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

#### karush

##### Well-known member
#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

Staff member
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
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
$(t^4y)'=te^{-t}$
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
So now write down what the solution to the DE is and you're done.

#### MarkFL

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

Staff member
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:

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