reported you to the Department of Homeland Security, and they say it didn't actually endanger the free world …

but they are keeping an eye on you …



Why are you going back to the original equation?

I thought that I was suppose

The question itself tells you exactly what to do … solve x' = bx.

I don't know if reason why we are talking past each other (I fear) is that my professor maybe is using his own reasoning (like talking about solutions for diff. eqn belong to their own respective sets??)))

Anyway trying to solve x' = bx I get that $$x(t) = e^{b(t)} + C$$ and since $$x \in C^{1}(\phi; \mathbb{C})$$ and that $$b \in C^{2}(\phi;\mathbb{C})$$

then since $$x' - bx = 0$$ gives a solution where b and x both are in C^1 and C^2, then those must also describe the solution for both x'(t) - bx = 0 and

x'' -2bx' + (b^2 - b')x = 0

?

Assuming that b(t) and x(t) are independent (you didn't say), that means:

x'(t)/x(t) = b(t), so … ?

Their curves cross each other?

(though I must confess I don't yet see how that gives you all the solutions )

If I am not dealing with a specific 'b' how else to I get the general solution for the original eqn?

Sincerely a confusses Alphaboy

 

[tiny-tim]

Hi Alphaboy!

… Anyway trying to solve x' = bx I get that $$x(t) = e^{b(t)} + C$$

No, it's $$x(t) = C e^{\int b(t) dt}$$

 

[Alphaboy2001]

Hi Alphaboy!


No, it's $$x(t) = C e^{\int b(t) dt}$$

Hello again Tim,

This is then the solution for x'(t) - bx = 0 don't I need to show here too that this is also the solution for

$$x''(t) - 2bx'(t) + (b^2- b'(t))x = 0*$$?? Sorry for the stupid question now. Is this then done by using the fact that there is analogy between every $$b \in C^{1}$$ and every possible x in $$C^2$$?

Is the solution for for x' - bx = 0 connected to finding the general solution for (*)??

Best Regards
Alphab

 

[tiny-tim]

This is then the solution for x'(t) - bx = 0 don't I need to show here too that this is also the solution for

$$x''(t) - 2bx'(t) + (b^2- b'(t))x = 0*$$?? Sorry for the stupid question now. Is this then done by using the fact that there is analogy between every $$b \in C^{1}$$ and every possible x in $$C^2$$?

No … you don't need to prove anything … you've already done it in part (1).

Is the solution for for x' - bx = 0 connected to finding the general solution for (*)??

Yes … it is the general solution (or at least a sort of sub-general solution, see below …)

Perhaps I'm missing something, but the hint in the question seems only to prove that every solution of x' = bx is a solution of the main equation, yet not the other way round.

So it produces a "family" of solutions … but not, I think, the most general family.

 

[Alphaboy2001]

No … you don't need to prove anything … you've already done it in part (1).


Yes … it is the general solution (or at least a sort of sub-general solution, see below …)

Perhaps I'm missing something, but the hint in the question seems only to prove that every solution of x' = bx is a solution of the main equation, yet not the other way round.

So it produces a "family" of solutions … but not, I think, the most general family.

What I don't get about question (2) Which says Find the general solution of (*) Isn't that in fact part of question(1) in the answer for question(2)?

...seems only to prove that every solution of x' = bx is a solution of the main equation, yet not the other way round.

So it produces a "family" of solutions … but not, I think, the most general family.

Is there a specific theorem here what I use to prove that

(2: answer) Such that The generalized solution of (*) is supposedly $$x(t) = C\cdot e^{\int b(t) dt}$$? because describes the 'family of solutions for (*) since b and x are members of the same family $$(\phi; \mathbb{C})$$??



Thank You for all Your help

Best Regards
Alpha

 

[tiny-tim]

What I don't get about question (2) Which says Find the general solution of (*) Isn't that in fact part of question(1) in the answer for question(2)?

(2: answer) Such that The generalized solution of (*) is supposedly $$x(t) = C\cdot e^{\int b(t) dt}$$?

Well, that's what I don't get either …

as I said at the end of my last post, C e^B(t), where B' = b, seems to be only one family of at least two families of a general solution, and I can't yet see any way of getting the other family.

(Don't forget, you haven't dealt with part (3) yet:

x'' - 2bx' + (b2 - b')x = e^B(t))

 

[Alphaboy2001]

Well, that's what I don't get either …

as I said at the end of my last post, C e^B(t), where B' = b, seems to be only one family of at least two families of a general solution, and I can't yet see any way of getting the other family.

(Don't forget, you haven't dealt with part (3) yet:

x'' - 2bx' + (b2 - b')x = e^B(t))

But what about limit on b(t)?? Since the original (3) says find the solution for

$$x''(t) - 2bx(t) + (b^2 - b'(t)) = e^{\int_{t_0}^{t} b(s) ds}$$

then $$x(t_0) = x'(t_0) = 0$$

Then $$x_{p}(t) = e^{B(t)} - e^{B(t_0)}$$

Is the solution here for (3) then

$$x(t) = C \cdot e^{\int b(t) dt} +x_{p}(t) = e^{B(t)} - e^{B(t_0)}$$?

