If I had to guess, I'd say E. (Given that it's late and I'm not checking my work on paper.)riggggghhhhhhht. I was working totally backwards. I just did the derivatives of the solution and got it. Answer is .... E I think, its either that one or D. Don't have the work in front of me, but the signs work on one of those.
I imagine that solution for the equation can be found from the information provided. It's just a matter of finding a general solution based on the relationship between the derivatives and then applying the boundary value to it.Do you know how to solve that equation? It's not too hard and if you are able to solve it then the solution to the limit is very easy, but I think you are supposed to do it without finding y(x)...
I'm just not sure how you'd get the answer without finding y(x).I know it can be done, it's actually pretty easy. But it's weird because it's such a jump in difficulty from the question in the OP. Also, the limit is trivial if you explicitly know y(x).
I think that's exactly what needs to be done. Technically, you still need to find C if they make a distinction between +∞ and -∞.Me neither.
So, how to solve it:
in equations of the type
diff eq=f(x), you can find a solution which is a linear combination of two solutions: the homogeneous solution, which is the solution of diff eq=0 (as in the OP), plus a particular solution found ad hoc so that the equality with f(x) is satisfied.
In this case, the homogeneous solution to y'-2y=0 is an exponential, y=e^Ax. to find A, we put the function into the equation: (A-2)*e^Ax=0 => A=2.=> y_h=exp(2*x)
The particular solution is very easy to find: since f(x) is a constant, it can just be a constant B the derivative of which is zero: 0-2B=4 => B=-2.
The general solution is, then, y(x)=C exp(2*x) - 2. To find C you only need to apply the initial condition. But you don't need to, you can already compute the limit from here (which makes me wonder if this is the correct answer or there is an easier one).
That type of equation, the relationship between derivatives, is part of the given for a differential equation problem. When you are analyzing a problem, sometimes it is not obvious what the actual function is, but a relationship between the function and one or more of its derivatives will be apparent.What I don't get is how you set y'-2y=0 like that.
Shit, I thought I had answered this, I'm sorry... If you look for int(1/sinx) instead of int(cscx) the tables will give you a simpler solution (ln(tan (x/2)) instead of ln(csc x + cot x))@ Fade- yep. You can spend hours trying to get the analytical solution for some differential equation, or just put it into the appropriate form, jump into excel (or some higher dimensional program like matlab depending), set your dx or dt or whatever to small, and run it. Sadly though, that's not a correct answer for my diff eq prof. Numerical methods for the win.
Anyways, hit a problem I am having some trouble with, thought yall could help again. I'm getting a lot better at this, but I am still rusty on my integral calc. Right now I am messing with linear diff eqs. The general form is :
y' + p(x) y = q(x)
here's the problem
y' + y * cot x = csc^2 x
p(x) = cot x
h(x) = ln(sin x)
e^h(x) = sin x
Multiply the starting equation by e^h(x) and get
sin x[y' + y cot x] = sin x * csc^2 x
[y*sin x]' = csc x = sin ^-1 x
now this is where I am wondering if I have done it right, because the integral for that is not simple:
y*sin x = integral(csc x) = - ln |csc x + cot x| + C
y = sin x * -ln(csc x + cot x) + C sin x
There has to be a more simple form of that. Any thoughts?