-2.2.27 - Analysis of first order IVP

[karush]

well each one is a little different so,,,
$$\dfrac{dy}{dt}=\dfrac{ty(4-y)}{3},\qquad y(0) =y_0$$
not sure if this is what they meant on the given expression

 

[Joppy]

Could you be clear on what you're struggling with? For instance a) requires first the solution of the DE. Can you identify the kind of DE that it is and suggest a solution approach? (e.g., separable)

 

[Alone]

You just need to integrate $$dy/(y(4-y))=tdt/3$$ by using the partial fractions $1/(y(4-y))=A/y+B/(4-y)$ where you find A and B by multiplying first by $y$ and then plugging y=0 to find A, as for B just try to multiply by 4-y and again plug $y=4$ to find the coefficients A and B. Thus you immediate integrals to solve.

 

[karush]

Could you be clear on what you're struggling with? For instance a) requires first the solution of the DE. Can you identify the kind of DE that it is and suggest a solution approach? (e.g., separable)

ok yes the 2,2 section exercises are all on separable equations
I have 10 more problems I want to post here
I'm retired and by myself on this. there is no real need for me me even for me to do this but it certainly has given me a sense of accomplishment
MHB has been a great place for me to spend all the free time I have

 

[MarkFL]

Yes, I see what looks like a separable first order ODE for which partial fractions will prove to be useful. :)

 

[karush]

You just need to integrate $$dy/(y(4-y))=tdt/3$$ by using the partial fractions $1/(y(4-y))=A/y+B/(4-y)$ where you find A and B by multiplying first by $y$ and then plugging y=0 to find A, as for B just try to multiply by 4-y and again plug $y=4$ to find the coefficients A and B. Thus you immediate integrals to solve.

$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$
like this?
do we really need this ugly brown box at the bottom!

 

[topsquark]

$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$
like this?
do we really need this ugly brown box at the bottom!

You are off by a minus sign.

\(\displaystyle \dfrac{1}{y(y - 4)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}\)

Watch the negative sign on the 4 - y.

-Dan

 

[karush]

$\displaystyle \dfrac{1}{y(4¬y)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}$ thus if so then $$\dfrac{1}{4} \displaystyle\int \dfrac{1}{y} \, dy- \dfrac{1}{4} \int\dfrac{1}{y-4}\, dy =\dfrac{1}{3} \int t\, dt $$

 

[skeeter]

$\displaystyle \dfrac{1}{y(4¬y)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}$ thus if so then $$\dfrac{1}{4} \displaystyle\int \dfrac{1}{y} \, dy- \dfrac{1}{4} \int\dfrac{1}{y-4}\, dy =\dfrac{1}{3} \int t\, dt $$

how about ...

$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy
=\dfrac{4}{3} \int t\, dt $$

 

[Joppy]

This is also a Bernoulli equation, if you wanted to take a different route :). But I'm not sure if your book covered linear ODE's before separable.

 

[Alone]

$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$
like this?
do we really need this ugly brown box at the bottom!

Which ugly box are you talking about?

 

[karush]

Which ugly box are you talking about?

on a tablet the brown footer is a huge retangle with ears on the sides
terrifying to stare at

 

[karush]

$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy
=\dfrac{4}{3} \int t\, dt$$

$$\ln |y|-\ln |y-4|=\frac{2t^2}{3}+C$$

ok assume find C from RHS and $y(0) =y_0$

does $y_0=t$??

 

[Alone]

on a tablet the brown footer is a huge retangle with ears on the sides
terrifying to stare at

OK, I am using a Desktop and this ugly box isn't shown there.

 

[Alone]

$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy
=\dfrac{4}{3} \int t\, dt$$

$$\ln |y|-\ln |y-4|=\frac{2t^2}{3}+C$$

ok assume find C from RHS and $y(0) =y_0$

does $y_0=t$??

y_0 is a constant and t is a variable, so this doesn't make so much sense.

 

[karush]

OK here is the book answer. #27 just not sure how it was derived!