Non-Linear DiffEq: Solve y''+(y')2+1=0

[frozenguy]

Homework Statement

Solve the given differential equation by using the substitution u=y^'
y^''+(y^')²+1=0

The Attempt at a Solution

If I got to this last integral correctly, I don't know how to solve it. I'm thinking I didn't get to that step correctly.. But I can't find my mistake..

 

[Je m'appelle]

Your work is absolutely correct, you just missed a little detail

\frac{dy}{dx} = (e^{-2y}-1)^{\frac{1}{2}}C_3 \Rightarrow \frac{dy}{\sqrt{(e^{-2y} - 1)}} = C_3 dx

You forgot the square root in the left side of the last equation

\frac{dy}{\sqrt{(e^{-2y} - 1)}} = C_3 dx

I believe you can take it from here right?

Hint: Trig substitution.

 

[frozenguy]

Hey thanks for replying Je m'appelle. I'm a little rusty on trig substitution. I know how to evaluate the normal three with x² as the function but I don't remember how to handle e^-x (or -y) function in that case..

And thanks for noting I didn't use the sqrt, sometimes I forget and move on with that mistake :/. I need to train new habits lol.

 

[frozenguy]

Ok.. So I went on to the next problem and I can't complete it either..

#5) x²y^''+(y^')²=0
disregard #7, I'm finishing that one up..

 

[Je m'appelle]

Hey thanks for replying Je m'appelle. I'm a little rusty on trig substitution. I know how to evaluate the normal three with x² as the function but I don't remember how to handle e^-x (or -y) function in that case..

Use v=e^{-y} and rewrite

\frac{dy}{\sqrt{(e^{-2y} - 1)}} = C_3 dx

As

-\frac{dv}{v\sqrt{(v^2 - 1)}} = C_3dx

Where

ln(v) = -y

\frac{-dv}{v} = dy

And remember that

\int \frac{dx}{x\sqrt{x^2 - 1}} = arcsec|x| + C

So back to our equation, and using the above we'll get to

arcsec|v| + C = -C_3x

v = sec(C_3x)

e^{-y} = sec(C_3x)

y = -ln(sec(C_3x))

Now find out C_3 by replacing y back into the original differential equation.

Ok.. So I went on to the next problem and I can't complete it either..

Your work is OK up to this point

\frac{-1}{u} = \frac{1}{x} + c_1

Now replace u = \frac{dy}{dx} and rewrite it as

\frac{-dx}{dy} = \frac{1}{x} + c_1

(\frac{1}{x} + c_1 )dx = -dy

Integrate both sides

\int (\frac{1}{x} + c_1)dx = -\int dy \Rightarrow \int \frac{1}{x} \ dx + \int c_1 \ dx = -\int dy

And work it out from here.

 

[frozenguy]

So how does this look? These two problems are the same 5 and 7 posted above. a->5, b->7

These look ok?

Use v=e^{-y} and rewrite

\frac{dy}{\sqrt{(e^{-2y} - 1)}} = C_3 dx

As

-\frac{dv}{v\sqrt{(v^2 - 1)}} = C_3dx

Where

ln(v) = -y

\frac{-dv}{v} = dy

And remember that

\int \frac{dx}{x\sqrt{x^2 - 1}} = arcsec|x| + C

So back to our equation, and using the above we'll get to

arcsec|v| + C = -C_3x

v = sec(C_3x)

e^{-y} = sec(C_3x)

y = -ln(sec(C_3x))

Now find out C_3 by replacing y back into the original differential equation.




Your work is OK up to this point

\frac{-1}{u} = \frac{1}{x} + c_1

Now replace u = \frac{dy}{dx} and rewrite it as

\frac{-dx}{dy} = \frac{1}{x} + c_1

(\frac{1}{x} + c_1 )dx = -dy

Integrate both sides

\int (\frac{1}{x} + c_1)dx = -\int dy \Rightarrow \int \frac{1}{x} \ dx + \int c_1 \ dx = -\int dy

And work it out from here.

Thank you for that!