Question

ODE's
Solve:

Answer 1

\[x (dy/dx) + y = 6xe^x\]

Answer 2

\[xy'+y=6xe^x\\
y'+\frac{1}{x}y=6e^x\]
You have here a linear first-order ODE. Find the integrating factor:
\[\mu(x)=e^{\int 1/x~dx}=e^{\ln|x|}=x\]
Multiply both sides of the ODE by \(\mu(x)\):
\[xy'+y=6xe^x\]
Note the LHS = \(\dfrac{d}{dx}[xy]\).
\[\frac{d}{dx}[xy]=6xe^x\\
xy=\int6xe^x~dx\\
~~~~~~~~~~\vdots\]

Answer 3

So there are two types of first order of ODE, right? Linear, and variable separable. How can I tell the difference?

Answer 4

They're not exactly two types of equations. For example,
\[xy'=1\]
is a linear, first order ODE, but it's also separable:
\[xy'=1~\Rightarrow~y'=\frac{1}{x}~\Rightarrow~dy=\frac{dx}{x}\]
Generally, a linear first order equation has the form
\[\large p(x)y'+q(x)y=r(x)\]
where \(p(x),q(x),\text{ and }r(x)\) are functions of the independent variable only (they can be 0, except for p(x), since that's what makes it an ODE in the first place).

The method used to solve these is the integrating factor, which I outlined above:
\[\large\mu(x)=e^{\int q(x)~dx}\]
A separable ODE has the form
\[p(y)q(x)y'=r(y)s(x)\]
(This is just my understanding; you might find a better explanation online.) Separable means you can move all the y's to one side and all the x's to the other:
\[\frac{p(y)}{r(y)}~dy=\frac{s(x)}{q(x)}~dx\]
An example would be
\[\sqrt{x^2-1}~(y-4)y'=e^x\]
You can separate the variables easily here:
\[(y-4)~dy=\frac{e^x}{\sqrt{x^2-1}}~dx\]
Does that kinda help?

Answer 5

Yeah, thanks man! Just need a little more practice I think. Thanks for the help.

Answer 6

You're welcome!