In dealing with differential equations

[Badgerspin]

Ok, I've been working this one out for a while, and I just can't seem to get it.

I'm looking for the condition such that x(t)=k (where in our class, K generally equals e^c)

The initial equation is as such:

7t^2(dx/dt)+3x+5=0

I've worked the equation down to two different forms:

x = (-e^((3/7t)-3c)-5)/3

and

x = c*e^(3/7t) - 3/5

For the life of me, I can't figure out what K (or C I suppose) equals in order so that x(t) = k

This is online work, and it only accepts a precise answer for credit.

It's also not letting me enter the variable t as an answer for c (or K), so I'm wondering if there's maybe an error in what was programmed as the correct answer.

I'm sure there's something I'm missing or another way that I need to work the problem. Any help would be appreciated.

 

[LCKurtz]

You didn't show all your steps but I assume one of your steps might have been:

$$\ln (3x+5) = \frac 3 {7t} + 3C$$

leading to

$$3x+5 = e^{\frac 3 {7t}+3C}$$

You would normally write this

$$3x+5 = e^{3C}e^{\frac 3 {7t}}$$

and rename the e^3C as K. Try that.

 

[Badgerspin]

It's still not letting me enter in a variable as a solution.

I worked it out as shown (thank you).

So:

k = (ke^(3/7t) - 5)/3

or working with k = e^c

((k^3)e^(3/7t)-5)/3

In either case, it's still not letting me enter in any variables. If there's a solution to this problem that's all constants, I can't seem to find it. The only solutions I can come up with all involve the variable t.

 

[fzero]

I'm a bit confused about what you're looking for. x(t)=k would only hold for some particular value of t, call it $$t_0$$ and you haven't stated such an initial condition $$x(t_0) =k$$. Has one been specified in the problem?

Secondly you refer to a "solution to this problem that's all constants," but you have already found one, which is the particular solution $$x_p = -5/3$$. This is of course different from an initial condition.

 

[Badgerspin]

fzero,

I punched in -5/3 and it said it was correct. When you say different initial conditions, what condition are you talking about? I'm not quite following that.

 

[fzero]

What I mean is when you have an equation y' = y and are given the initial condition y(0) =1. The general solution is y = c e^t, but the initial condition requires that c=1. In general you need an initial condition if you want to fix an integration constant.