# solve the differential equation by integrating directly

I am trying to solve a differential equation and I don't know how to solve it when it comes to integrating directly. I'd like to know how to do this so I can start doing other problems. Thanks in advance.

Solve the differential equation by integrating directly

$${{\rm d}y \over {\rm d} t} = {4t + 4 \over \left(t + 1\right)^{2}}$$

• I think you mean $y_t$, although it is kind of weird nottation for a simple ODE. – Ron Gordon Feb 19 '14 at 18:05
• @RonGordon thanks..sorry that's what I meant!... – manual Feb 19 '14 at 18:12
• can you please clarify whether you mean (4t + 4) / (t + 1)^2, or 4t + (4 / (t + 1)^2)? I just rolled back an edit which implied you meant the former. But I'm not sure so I'm putting your earlier post back and hopefully you'll clarify. – TooTone Feb 19 '14 at 18:19
• ok I edited ...it is (4t + 4) / (t + 1)^2 – manual Feb 19 '14 at 18:20
• OK, in that case I was mistaken in rolling back the edit, and I will reinstate it. – TooTone Feb 19 '14 at 18:24

$$y'(t)=4t+\frac4{(t+1)^2}\implies y(t)=2t^2+\frac{4t}{t+1}+y(0)$$ $$y'(t)=\frac{4t+4}{(t+1)^2}\implies y'(t)=\frac{4}{t+1}\implies y(t)=4\log(t+1)+y(0)$$ The solution of the second version on the interval $(-\infty,-1)$ would be $$y(t)=4\log|t+1|+y(-2).$$
• Deleting other comments. I realized why you choose a more complicated form of the answer - to make evaluating at $0$ nice. – John Habert Feb 19 '14 at 18:33
• If $t=-2$ then $\log|t+1|=0$ – Did Feb 19 '14 at 21:49