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DE 27 - Analysis of first order IVP

karush

Well-known member
Jan 31, 2012
2,622
Wahiawa, Hawaii
2020_05_11_12.41.40~2.jpg
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

Well-known member
MHB Math Helper
Mar 17, 2016
256
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)
 

Alan

Member
Jul 21, 2012
58
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

Well-known member
Jan 31, 2012
2,622
Wahiawa, Hawaii
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

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,684
St. Augustine, FL.
Yes, I see what looks like a separable first order ODE for which partial fractions will prove to be useful. :)
 

karush

Well-known member
Jan 31, 2012
2,622
Wahiawa, Hawaii
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,104
The Astral plane
$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$
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

Well-known member
Jan 31, 2012
2,622
Wahiawa, Hawaii
$\displaystyle \dfrac{1}{y(4¬y)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}$
thus
if so then
$$\dfrac{1}{4} \displaystyle\int \dfrac{1}{y} \, dy- \dfrac{1}{4} \int\dfrac{1}{y-4}\, dy
=\dfrac{1}{3} \int t\, dt $$
 
Last edited:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
$\displaystyle \dfrac{1}{y(4¬y)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}$
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

Well-known member
MHB Math Helper
Mar 17, 2016
256
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.
 

Alan

Member
Jul 21, 2012
58
$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$
like this?
do we really need this ugly brown box at the bottom!
Which ugly box are you talking about?
 

karush

Well-known member
Jan 31, 2012
2,622
Wahiawa, Hawaii
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

Well-known member
Jan 31, 2012
2,622
Wahiawa, Hawaii
$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy
=\dfrac{4}{3} \int t\, dt$$

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

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

does $y_0=t$??
 
Last edited:

Alan

Member
Jul 21, 2012
58
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.
 

Alan

Member
Jul 21, 2012
58
$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy
=\dfrac{4}{3} \int t\, dt$$

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

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

Well-known member
Jan 31, 2012
2,622
Wahiawa, Hawaii
2020_05_06_10.28.47~2.jpg
OK here is the book answer. #27 just not sure how it was derived!