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[SOLVED] DE 19 t^3y'+4t^2y=e^{-t} y(-1)=1,t>0

karush

Well-known member
Jan 31, 2012
2,678
DE - 2.1.1
#19
$t^3y'+4t^2y=e^{-t},\quad y(-1)=1,\quad t>0$
rewrite
$y'+\dfrac{4}{t}y=\dfrac{e^{-t}}{t^3}$
$u(t)=e^{\int 4/t \, dt}=e^{4\ln \left|t\right|}=t^4$


first bold steps...
typos???
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
It appears that you've correctly computed the integrating factor, which then gives you:

\(\displaystyle t^4y'+4t^3y=te^{-t}\)

\(\displaystyle \frac{d}{dt}(t^4y)=te^{-t}\)

Now integrate w.r.t \(t\) :)
 

karush

Well-known member
Jan 31, 2012
2,678
It appears that you've correctly computed the integrating factor, which then gives you:

\(\displaystyle t^4y'+4t^3y=te^{-t}\)

\(\displaystyle \frac{d}{dt}(t^4y)=te^{-t}\)

Now integrate w.r.t \(t\) :)
$(t^4y)'=te^{-t}$
IBP


$t^4y=-e^{-t}t-e^{-t}+C$
isolate y


$y=\dfrac{-e^{-t}t}{t^4}-\dfrac{e^{-t}}{t^4}+C$


$y(-1)=1$ so
$y(-1)=\cancel{\dfrac{e^{1}}{1}}-\cancel{\dfrac{e^{1}}{1}}+C=1$


hopefully...
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,121
$(t^4y)'=te^{-t}$
IBP


$t^4y=-e^{-t}t-e^{-t}+C$
isolate y


$y=\dfrac{-e^{-t}t}{t^4}-\dfrac{e^{-t}}{t^4}+C$


$y(-1)=1$ so
$y(-1)=\cancel{\dfrac{e^{1}}{1}}-\cancel{\dfrac{e^{1}}{1}}+C=1$


hopefully...
Yup.

-Dan
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
So now write down what the solution to the DE is and you're done.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
$(t^4y)'=te^{-t}$
IBP


$t^4y=-e^{-t}t-e^{-t}+C$
isolate y


$y=\dfrac{-e^{-t}t}{t^4}-\dfrac{e^{-t}}{t^4}+C$


$y(-1)=1$ so
$y(-1)=\cancel{\dfrac{e^{1}}{1}}-\cancel{\dfrac{e^{1}}{1}}+C=1$


hopefully...
When you isolated \(y\), why didn't you divide the parameter (constant of integration) by \(t^4\)? Also, the domain is \(t<0\) for #19. :)
 

karush

Well-known member
Jan 31, 2012
2,678
So like this
$y=\dfrac{-e^{-t}t}{t^4}-\dfrac{e^{-t}}{t^4}+\dfrac{c}{t^4}$
$$y(-1)=\cancel{\dfrac{e^{1}}{1}}-\cancel{\dfrac{e^{1}}{1}}+\dfrac{c}{1}=1$$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
I would likely have written:

\(\displaystyle t^4y=c_1-e^{-t}(t+1)\)

\(\displaystyle y(t)=\frac{c_1-e^{-t}(t+1)}{t^4}\)

And then:

\(\displaystyle y(-1)=c_1=1\)

Hence, the solution to the given IVP is:

\(\displaystyle y(t)=\frac{1-e^{-t}(t+1)}{t^4}\)
 

karush

Well-known member
Jan 31, 2012
2,678
Untitled69_20200426175428.png

Here is the book answers
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
Here is the book answers
You incorrectly stated the initial value, so that's why the difference in the result I posted vs. that of the book.