cepheid
Sep12-04, 12:47 PM
Hi,
Please bear with me, I've only had the first sort of "pseudo-lecture" in ordinary d.e.'s this past week, and I was doing some reading ahead. It occurred to me that if linear first-order differential equations are those that can be written in the general form:
\frac{dy}{dx} + P(x)y = Q(x)
and if I understood my prof's remark correctly that homogeneous linear first-order d.e.'s are those for which Q(x) = 0 , then all homogeneous first-order linear differential equations are actually separable because:
\frac{dy}{dx} + P(x)y = 0
\frac{dy}{dx} = -P(x)y
Which can be solved as follows:
\int{\frac{dy}{y}} = -\int{P(x)dx}
^^Looks separable to me. I just wanted to make sure I was getting the basics before I moved on. So is the thread title statement true then? Oh yeah, and I'm guessing that the converse is not always true right? A separable first-order d.e. is not necessarily a homogeneous linear d.e. right? After all, \frac{dy}{dx} = \frac{6x^2}{2y + cosy} is not linear in y right? But it looks like it could be homogeneous non-linear (if such a thing exists, I don't know???). In that case, would the statement homogeneous = separable be true in the most general sense? Moving on with the original example (I'm just solving while "LaTeXing" to see what comes out) ...
\ln|y| = -\int{P(x)dx}
|y| = e^{-\int {P(x)dx}}
Now, the most general solution for y must include the most general antiderivative, so we'll have a C stuck in there if and when we solve the integral:
y = \pm e^{-\int {P(x)dx} + C} = \pm e^{-\int {P(x)dx}}e^C = \pm Ae^{-\int {P(x)dx}}
Whoah, cool! So in a homogeneous first order d.e. the solution takes the form of some constant \pm A times the reciprocal of the integrating factor I(x) ?! Is this always true?
EDIT: Yeah actually I can see why that is. Take the usual statement about the integrating factor:
(I(x)y)^{\prime} = I(x)Q(x)
If Q(x) = 0 , then
(I(x)y)^{\prime} = 0
I(x)y = \pm A
y = \frac{\pm A}{I(x)}
Awesome! I discovered a lot more writing this post than I expected. I hope someone will correct me if I've made any errors.
Please bear with me, I've only had the first sort of "pseudo-lecture" in ordinary d.e.'s this past week, and I was doing some reading ahead. It occurred to me that if linear first-order differential equations are those that can be written in the general form:
\frac{dy}{dx} + P(x)y = Q(x)
and if I understood my prof's remark correctly that homogeneous linear first-order d.e.'s are those for which Q(x) = 0 , then all homogeneous first-order linear differential equations are actually separable because:
\frac{dy}{dx} + P(x)y = 0
\frac{dy}{dx} = -P(x)y
Which can be solved as follows:
\int{\frac{dy}{y}} = -\int{P(x)dx}
^^Looks separable to me. I just wanted to make sure I was getting the basics before I moved on. So is the thread title statement true then? Oh yeah, and I'm guessing that the converse is not always true right? A separable first-order d.e. is not necessarily a homogeneous linear d.e. right? After all, \frac{dy}{dx} = \frac{6x^2}{2y + cosy} is not linear in y right? But it looks like it could be homogeneous non-linear (if such a thing exists, I don't know???). In that case, would the statement homogeneous = separable be true in the most general sense? Moving on with the original example (I'm just solving while "LaTeXing" to see what comes out) ...
\ln|y| = -\int{P(x)dx}
|y| = e^{-\int {P(x)dx}}
Now, the most general solution for y must include the most general antiderivative, so we'll have a C stuck in there if and when we solve the integral:
y = \pm e^{-\int {P(x)dx} + C} = \pm e^{-\int {P(x)dx}}e^C = \pm Ae^{-\int {P(x)dx}}
Whoah, cool! So in a homogeneous first order d.e. the solution takes the form of some constant \pm A times the reciprocal of the integrating factor I(x) ?! Is this always true?
EDIT: Yeah actually I can see why that is. Take the usual statement about the integrating factor:
(I(x)y)^{\prime} = I(x)Q(x)
If Q(x) = 0 , then
(I(x)y)^{\prime} = 0
I(x)y = \pm A
y = \frac{\pm A}{I(x)}
Awesome! I discovered a lot more writing this post than I expected. I hope someone will correct me if I've made any errors.