- #1
Saladsamurai
- 3,020
- 7
I have always gone along with this, but now I am not so sure:
Take the DE:
[tex]y' +4y = 0[/tex]
If you carry out the integration, we get
[tex]y = e^{-4x} +C\qquad(1)[/tex]
but we always pull some stuff like "well since we could write C as ec1, since it's arbitrary", we can write the solution like
[tex]y = c_1e^{-4x}\qquad(2)[/tex].
I am not buying that they are the same. Yes, I know... the solution in (2) satisfies the original DE. Here is my problem, (1) gives a family of curves that differ by a vertical shift whereas (2) gives a family of curves that differ by a "stretch" or compression about the y-axis. How can these be the same thing?
So why do I bring this up anyway? I was doing the following initial value problem
[tex]y' = y^2 + y -6 \qquad(3)[/tex]
I got a solution of
[tex]\ln\frac{y-2}{y+3} = 5x + e^{C}\qquad(4)[/tex]
you can see that I used "the trick" in (4), letting co =eC so that when I exponentiate, I get a nice
[tex]y(x) =\frac{3Ce^{5x}+2}{1-Ce^{5x}}\qquad(5)[/tex]
now comparing this to the book solution, this time they did not use the trick and got
[tex]y(x) =\frac{3e^{5x + c}+2}{1-e^{5x+c}}\qquad(6)[/tex]
Naturally, I figured that (5) and (6) are the same. But when I go to use the initial value of y(5) = 10, if I use my version (5), I get that C is virtually zero since it is given from
[tex]10 - 10Ce^{25} = 3Ce^{25} +2\qquad(7)[/tex]
However, the book plugs it into (4) of all god damn places and gets that C = -25.
What am screwing up conceptually here?
Take the DE:
[tex]y' +4y = 0[/tex]
If you carry out the integration, we get
[tex]y = e^{-4x} +C\qquad(1)[/tex]
but we always pull some stuff like "well since we could write C as ec1, since it's arbitrary", we can write the solution like
[tex]y = c_1e^{-4x}\qquad(2)[/tex].
I am not buying that they are the same. Yes, I know... the solution in (2) satisfies the original DE. Here is my problem, (1) gives a family of curves that differ by a vertical shift whereas (2) gives a family of curves that differ by a "stretch" or compression about the y-axis. How can these be the same thing?
So why do I bring this up anyway? I was doing the following initial value problem
[tex]y' = y^2 + y -6 \qquad(3)[/tex]
I got a solution of
[tex]\ln\frac{y-2}{y+3} = 5x + e^{C}\qquad(4)[/tex]
you can see that I used "the trick" in (4), letting co =eC so that when I exponentiate, I get a nice
[tex]y(x) =\frac{3Ce^{5x}+2}{1-Ce^{5x}}\qquad(5)[/tex]
now comparing this to the book solution, this time they did not use the trick and got
[tex]y(x) =\frac{3e^{5x + c}+2}{1-e^{5x+c}}\qquad(6)[/tex]
Naturally, I figured that (5) and (6) are the same. But when I go to use the initial value of y(5) = 10, if I use my version (5), I get that C is virtually zero since it is given from
[tex]10 - 10Ce^{25} = 3Ce^{25} +2\qquad(7)[/tex]
However, the book plugs it into (4) of all god damn places and gets that C = -25.
What am screwing up conceptually here?