Homework assignment that I was able to solve using EDO, but need to use Laplace now

[Diego]

Homework Statement

Find the currents of the circuit after the switch S is closed.

Relevant Equations

Use laplace.

The statement is: Find the currents of the circuit after the switch S is closed. I solved it by EDO (ordinary differential equations) and then I tried to solve it by Laplace and they gave different answers.

 

[berkeman]

Welcome to PF.

What's EDO? Also, please attach a clear image of the circuit in JPEG format (not sideways like in the PDF) to help us understand the problem. Please show the work you've done so far, and say how you want to solve it using Laplace methods. Thanks.

(I'll send you a separate message with tips for how to use LaTeX to post math at PF.)

 

[Diego]

EDO is an abbreviation for ordinary differential equations.
I'll send what I've already done, by EDO. and then I will send as I tried to solve by Laplace and quote what I think is wrong.

 

[Orodruin]

EDO is an abbreviation for ordinary differential equations.
I'll send what I've already done, by EDO. and then I will send as I tried to solve by Laplace and quote what I think is wrong.

The English abbreviation is ODE (for obvious reasons). I am guessing your native language has Latin roots?

Regarding the actual question, please show your work.

 

[Diego]

I'm sorry for the delay, I was putting the pdf in English, if there is anything you don't understand you can ask.
Below is the resolution in ODE and the doubts about Laplace.

 

[Gavran]

The first mistake is the solution of the differential equation provided in the “solved by ode.pdf” file. The correct solution is $Q_2(t)=C(8,57143-8,57143e^{-0,225806t/C})$.
The second mistake is badly done numbers rounding. It is okay to round numbers, but sometimes, rounding numbers can introduce a significant roundoff error when the final result can not be accepted as a correct result. The solution in the “solved by ode.pdf” file shows how rounding numbers can be done badly and produce the wrong answer. It is not true that the current $i_3$ through the resistor $R=5$ does not depend on time. It can be easily checked that $i_3=1,17$ for $t=0$ and $i_3\to1,71$ for $t\to\infty$.

The second solution provided in the “Solving by laplace.pdf” file is wrong too. The voltage source ($12$) must also be converted to the Laplace transform ($12/s$).