Integrating factor problems (differential equation with initial conditions)

[dwilmer]

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

 

[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!

 

[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?

 

[Mark44]

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.