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#### karush

##### Well-known member

- Jan 31, 2012

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- Thread starter karush
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$(t^4y)'=te^{-t}$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 \(t\)

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...

- Aug 30, 2012

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

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...

-Dan

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- #6

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

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...

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- Jan 31, 2012

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

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

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- #8

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

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

And then:

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

Hence, the solution to the given IVP is:

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

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- Jan 31, 2012

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- #10

You incorrectly stated the initial value, so that's why the difference in the result I posted vs. that of the book.Here is the book answers