Solving Linear ODEs: How to Obtain the Highlighted 1?

[yecko]

Homework Statement

How to obtain the "1" highlighted?

Homework Equations

multiply by μ then by dt (integration )to both sides

The Attempt at a Solution

[/B]
lets just consider part "2y/t":

∫2y/t from pi/2 to t
=ln(t^2/(pi^2/4)) from pi/2 to t
=ln1-ln(t^2/(pi^2/4))
= -ln(t^2/(pi^2/4))

how can I obtain the "1" highlighted?
Thank you very much

 

[Orodruin]

2y/t from pi/2 to t
=ln(t^2/(pi^2/4)) from pi/2 to t
=ln1-ln(t^2/(pi^2/4))
= -ln(t^2/(pi^2/4))

This is not correct. Where did your $y$ go? You need to use the integrating factor.

 

[yecko]

Isnt the I.F. used in the front as a multiple of whole thing, and the back thing? How is it related to part 2y/t? Also, why is it the initial value which is 1? Thanks

 

[Orodruin]

as a multiple of whole thing, and the back thing

Please be more specific. It is impossible to understand what you mean by this.

An integrating factor $\mu(t)$ for a linear differential equation of the form $y'(t) + g(t) y(t) = h(t)$ is a function such that $d(y\mu)/dt = \mu(t) y'(t) + \mu'(t) y(t) = \mu(t) [y'(t) + g(t) y(t)]$. This allows you to rewrite the entire differential equation as $z'(t) = \mu(t) h(t)$, where $z(t) = \mu(t) y(t)$ and the left-hand side is trivial to integrate. Your full solution then follows from integrating the right-hand side and dividing by $\mu(t)$. Do not forget any integration constants that will later be needed in order to adapt your solution to the boundary conditions.

 

[yecko]

d(yμ)/dt=μ(t)y′(t)+μ′(t)y(t)=μ(t)[y′(t)+g(t)y(t)]d(yμ)/dt=μ(t)y′(t)+μ′(t)y(t)=μ(t)[y′(t)+g(t)y(t)]d(y\mu)/dt = \mu(t) y'(t) + \mu'(t) y(t) = \mu(t) [y'(t) + g(t) y(t)].

z′(t)=μ(t)h(t)

ok, let me be more specific...
after looking at more examples, I found that almost the whole chapter use equation like this:
for y'+f(t)y=g(t), y(t_o)=y_o
calculation: y(t)=(1/μ(t)) ∫ {from t_o to t} [y_o+∫(g(t)μ(t))dt]
why is it y_o?
Thanks

 

[Ray Vickson]

ok, let me be more specific...
after looking at more examples, I found that almost the whole chapter use equation like this:
for y'+f(t)y=g(t), y(t_o)=y_o
calculation: y(t)=(1/μ(t)) ∫ {from t_o to t} [y_o+∫(g(t)μ(t))dt]
why is it y_o?
Thanks

I hope that no book you are using wrote that equation; it is 100% wrong.

 

[yecko]

I hope that no book you are using wrote that equation; it is 100% wrong.

this is the relationship i have found from all those example, but i have got no clues on how to prove out the part regarding f(t)...

multiply by μ then by dt (integration )to both sides

this is my method on trying to prove out for y'+f(t)y=g(t), y(to)=yo ==> y(t)=(1/μ(t)) ∫ {from to to t} [yo+∫(g(t)μ(t))dt]
is it wrong?
thanks

 

[Orodruin]

is it wrong?

You are most likely copying it wrong. You have
$$
\frac{d(y\mu)}{dt} = \mu(t) g(t).
$$
Integrating this from $t = t_0$ to $t$ leads to
$$
y(t)\mu(t) - y(t_0) \mu(t_0) = \int_{t_0}^t \mu(s) g(s) ds.
$$
Solving for $y(t)$ leads to
$$
y(t) = \frac{1}{\mu(t)} \left[y_0 \mu(t_0) + \int_{t_0}^t \mu(s)g(s) ds\right].
$$

 

[yecko]

Thank you for your help.. can I ask one more question on the topic?

Question:
A tank originally contains 100 gal of fresh water. Then water containing 1/2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount salt in the tank at the end of an additional 10 min.

How can the two highlighted part obtain?

y(t)=1μ(t)[y0μ(t0)+∫tt0μ(s)g(s)ds].

My attempt: I have tried to substitute the formula, S₁(t)=e^(-0.02t) * ∫ {from 0 to t} (S₁(t)/50)e^(0.02t) dt
which seems wrong...

Thanks

 

[Orodruin]

can I ask one more question on the topic?

That is a new problem. You should start a new thread for each problem.