second grade differential equation y'' = y'² + 1

[Ritch]

second grade differential equation y'' = y'² + 1

Hello, after some doubt I have decided to try and post a problem here to come to a solution.

It's about teh differential equation y'' = (y')^2 + 1

I've worked out a part already which may be just right but when I meet up with a second grade differential equation i'm getting stuck kind of.


to solve I've used this method (because x is not mentioned):
y' = p
then y'' = d(p)/dx = d(p)/dy * dy/dx = p'*p

Starting then:

p'*p = p² +1
<=> p' - p = 1/p
<=> dp = [(1+p²)/p]*dy

gives eventually
<-> 1+ p² = e^(2y+C)

<=> 1+y'² = e^(2y+C)
<=> dy = sqrt{e^(2y+C) -1}*dx

<=> [1/sqrt{e^(2y+C) -1}]*dy = dx

Integrating this seems like a mistery for me.

This last step gets me stuck in something I don't have experience with.

The answer to this problem is y = -ln|C₂*cos(x+C₁)|, I just can't figure out the second part of the differentialequation.

Done some other excercices of similar kind and these work out for half of the time but usually I don't meet the y'² issue i'm having here.


Any hits would help on this matter.

 

[MarkFL]

I would rewrite the ODE as:

\(\displaystyle \dfrac{1}{(y')^2+1}\,d(y')=dx\)

Integrating, we get:

\(\displaystyle \tan^{-1}(y')=x+c_1\)

or:

\(\displaystyle y'=\tan(x+c_1)\)

Integrating again, we find:

\(\displaystyle y=-\ln|\cos(x+c_1)|+c_2=-\ln|c_2\cos(x+c_1)|\)

 

[Deleted member 4993]

Hello, after some doubt I have decided to try and post a problem here to come to a solution.

It's about teh differential equation y'' = (y')^2 + 1

I've worked out a part already which may be just right but when I meet up with a second grade differential equation i'm getting stuck kind of.


to solve I've used this method (because x is not mentioned):
y' = p
then y'' = d(p)/dx = d(p)/dy * dy/dx = p'*p

Starting then:

p'*p = p² +1
<=> p' - p = 1/p
<=> dp = [(1+p²)/p]*dy

gives eventually
<-> 1+ p² = e^(2y+C)

<=> 1+y'² = e^(2y+C)
<=> dy = sqrt{e^(2y+C) -1}*dx

<=> [1/sqrt{e^(2y+C) -1}]*dy = dx

Substitute

sec(Θ) = C₁e^y ← C₁ = √(e^C)

This is what you should do - if you want to continue in the path you started.

However, method shown by Mark is much simpler.

Integrating this seems like a mistery for me.

This last step gets me stuck in something I don't have experience with.

The answer to this problem is y = -ln|C₂*cos(x+C₁)|, I just can't figure out the second part of the differentialequation.

Done some other excercices of similar kind and these work out for half of the time but usually I don't meet the y'² issue i'm having here.


Any hits would help on this matter.

.