Integrating factor problems (differential equation with initial conditions)

1. Nov 27, 2009

dwilmer

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 27, 2009

Dunkle

You forgot to divide both sides of the equation by the leading term!

Your ODE in standard form should be

$$y' + (4t^{-1})y = e^{-t}t^{-3}$$

The good news is that you found the correct integrating factor and you now have an easier integral to evaluate!

3. Nov 27, 2009

dwilmer

wow that was stupid. yup, when i multiply RHS by t^4 it works out. thanks

also, what does quetion mean when it says t must be less than zero, when they already specified that t = -1?

4. Nov 27, 2009

Staff: Mentor

I haven't worked this all the way through, so can't say for sure, but it could be that your solution might involve a square root in some way. Knowing whether t > 0 or t < 0 would enable you to identify which square root you should choose for your solution.