Nicholas Hunter

2022-11-21

Is $(2x+y)dx-xdy=0$ a separable differential equation?

I was given the following differential equation in an assignment the other day:

$(2x+y)dx-xdy=0$

The problem specified to solve the equation using the method of separation of variables. My problem was setting the integral, I tried multiple manipulations with but nothing seemed to work. So, I have to ask can this equation be solved using separation of variables?

I was given the following differential equation in an assignment the other day:

$(2x+y)dx-xdy=0$

The problem specified to solve the equation using the method of separation of variables. My problem was setting the integral, I tried multiple manipulations with but nothing seemed to work. So, I have to ask can this equation be solved using separation of variables?

Liehm1mm

Beginner2022-11-22Added 13 answers

Write it as

${y}^{\prime}-y/x=2$

then use the change of variables

$y=xu\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}=x{u}^{\prime}+u$ to transformed it to the separable ode

$x{u}^{\prime}=2$

which is the required method in the question!

${y}^{\prime}-y/x=2$

then use the change of variables

$y=xu\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}=x{u}^{\prime}+u$ to transformed it to the separable ode

$x{u}^{\prime}=2$

which is the required method in the question!

Annie French

Beginner2022-11-23Added 4 answers

The equation can be rewritten as

$\frac{dy}{dx}-\frac{1}{x}y=2$

That means it is not a separable equation. However, it is a first-order linear ordinary differential equation, so it can be solve by the usual means. You will end up with at least one "special function" however, since part of the solution needs $\int {e}^{-1/x}\phantom{\rule{thinmathspace}{0ex}}dx$ which cannot be expressed in elementary functions.

$\frac{dy}{dx}-\frac{1}{x}y=2$

That means it is not a separable equation. However, it is a first-order linear ordinary differential equation, so it can be solve by the usual means. You will end up with at least one "special function" however, since part of the solution needs $\int {e}^{-1/x}\phantom{\rule{thinmathspace}{0ex}}dx$ which cannot be expressed in elementary functions.

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