# Give general solution to differential equation

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fuzzybear

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#1

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?

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?

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lazy_fish

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#2

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#2

Can solve as a homogeneous equation with constant coefficients.

Rewrite in form

Solve auxillary equation for m of form

There should be two solutions. If we call them and .

Then general equation is where and are arbitrary constants.

Good luck.

Rewrite in form

Solve auxillary equation for m of form

There should be two solutions. If we call them and .

Then general equation is where and are arbitrary constants.

Good luck.

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fuzzybear

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#3

(Original post by

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.

Need to learn to use Latex...

**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.

Need to learn to use Latex...

- 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?

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lazy_fish

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#4

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#4

Homogeneous equations are such that all terms contain the dependent variable or derivatives of it. In your case, the dependent variable is . When you are only dealing with second order differentials, they have the form:

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:

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:

.

Good luck. Hope this helps.

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:

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:

.

Good luck. Hope this helps.

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fuzzybear

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#5

(Original post by

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,

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.

**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.

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

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lazy_fish

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

Good luck.

Good luck.

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james22

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#7

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#7

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.

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atsruser

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#8

(Original post by

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?

**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?

is then separable and can be solved for and then for

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Mr M

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#9

(Original post by

...

**fuzzybear**)...

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#10

(Original post by

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.

**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.

**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

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#11

(Original post by

If you introduce a new variable then and by thinking of as a composite function then we have:

is then separable and can be solved for and then for

**atsruser**)If you introduce a new variable then and by thinking of as a composite function then we have:

is then separable and can be solved for and then for

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?

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#12

(Original post by

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.

**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.

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Mr M

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#14

(Original post by

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.

**fuzzybear**)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.

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atsruser

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#15

The next step is:

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 which has sin/cos solutions. (It's the equation for SHM).

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srikanthsrnvs123

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#16

can somone tell me if this is core 4?

im ****ting myself if ive gotta know this for next months a2s

im ****ting myself if ive gotta know this for next months a2s

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guyag

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#18

(Original post by

I'm not sure what you mean by "seems unlikely".

**atsruser**)I'm not sure what you mean by "seems unlikely".

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james22

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#19

(Original post by

Yes. Didn't think it would be expected of an A-level student but then again I'm not familiar with the FP modules

**lazy_fish**)Yes. Didn't think it would be expected of an A-level student but then again I'm not familiar with the FP modules

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Mr M

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(Original post by

can somone tell me if this is core 4?

im ****ting myself if ive gotta know this for next months a2s

**srikanthsrnvs123**)can somone tell me if this is core 4?

im ****ting myself if ive gotta know this for next months a2s

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