1. ## Help =]

dx/dt + (x)/(20+t) = 3

just want to make sure im doing the right thing

ty

2. Hello,

$x'(t)+\frac{x(t)}{20+t}=0\Rightarrow \frac{x'(t)}{x(t)}=-\frac{1}{20+t}\Rightarrow \ln\left(\frac{x(t)}{C}\right)=-\ln(20+t)\Rightarrow x(t)=\frac{C}{20+t}$

And notice that $t\mapsto \frac{3}{2}(20+t)$ is solution.

Therefore solutions of this differential equation are :

$\color{blue}\boxed{x(t)=\frac{C}{20+t}+\frac{3}{2} (20+t)}$

3. Hello, Charbel!

$\frac{dx}{dt} + \frac{1}{t+20}\,x \:=\: 3$

Integrating factor: . $I \;=\;e^{\int\frac{dt}{t+20}} \;=\;e^{\ln(t+20)} \;=\;t+20$

We have: . $(t+20)\frac{dx}{dt} + x \:=\:3(t+20) \quad\Rightarrow\quad \frac{d}{dt}\bigg[(t+20)x\bigg] \:=\:3(t+20)$

Integrate: . $(t+20)x \;=\;3\left(\frac{t^2}{2} + 20t\right) + C \;=\;\frac{3}{2}(t^2 + 40t + C)$

Therefore: . $x \;=\;\frac{3(t^2 + 40t + C)}{2(t+20)}$

4. Our results are the same

5. Originally Posted by Infophile
Our results are the same
Yes but for linear first order ODE's, one typical goes the route of finding an integrating factor as Soroban did.

6. I've just learn it, in France we don't use this method, that's funny

7. Originally Posted by danny arrigo
Yes but for linear first order ODE's, one typical goes the route of finding an integrating factor as Soroban did.
One may, but one doesn't necessarily. Most people learn to integrate "separable equations" before learning about "integrating factors" for linear equations.

8. Originally Posted by Charbel
dx/dt + (x)/(20+t) = 3

just want to make sure im doing the right thing

ty
The best way to do what you say you want is for you to post what you have done for us to comment on.

CB