Deleted User's Maths Thread

[b^3]

I don't think it's every even set; for something which has sinh^4 x, no sign flip would occur; imagine it as sinh^2 x then squared; the sign flip is negated by the square.

Probably bad wording on my part. What I meant was, well for , we have two sets of , so we would flip the sign twice (flip the flip), ending up with what we started with. In other words, contains two implied products, so we'd get .

[Deleted User]

Thanks guys

What about tanh(x)sech²(x)? How would I go about integrating this?

antidiff [tanh(x)sech²(x) dx]
= antidiff [ sinh(x)/cosh³(x) dx]

And I'm not sure how to proceed from here. Do I use substitution and let u=cosh(x)?

[b^3]

We want to try and get rid of either the sinh or cosh, since we have sinh in the first power only, we will look for a u substitution that will allow us to cancel this out. (Note: That won't always work but it's a good place to try and start.)

Spoiler

 

[lzxnl]

Thanks guys

What about tanh(x)sech²(x)? How would I go about integrating this?

antidiff [tanh(x)sech²(x) dx]
= antidiff [ sinh(x)/cosh³(x) dx]

And I'm not sure how to proceed from here. Do I use substitution and let u=cosh(x)?

Well...derivative of tanh x = d/dx (sinh x/cosh x) = (cosh^2 x - sinh^2 x)/(cosh^2 x) = sech^2 x
Just sub u=tanh x. You should get 1/2*tanh^2 x + c
As 1 = cosh^2 x - sinh^2 x, sech^2 x = 1 - tanh^2 x, my solution differs from b^3's solution by a constant only.

[Deleted User]

Thanks!

How do I solve the differential equation dx/dt + 10*x/(20000-5*t) = 5?

[Hancock]

Is this for an assignment?

[lzxnl]

Thanks!

How do I solve the differential equation dx/dt + 10*x/(20000-5*t) = 5?

I know where this comes from. It's from one of those spesh questions with the outflow rate greater than that of the inflow rate.

So, note that this equation is of the form dx/dt + x*f(t) = constant
The left hand side, if multiplied by g(t), could be written as the derivative of x*g(t) if one restriction is met.
If d(x*g(t))/dt = g(t)*dx/dt + x*g'(t) = g(t)*dx/dt + x*f(t)*g(t), we can see that g'(t) = f(t)*g(t)
Here, f(t) = 10/(20000-5t)
So dg/dt = 10g/(20000-5t)
1/10g * dg/dt = 1/(20000-5t)
1/10 ln g = -1/5 ln(20000-5t)                            we don't need an integration constant as we just need a SINGLE function g(t) that              ln g = -2 ln(20000-5t)                                       satisfies the above DE
g=(20000-5t)^-2
So we can rewrite our equation as dx/dt + x*f(t) = c
g(t) dx/dt + x*f(t)*g(t) = c*g(t)
Using the fact that d(x*g(t))/dt = the left hand side by definition
d(x*g(t))/dt = c*g(t)
Integrating both sides with respect to t
x*g(t) = integral of c*g(t) dt            c=5 remember from the RHS of the question?
=integral of 5*(20000-5t)^-2=(20000-5t)^-1 + constant
g=(20000-5t)^-2 from above
so x=(20000-5t)+constant*(20000-5t)^2

I don't have the initial conditions so I can't find the constant. But yeah; this is how I would solve it. The aim is to rewrite the left hand side as a derivative of a product as we have a dx/dt + x*f(t) term on the right. My calculator is happy with my answer.

[Deleted User]

Thanks!

How do I know if the following improper integral converges or not?
Antidiff {1/(x*ln(x))^p dx} for the interval 1 to infinity where p is a positive real number.

[lzxnl]

Thanks!

How do I know if the following improper integral converges or not?
Antidiff {1/(x*ln(x))^p dx} for the interval 1 to infinity where p is a positive real number.

It will matter whether p is above or lower than 1. At p=1, the integral diverges as your integral becomes 1/(x*ln x) from 1 to inf, aka ln(ln(x)) between 1 and inf. Not defined.

However, if p is any smaller than one, you can rewrite the integral as 1/(x^p*ln^p x) dx<1/(x^p*x) as ln x < x for all x, and (ln x)^p< ln x as p<1. You can integrate (1/(x^(p+1)) for 0<p<1 between x=1 and x= infinity and the integral will converge. So by the comparison test, the integral converges for 0<p<1 (obviously if p=0 you get a divergent integral)

Correct me if I'm wrong somewhere; I lack the mathematical rigour of more seasoned mathematicians.

[Deleted User]

Mixing problem question here:

How do I find the initial condition of a differential equation?

If the differential equation is:

dx/dt = Q_inC - Q_out*x/[V₀+(Q_in-Q_out)*t], how would I find the initial condition?

[lzxnl]

Initially, how much x do you have? That's your initial condition; the question should state what x(0) is.

[Deleted User]

The question asks me to find the initial condition, it's not given.

[Thu Thu Train]

What is the full question? That is a bit strange for a question to ask you to find initial conditions