# Differential Equations. Solve in two ways.

1. Sep 13, 2012

### Gummy Bear

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

Solve y''=1+(y')2 in two ways, for x is missing and y is missing.

2. Relevant equations

Integration, and reduction of order.

3. The attempt at a solution

First method: (this is correct)
y''=1+(y')2 let y'=p and y''=p'
p'=1+p2
p'/(1+p2)=1
∫p'/(1+p2)dx=∫1dx
arctan(p)=x+c
solve for p: p=tan(x+c)
Substitute back for p=y'
y'=tan(x+c)
y=-ln(cos(x+c))+d

Second method: (I can't get it to come out as same answer from first method)
y''=1+(y')2 let y'=p and y''=p*(dp/dy)
p*(dp/dy)=1+p2

From here I can already tell i'm not going to get y=-ln(cos(x+c))+d but this is the method I am told to use..

Last edited: Sep 13, 2012
2. Sep 13, 2012

### voko

The second method will give you some P(p) = P(y') = y. So you have reduced the order. Then you transform it to y' = Q(y) and solve that.

3. Sep 14, 2012

### Gummy Bear

I'm not following. What do you mean by P(p)=P(y')=y? And transforming it?

4. Sep 14, 2012

### voko

The second method will have dp/dy R(p) = 1. Integrated, that becomes P(p) = y, which means P(y') = y. This is a differential equation. To solve it, you need to find Q(y) that is the inverse of P(p), then Q(P(y')) = y' = Q(y).

5. Sep 14, 2012

### Gummy Bear

Could you show me what it would look like?

For example, my first equation looks like p'=1+p2 after reduction of order.

6. Sep 14, 2012

### Gummy Bear

Because I'm not sure where these R's and Q's are coming from. I know that typically in the second method y'=p and y''=p*(dp/dy)

7. Sep 14, 2012

### voko

$p' = 1 + p^2 \Rightarrow R(p) = \frac 1 {1 + p^2} \Rightarrow P(p) = \int R(p) dp$

8. Sep 14, 2012

### Gummy Bear

Doesn't this give y=∫1/(1+p2)dp=arctan(p)+c

The way I understood it is that both methods should get you the same answer.

I know that the answer should be -ln(cos(x+c))+d

9. Sep 14, 2012

### voko

arctan p + C = P(p). Now you need the inverse function Q(y).

10. Sep 14, 2012

### Gummy Bear

This makes no sense. The inverse is -tan(c-p)

11. Sep 14, 2012

### voko

Again. You have integrated dp/dy R(p) = 1. That gave you P(p) = y. P(p) = arctan p + c. You then obtain p = Q(y). Q(y) is tan (y - c). Since p = y', you get y' = p = tan (y - c). Integrate it.

12. Sep 14, 2012

### Gummy Bear

-ln(cos(c-y))+d That's what I keep getting. Maybe that's right, I just assumed the answer was supposed to be -ln(cos(x+c))+d

13. Sep 14, 2012

### voko

That's because you made a mistake. It must be R(p) = p/(1 + p^2), not 1/(1 + p^2).