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

just want to make sure im doing the right thing

ty

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- May 3rd 2009, 02:21 AM #1

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- May 3rd 2009, 02:45 AM #2
Hello,

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

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

Therefore solutions of this differential equation are :

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

- May 3rd 2009, 04:10 AM #3

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Hello, Charbel!

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

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

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

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

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

- May 3rd 2009, 04:48 AM #4

- May 3rd 2009, 06:15 AM #5

- May 3rd 2009, 07:47 AM #6

- May 3rd 2009, 08:13 AM #7

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- May 3rd 2009, 10:13 AM #8

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