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

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In summary, the given differential equation can be rewritten as $y'+\dfrac{4}{t}y=\dfrac{e^{-t}}{t^3}$ using the integrating factor method. The solution to the initial value problem $y(-1)=1$ is $y(t)=\dfrac{1-e^{-t}(t+1)}{t^4}$ for $t>0$.
  • #1
karush
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#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?
 

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  • #2
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 \(t\) :)
 
  • #3
MarkFL said:
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...
 
  • #4
karush said:
$(t^4y)'=te^{-t}$
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
 
  • #5
So now write down what the solution to the DE is and you're done.
 
  • #6
karush said:
$(t^4y)'=te^{-t}$
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. :)
 
  • #7
So like this
$y=\dfrac{-e^{-t}t}{t^4}-\dfrac{e^{-t}}{t^4}+\dfrac{c}{t^4}$
$$y(-1)=\cancel{\dfrac{e^{1}}{1}}-\cancel{\dfrac{e^{1}}{1}}+\dfrac{c}{1}=1$$
 
  • #8
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:

\(\displaystyle y(t)=\frac{1-e^{-t}(t+1)}{t^4}\)
 
  • #9

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  • #10
karush said:
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.
 

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

What is the meaning of the variables in the given differential equation?

The variable t represents time, while y represents the dependent variable. The function y(t) is the solution to the differential equation.

What is the order of this differential equation?

The order of a differential equation is determined by the highest derivative present. In this case, the highest derivative is t^3y', making this a third-order differential equation.

What is the initial condition given in this differential equation?

The initial condition is y(-1)=1, which means that the value of the dependent variable at time t=-1 is 1.

How can this differential equation be solved?

This differential equation can be solved using various methods such as separation of variables, integrating factors, or using a series solution. The specific method used will depend on the form of the equation.

What is the significance of the restriction t>0 in this differential equation?

The restriction t>0 indicates that the solution to this differential equation is only valid for positive values of t. This could be due to the specific context or application of the equation.

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