I'm trying to do exercise 19, chapter 2.1, of the book on elementary differential equations from Boyce and DiPrima. Basically to solve: $t^{3} y' + 4 t^{2} y = e^{-t}, \quad y(-1)=0, \quad t<0$. I try to solve this problem in the following way:
1). Since $t \neq 0$, then the above equation can be written as $y' + \frac{4}{t}y = \frac{e^{-t}}{t^{3}}$.
2). Multiply the above equation by an integrating factor $\mu(t)$: $\mu(t) y' + \mu(t) \frac{4}{t} y = \mu(t) \frac{e^{-t}}{t^{3}}.$
3). Find $\mu(t)$ such that the left side satisfies: $\frac{d}{dt}(\mu(t) y) = \mu(t) y' + \mu(t) \frac{4}{t} $. Hence $\frac{d \mu(t)}{dt} = \mu(t) \frac{4}{t}$. Assuming $\mu(t)>0$, and after solving this differential equation, I reach that $\mu(t) = t^{4}$.
4). My original diff. equation becomes: $\frac{d}{dt}\left( t^{4} y \right) = t e^{-t}$ and hence $y(t) = \frac{1}{t^{4}} \left( \int_{t}^{0}s e^{-s}ds + c \right)$, for some constant $c$. I took the limit of integration between $t$ and $0$ since $t<0$.
5). I solve by parts the integral $\int_{t}^{0}s e^{-s}ds = e^{-t}(1+t)-1$. In order to satisfy the initial condition, $c=1$.
6). My solution becomes $y = \frac{e^{-t}}{t^{4}} \left( 1 + t \right)$, which is not exactly the solution given in the book. There the solution is just as mine, but with an additional negative sign, i.e. $y = - \frac{e^{-t}}{t^{4}} \left( 1 + t \right)$.
Can someone please shed some light where I might have missed this negative sign? Given that $t<0$, should I have taken extra considerations in the above example? Thanks.
You need to correctly apply the fundamental theorem, as the derivative of $$ \int_t^0f(s)ds=F(0)-F(t) $$ is $-F'(t)=-f(t)$. Thus in your formula for $y$ you have to use $-\int_t^0f(s)ds$, thus the extra minus sign. That is, changing the orientation of the integration changes the sign of the integral.