Give general solution to differential equation

I'm asked to give the general solution to this differential equation:

With all the other previous equation where it was just a first order derivative (i.e. dx/dt = 6x), it was jusr a simple matter of getting the x's and t's are separate sides (which I think is what you're supposed to do, right?), and then integrating both sides, so with dx/dt = 6x:

(1/6x) dx = dt

becomes:

(1/6)lnx + constant = t + constant

which is then simple to re-write and get: x = .......

I tried the same method for the d^2 x / dt^2 = 8x, but it doesn't seem to work. How do you find the general solution?

(edited 9 years ago)

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Reply 1

lazy_fish

Can solve as a homogeneous equation with constant coefficients.
Rewrite in form

a\frac{d^2x}{dt^2}+b\frac{dx}{dt}+cx=0

Solve auxillary equation for m of form

am^2+bm+c=0

There should be two solutions. If we call them

d

and

f

.
Then general equation is

x(t)=Ae^{dt}+Be^{ft}

where

A

and

B

are arbitrary constants.

Good luck. :smile:

(edited 9 years ago)

Reply 2

fuzzybear

Original post by lazy_fish

Can solve as a homogeneous equation with constant coefficients.
Rewrite in form a*d2x/dt2 +b*dx/dt +c*x = 0
Solve auxillary equation for m of form a*m^2 +b*m +c = 0
There should be two solutions. If we call them d and f.
Then general equation is x = A*e^(d*t) +B*e^(f*t) where A and B are arbitrary constants.

Good luck. :smile:

Need to learn to use Latex...

a few questions,

- how can you just assume its a homogeneous equation rather than a non-homogeneous?

- I don't quite understand the part, ''Solve auxillary equation for m of form a*m^2 +b*m +c = 0'', what is an auxillary equation?

- so was my method of having the x's and t's on separate sides and then integrating each side (in the OP), wrong?

(edited 9 years ago)

Reply 3

lazy_fish

Homogeneous equations are such that all terms contain the dependent variable or derivatives of it. In your case, the dependent variable is

x

. When you are only dealing with second order differentials, they have the form:

a\frac{d^2x}{dt^2}+b\frac{dx}{dt}+cx=0

where a,b and c are constants.

After placing our differential equation into this form, we can take from it an auxiliary equation. That is an equation that uses the constants a, b and c and puts them into a different quadratic equation of form:

am^2+bm+c=0

Then we can simply solve for m using whichever method you prefer.

Yeah. I am afraid you cannot use that method to solve second order differential equations. Or at least not directly. They are for solving those of the form:

\frac{dy}{dx}=g(x) h(y)

.

Good luck. Hope this helps. :biggrin:

(edited 9 years ago)

Reply 4

fuzzybear

Original post by lazy_fish

Homogeneous equations are such that all terms contain the dependent variable or derivatives of it. In your case, the dependent variable is x. When you are only dealing with second order differentials, they have the form a*d2x/dt2 +b*dx/dt +c*x = 0 where a,b and c are constants.

After placing our differential equation into the form a*d2x/dt2 +b*dx/dt +c*x = 0, we can take from it an auxiliary equation. That is an equation that uses the constants a, b and c and puts them into a different quadratic equation of form a*m^2 +b*m +c = 0. Then we can simply solve for m using whichever method you prefer.

Yeah. I am afraid you cannot use that method to solve second order differential equations. Or at least not directly. They are for solving those of the form dy/dx = g(x) h(y).

Good luck. Hope this helps. :biggrin:

what if I had a first order or third order differential to begin with?

would the auxiliary equation be a linear for first order, and cubic for third order?

Reply 5

lazy_fish

Pretty much. Although homogeneous equations mean a different thing for first order differential equations.

Good luck. :smile:

(edited 9 years ago)

Reply 6

james22

When dealing with second order differential equations you cannot just multiply by dx or dt, as you would for first order ones. Have you been shown how to solve second order equations yet? If so, then start there.

