# Differential Equations: Find the Solutions and Limiting Factos

1. Feb 4, 2013

### Northbysouth

1. The problem statement, all variables and given/known data
Determine the solution and limiting factor

(t+1)y' + y = 6

y(1) = -2

2. Relevant equations

3. The attempt at a solution

So I started off by isolating y'

y' + y/(t+1) = 6(t+1)

Then I found u(t)

u(t) = e∫ t+1 dt = et2/2 +t

y' =(et2/2 +t) = y(et2/2 +t) = 6(et2/2 +t)/(t+1)

The difficulty I'm having is integrating 6(et2/2 +t)/(t+1)

u-substitution doesn't help. Suggestions are appreciated

2. Feb 4, 2013

### LCKurtz

That should be $u(t) = e^{\int \frac 1 {t+1}}\, dt$.

3. Feb 4, 2013

### Northbysouth

Ahh, thank you.

Can you explain to me what is meant by the limiting factor? Is it the largest range of t values for the given information of y(1) = -2?

4. Feb 4, 2013

### LCKurtz

I am not familiar with the term "limiting factor". Perhaps it refers to disallowed values of t in your answer.