- #1

dwilmer

- 11

- 0

## Homework Statement

Find general solution of equation

(t^3)y' + (4t^2)y = e^-t

with initial conditions:

y(-1) = 0 and t<0

book answer gives y = -(1+t)(e^-t)/t^4 t not = 0

## Homework Equations

## The Attempt at a Solution

(t^3)y' + (4t^2)y = e^-t

get integrating factor...

u(t) = e^integ (4t^2)/t^3

u(t) = t^4

(t^4)y = integ (e^-t)(t^4)

(t^4)y = (-t^4)e^-t - (4t^3)e^-t - (12t^2)e^-t - 24te^-t - 24e^-t + c

y = [ (-t^4)e^-t - (4t^3)e^-t - (12t^2)e^-t - 24te^-t - 24e^-t + c] / (t^4)

simplifying...

y = [ e^-t ((-t^4) - (4t^3) - (12t^2) - (24t) - 24 + c) ] / (t^4)

when i put in inititial condition y(-1) = 0

i end up with

0 = (-9e^1) + c

so, c = 9e

so solution to general solution is:

y = [ (-t^4)e^-t - (4t^3)e^-t - (12t^2)e^-t - 24te^-t - 24e^-t + 9e] / (t^4)

but this is really wrong.

also, I am confused about the other initial condition.. What does it mean that t must be less than zero with repect to the original equation?? i mean , t already is < 0 because y(-1) = 0

thanks for any help