Your second-order equation
y^{\prime \prime} - \frac{y^{\prime 2}}{y} + \frac{y^3}{C^2} = 0
probably has many solutions, but there is a simple solution of the form
y(\lambda) = \pm i \alpha C \sec{(\alpha \lambda + \beta)}
where \alpha and \beta are constants. You can show this very...
Sorry, I misunderstood -- the equation I quoted has no known general solution (analytical), but I suspect there are many numerical methods already associated with it. If you apply your method, you should then do a survey on the web of other numerical techniques applied this class of equations...
How about this Riccati equation:
y^{\prime} + y^2 + \alpha(x) = 0
(where alpha is an arbitrary function of x, and y = y(x) as well). This has no general solution (as far as I know) -- and it is very important. If you can provide an analytic solution to this, then fame and fortune is yours. ;-)
Take your equation, and make the change of variable
\tau = 2 x
This means that
y^{\prime}_{x} = 2 y^{\prime}_{\tau}
and
y^{\prime \prime}_{xx} = 4 y^{\prime \prime}_{\tau \tau}
Substitute these into your equation, and it becomes
\tau y^{\prime \prime}_{\tau \tau} + (b -...
Use your first equation to isolate y, namely,
y = \sin{\omega t} - x^{\prime} - x
Now, differentiate this to get y prime,
y^{\prime} = \omega \cos{\omega t} - x^{\prime \prime} - x^{\prime}
and substitute these into your second equation to get...
\omega \cos{\omega t} - x^{\prime...
I see what you're saying -- and you can also repeat this process and provide a sort of superposition of these solutions, for example:
For any solution, v_0, of the Ricatti equation
v^{\prime } + v^2 + \Psi = 0
we can show, through differentiation, that there will always be another...
Any second-order equation of the form
y^{\prime \prime} + \alpha(y) y^{\prime 2} + \beta(y) = 0
(where the derivative is with respect to 'x') may be converted into a first order equation of the form
\frac{du}{dy} + 2 \alpha(y) u + 2 \beta(y) = 0
with the simple substitution
u =...
You're missing two partial symbols. Are they supposed to be:
\lambda f(y)= i m y \frac{\partial f(y)}{\partial y} + \frac{\partial g(y)}{\partial y} -\frac{k}{y}g
\lambda g(y)= i m y \frac{\partial g(y)}{\partial y} - \frac{\partial f(y)}{\partial y} -\frac{k}{y}f
?
Also, if f and g only...
Re-arrange your equation as
y^{\prime \prime \prime} = \frac{y^{\prime \prime}}{y^{\prime}}
Now integrate with respect to x to get
y^{\prime \prime} = \kappa + \ln{y^{\prime}}
where \kappa is a constant of integration. Now re-arrange and integrate to get
\int{\frac{d...
There is a solution that does not involve a substitution... if that's any help...
First, multiply through by x + y^2, to get
x y^{\prime} + y^2 y^{\prime} = y
rearrange to get
x y^{\prime} - y = -y^2 y^{\prime}
but
x y^{\prime} - y = y^2 ( \phi - \frac{x}{y})^{\prime}
(where \phi is a...