Reply 7

atsruser

Original post by fuzzybear

I tried the same method for the d^2 x / dt^2 = 8x, but it doesn't seem to work. How do you find the general solution?

If you introduce a new variable

u=\frac{dx}{dt}

then

\frac{du}{dt} = \frac{d^2x}{dt^2}

and by thinking of

u

as a composite function

u = u(x(t))

then we have:

\frac{d^2x}{dt^2} = \frac{du}{dt} = \frac{du}{dx}\frac{dx}{dt} = u\frac{du}{dx} = 8x

u\frac{du}{dx}=8x

is then separable and can be solved for

u

and then for

x

Reply 8

Mr M

Study Forum Helper

Original post by fuzzybear

...

If this is not an A2 Further Maths question then you probably have not shared the whole thing with us. I think Atsruser has probably hit on the expected approach.

Reply 9

fuzzybear

Original post by james22

yeah, but for the other second order equations I've come across, the right hand side contains the bottom variable of the dx/dt part, for example

d^2 x / dt^2 = t^2 - 2t

with this, I just differentiate the right hand side to get 2t - 2, and then for the left hand side, I reduce it down to a first order differential: dx/dt. And its relatively simple

but when I'm given a second order equation in which the right hand side contains the top variable of the the dx/dt part, like in the question of this thread, I'm stuck

Reply 10

fuzzybear

Original post by atsruser

If you introduce a new variable

u=\frac{dx}{dt}

then

\frac{du}{dt} = \frac{d^2x}{dt^2}

and by thinking of

u

as a composite function

u = u(x(t))

then we have:

\frac{d^2x}{dt^2} = \frac{du}{dt} = \frac{du}{dx}\frac{dx}{dt} = u\frac{du}{dx} = 8x

u\frac{du}{dx}=8x

is then separable and can be solved for

u

and then for

x

separating the u's and x's on different sides, and then integrating, I get:

(u^2 / 2) + constant = (8x^2 / 2) + constant

what next?

Reply 11

fuzzybear

Original post by Mr M

If this is not an A2 Further Maths question then you probably have not shared the whole thing with us. I think Atsruser has probably hit on the expected approach.

Its in one of the non-official A level textbooks, the whole question is pretty much as I've put on the OP. It also asks to state how many arbitrary constants you expect to find in the general solution and whether the number is expected from the given differential equation (because you usually have one arbitrary constant for a first order diff equation, two arbitary constants for second order diff, and so on...). But this is not what I'm having trouble with so I didn't include it in the OP.

Reply 12

lazy_fish

If you go down this route:

\displaystyle u \frac {du}{dx} = 8x \Rightarrow u^2 = (\frac{dx}{dt})^2 = 8x^2 + c \Rightarrow \frac{dx}{dt} = \pm \sqrt {8x^2 + c}

Seems unlikely. :s-smilie:

Reply 13

Mr M

Study Forum Helper

Original post by fuzzybear

As I said before this can only be solved using techniques from beyond Core 4. If you want us to make absolutely sure that you don't need to know this you need to confirm the examinations you are studying for and provide the complete question without amendments, omissions or alterations (scanning the page of the textbook would help).

Reply 14

atsruser

Original post by lazy_fish

If you go down this route:

\displaystyle u \frac {du}{dx} = 8x \Rightarrow u^2 = (\frac{dx}{dt})^2 = 8x^2 + c \Rightarrow \frac{dx}{dt} = \pm \sqrt {8x^2 + c}

Seems unlikely. :s-smilie:

I'm not sure what you mean by "seems unlikely". Unlikely that the OP would have to perform the second integral? If so, possibly, but that's the correct result regardless.

The next step is:

\int dt = \pm \frac{1}{2\sqrt{2}} \int \frac{dx}{\sqrt{x^2+k^2}} \Rightarrow t = \pm \frac{1}{2\sqrt{2}} \sinh^{-1}(x/k) + d

which I think is about all we can say without knowing more about the initial conditions. This type of integral only comes up in the FP syllabus, though.

It's just occurred to me though that the question may make more sense if it was meant to be