# Integrating factor problems (differential equation with initial conditions)

• dwilmer
In summary, the conversation discussed finding the general solution of the equation (t^3)y' + (4t^2)y = e^-t with initial conditions y(-1) = 0 and t<0. The correct integrating factor was found and the correct solution was obtained by dividing both sides of the equation by the leading term. The discussion also touched on the significance of t < 0 in relation to the solution involving a square root.
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

## 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

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!

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?

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.

## 1. What is an integrating factor in a differential equation?

An integrating factor is a function that is used to simplify a differential equation and make it easier to solve. It is multiplied by both sides of the equation to transform it into an exact differential equation.

## 2. How do you determine the integrating factor for a given differential equation?

The integrating factor can be determined by first checking if the given differential equation is exact. If it is not, then the integrating factor can be found by dividing the coefficient of the derivative term by the coefficient of the function term.

## 3. Can an integrating factor be used for any type of differential equation?

No, an integrating factor can only be used for first-order linear differential equations or nonlinear differential equations that can be transformed into exact equations.

## 4. How do you solve a differential equation using an integrating factor?

To solve a differential equation using an integrating factor, first multiply both sides of the equation by the integrating factor. This will transform the equation into an exact equation, which can then be solved using standard integration techniques.

## 5. What are initial conditions in the context of integrating factor problems?

Initial conditions refer to the specific values of the dependent variable and its derivative at a given point. These values are used to find the particular solution to a differential equation and are necessary when solving problems involving integrating factors.

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