Best Regards
Alphaboy

 

[tiny-tim]

But what about limit on b(t)?? Since the original (3) says find the solution for

$$x''(t) - 2bx(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} b(s) ds}$$

then $$x(t_0) = x'(t_0) = 0$$

Then $$x_{p}(t) = e^{B(t)} - e^{B(t_0)}$$

Is the solution here for (3) then

$$x(t) = C \cdot e^{\int b(t) dt} +x_{p}(t) = e^{B(t)} - e^{B(t_0)}$$?

No, that doesn't work at all …

you need to find a "particular solution" for $$x''(t) - 2bx(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} b(s) ds}$$ …

which is basically guesswork …

just spend a few hours trying out guesses!

 

[Alphaboy2001]

No, that doesn't work at all …

you need to find a "particular solution" for $$x''(t) - 2bx(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} b(s) ds}$$ …

which is basically guesswork …

just spend a few hours trying out guesses!

Hi again,

If I set x t to be $$x = e^{\int b(s) \ ds}$$ ?? what do I regarding the bound on rhs of inhomogenous equation??

since x(t_0) - x'(t_0) = 0 then x(t_0) - x'(t_0) = 0??

remove it??

Sincerely Alphaboy

 

[tiny-tim]

Hi again,

If I set x t to be $$x = e^{\int b(s) \ ds}$$ ?? what do I regarding the bound on rhs of inhomogenous equation??

since x(t_0) - x'(t_0) = 0 then x(t_0) - x'(t_0) = 0??

remove it??

Hello Alphaboy!

I'm sorry … I'm not following you at all …

the question gives the bounds (for the ∫) of t₀ and t …

if you change the lower bound from t₀ to t₁, say, then that's the same as multiplying by

$$C\ =\ e^{-\int_{t_0}^{t_1} b(s)\,ds}$$

your next line just repeats itself …

and remove what?

 

[Alphaboy2001]

Hello Alphaboy!

I'm sorry … I'm not following you at all …

the question gives the bounds (for the ∫) of t₀ and t …

if you change the lower bound from t₀ to t₁, say, then that's the same as multiplying by

$$C\ =\ e^{-\int_{t_0}^{t_1} b(s)\,ds}$$

your next line just repeats itself …

and remove what?

Hello again tim, and thanks for your answer.

I meant the bound, sorry I am messed up.

Anyway if I choose $$x = e^{\int b(t) \ ds}$$ and insert this to obtain the expression

$$(e^{\int b(t) \ ds})^2 -2b(e^{\int b(t) \ ds}) + (b^2 - b) \cdot (e^{\int b(t) \ ds}) = \ e^{-\int_{t_0}^{t_1} b(s)\,ds}$$

But don't I need to derive b for a complete solution?

then condition for the eqn is x(t_0) = x'(t_0) =0...

Best Regards
Alphaboy

 

[tiny-tim]

Anyway if I choose $$x = e^{\int b(t) \ ds}$$ and insert this to obtain the expression

$$\left(e^{\int b(t) \ ds}\right)^2 -2b\left(e^{\int b(t) \ ds}\right) + (b^2 - b) \cdot \left(e^{\int b(t) \ ds}\right) = \ e^{-\int_{t_0}^{t_1} b(s)\,ds}$$

But the LHS is zero, isn't it?

 

[Alphaboy2001]

But the LHS is zero, isn't it?

Hi Tiny Tim,

Its here that I am confused. Because if I set the RHS to zero, then I the homogenous solution from (2) and solution for inhomogenous system?

If $$e^\int(b(s) ds)$$ is a solution for the homogenous system. How do I find solution for the inhomogenous without having suitable b?

since the inhomogenous system being

$$x''(t) - 2b x'(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} b(s) \ ds$$ which needs to be solved. And if I reuse our solution for (2) and our choosen x = ... which I insert in place of x in the inhomogenous equation. But then I arrive a equation where rhs is still dependent the interval [t0,t] and lhs is only dependent on t ?

Sincerely Alphaboy

 

[tiny-tim]

Its here that I am confused. Because if I set the RHS to zero, then I the homogenous solution from (2) and solution for inhomogenous system?

If $$e^\int(b(s) ds)$$ is a solution for the homogenous system. How do I find solution for the inhomogenous without having suitable b?

since the inhomogenous system being

$$x''(t) - 2b x'(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} b(s) \ ds$$ which needs to be solved. And if I reuse our solution for (2) and our choosen x = ... which I insert in place of x in the inhomogenous equation. But then I arrive a equation where rhs is still dependent the interval [t0,t] and lhs is only dependent on t ?

Hi Alphaboy!

I don't know why you're worrying about the RHS being "dependent the interval [t₀,t]".

If you write $$B(t)\ =\ \int_{t_0}^{t} b(s) \ ds$$, then it's just

$$x''(t) - 2b x'(t) + (b^2 - b'(t))x = e^{B(t)}\ \text{with}\ B' \ =\ b$$

(or $$x''(t) - 2B' x'(t) + (B'^2 - B''(t))x = e^{B(t)}$$)

with no interval to worry about

You now need a "particular solution" …

I must admit I can't yet see one …

just make intelligent guesses, starting with things like x = te^B(t), and then trying the general substitution x = y(t) e^B(t)

 

[tiny-tim]

ping!

ah … I see how to do it now … and also how to prove that it's the most general solution …

definitely try the general substitution x(t) = y(t) e^B(t)