Separable equations: How do you know which variable to solve for?

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SUMMARY

This discussion focuses on solving separable equations, specifically determining which variable to solve for in the context of given initial conditions. The first question emphasizes solving for the variable u in the equation du/dt, while the second question addresses the conventions of handling absolute values and signs when simplifying expressions. The final conclusion suggests that the absolute value signs can be omitted in certain contexts, particularly when the variable x is restricted to the interval (0, π/2).

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Separable equations: How do you know which variable to solve for? + extra question

Homework Statement


I attached a sample problem with variables u and t. How do I know what the answer should be at the end? In terms of u or in terms of t or it doesn't matter?

Homework Statement


I also have another question, so to prevent spam I will also post it in this thread:

The second picture I attached is the question and I ended up with the answer (as shown in my picture):
y = ±(|2a|sinx/(√3/2))-a

For this question I am confused about the plug or minus sign conventions when you remove absolute value brackets. The answer key shows:
y = (4asinx/(√3/2))-a
It doesn't even have the plus or minus sign or the absolute value brackets for 2a.

Is my answer correct as well?

Thanks!
 

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Your second question is usually handled like this. Before you put in the initial condition you have$$
\ln|a+y| = \ln(\sin x) +C$$You don't need absolute value signs in the sine because x is restricted to ##(0,\pi/2)##.$$
\ln\frac{|a+y|}{\sin x} = C$$ $$
|a+y| = (\sin x) e^C$$ $$
a+y = \pm e^c \sin x$$which is similar to your steps. At this point ##C## can be any real number so ##e^C## can be any positive number, so ##\pm e^C## can be any nonzero number. So just call it a new constant ##K## and you get$$
a+y = K\sin x$$Now put in your initial conditions.

For your first question, you are given an equation with ##du/dt## and an initial condition for u as a function of t. That suggest solving for u if you can.
 
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