I did a bit of simplification, but I may have done something wrong in between:
\int -csc(u)du = -ln|csc(u) - cot(u)| = ln|csc(2t) - cot(2t)|
=-ln|\frac{csc(2t)}{cot(2t)}| = -ln|\frac{1}{sin(2t)}*\frac{sin(2t)}{cos(2t)}|
= -ln|sec(2t)| = ln|sec(2t)|^-1 = ln|cos(2t)|
When...
That's a neat trick. So basically, my equation
ln|t| = ln|\frac{y-1}{y}| + c
turns into
ln|t| + b = ln|\frac{y-1}{y}| with b = -c
Then ln|t| + b can be thought of as ln|t| + ln|C|, which is then ln|Ct|. If I'm getting the logic right.
Nice. Thanks for the tip.
Homework Statement
y'sin(2t) = 2(y+cos(t))
y(\frac{∏}{4}) = 0
Homework Equations
\frac{dy}{dx} + p(x)y = q(x)
y = \frac{\int u(x) q(x) dx + C}{u(x)}
where
u(x) = exp(\int p(x)dx)
The Attempt at a Solution
I've set the equation in the form above, simplified the RHS and solved for...
Good point that my answer isn't even defined there.
I end up with:
ln|t| = ln|\frac{y-1}{y}| + c
c ends up being 0 with the initial value, so that simplifies to:
|t| = |\frac{y-1}{y}|
I suppose here is where I need to look at t equal to the positive or negative value of the...
Hello. I need some help solving a differential equation. I think where I'm going wrong is integrating one side via partial fractions, but I'm not quite sure where my mistake is. Using Wolfram, I found the correct answer, which is below. Thanks.
Homework Statement
Solve the following initial...
Hi,
Is the circuit in the attached image essentially one with three parallel branches, each with a different source of emf and two resistors? ie, ε1, R1, R2 on one branch, ε2, R3, R4 on another, and similarly for the last one?
Thanks!