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

[karush]

View attachment 9708​

#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]

It appears that you've correctly computed the integrating factor, which then gives you:

$$t^4y'+4t^3y=te^{-t}$$

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

Now integrate w.r.t \(t\) :)

 

[karush]

It appears that you've correctly computed the integrating factor, which then gives you:

$$t^4y'+4t^3y=te^{-t}$$

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

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]

$(t^4y)'=te^{-t}$
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]

So now write down what the solution to the DE is and you're done.

 

[MarkFL]

$(t^4y)'=te^{-t}$
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]

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]

I would likely have written:

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

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

And then:

$$y(-1)=c_1=1$$

Hence, the solution to the given IVP is:

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

 

[karush]

View attachment 9710

Here is the book answers

 

[MarkFL]

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.