Answer
Question: Solve the initial value problem.
$y't^3+4yt^2=e^{-t}$ , $y(-1)=0$ , $t<0$
Answer: $y=\frac{-(t+1)e^{-t}}{t^4}$, $t\ne0$, $c=0$ where c is the arbitrary constant resulting from integrating
Work Step by Step
$y't^3+4yt^2=e^{-t}$ , $y(-1)=0$ , $t<0$
recognize that
$\frac{d}{dx}(At^ny)=nAyt^{n-1}+Ay't^n$
where $A$ is an arbitrary constant, $n$ is an arbitrary exponent (non-variable), $t$ and $y$ are variables. how can we make this fit our equation? If the 4 was a 3 then our equation would fit perfectly, but since its is a four, integrating t must require $n$ in
$\frac{d}{dx}(At^ny)$
to be 4, so with
$n=4$
we have
$\frac{d}{dx}(At^4y)=4Ayt^{3}+Ay't^4$,
this is almost our equation and can easily be made our equation by simply setting $A=1$ and multiplying our equation by t. If we multiply both sides of our given equation by t then we get
$y't^4+4yt^3=te^{-t}$
which condenses to
$\frac{d}{dx}yt^4=te^{-t}$
This implies that
$d(yt^4)=(te^{-t})dt$
Integrating both sides (using integration by parts) yields $t^4y=-te^{-t}+\int(e^{-t})dt$
which evaluates to
$t^4y=-te^{-t}+(-e^{-t}+c)$
At this point, it is easiest to use our initial value and solve for c rather than solve for y, then for c, then correct.
$t^4y=-te^{-t}+(-e^{-t}+c)$–> $0=e-e+c$ therefore $c=0$
therefore
$t^4y=-te^{-t}+(-e^{-t})$
which can be rearranged to
$y=\frac{-te^{-t}+(-e^{-t})}{t^4}$ or
$y=\frac{-(t+1)e^{-t}}{t^4}$
Finally, $t=0$ is undefined for this solution, so we write $t\ne0$
