# First order differentials

## Homework Statement

Solve: (2t+x) dx/dt + t = 0

## Homework Equations

y' +p(X)y = q(x)
and y(x) = ($$\int$$u(x)q(x) + c)/u(x)
where u(x) = e$$\int$$p(x)dx
Note this u(x) is 2 to the power of the integral of p(x)

## The Attempt at a Solution

(2t+x) dx/dt + t = 0 becomes:
dx/dt + t/(2t+x) = 0 by dividing through by (2t+x)

However, now I don't know how to separate t and x in the second term.

## Answers and Replies

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It looks like it is first-order homogeneous. Make the substitution z=x/t by dividing by t throughout your t/(2t+x), and then x=tz and x'=z+tz' by the product rule. Now it's in a form you want it for the relevant equations you listed. Just make the substitution the other way when you've solved for z.

HallsofIvy
Homework Helper

## Homework Statement

Solve: (2t+x) dx/dt + t = 0

## Homework Equations

y' +p(X)y = q(x)
and y(x) = ($$\int$$u(x)q(x) + c)/u(x)
where u(x) = e$$\int$$p(x)dx
Note this u(x) is 2 to the power of the integral of p(x)
These apply only to a linear first order d.e. And this d.e. is NOT linear because of the x multiplying dx/dt.

## The Attempt at a Solution

(2t+x) dx/dt + t = 0 becomes:
dx/dt + t/(2t+x) = 0 by dividing through by (2t+x)

However, now I don't know how to separate t and x in the second term.

Okay so does that mean that what jeffreydk would not work? If so what can i use instead? I have only been taught about linear first order d.e so really don't know what to do!

Yea it is non-linear I didn't notice that. What I said wont work.

HallsofIvy