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...
Multiply both sides by \sec^2{y}, to get
y^{\prime} \sec^2{y} = x
Now, integrate both sides w.r.t. x to get
\tan{y} = \frac{x^2}{2} + \kappa
where \kappa is a constant.
So, as
r^{\prime}(t) = \frac{1}{2} \frac{(2 x(t) x^{\prime}(t) + 2 y(t) y^{\prime}(t))}{\sqrt{x^2(t) + y^2(t)}} = \frac{x(t)x^{\prime}(t) + y(t) y^{\prime}(t)}{\sqrt{x^2(t) + y^2(t)}}
You can write
m z^{\prime \prime}(t) = -q r^{\prime}(t)
Hence, for z you get
z^{\prime}(t) =...
Hm. All I can think of is that your equation may be derived from
y y^{\prime \prime} + A x + B - \frac{y^{\prime 2}}{2} = 0
where B is a constant. But
y y^{\prime \prime} - \frac{y^{\prime 2}}{2} = 2 y^{\frac{3}{2}} (\sqrt{y})^{\prime \prime}
Therefore you can re-write your equation...
Fair enough, but it wasn't a guess -- I got there by trying to find out what function, f(x), when divided by (x \sin{x} + \cos{x}) yields the integrand in part of its derivative. (The answer of course is f(x) = -x \sec{x}). Then it's a question of seeing if you can integrate the other part(s) of...
To solve your integral, you can start by differentiating
-\frac{x \sec{x}}{(x \sin{x} + \cos{x})}
this will give you
-(\frac{x \sec{x} }{x \sin{x} + \cos{x}})^{\prime} = \frac{x^2}{(x \sin{x} + \cos{x})^2} - \frac{(\sec{x} + x \sec{x} \tan{x})}{(x \sin{x} + \cos{x})}
Now, you'll recognize...
I'm not sure that integrating factors will be of any use here. But you can still solve it, i.e. by making the substitution
y = x + \sqrt{v}
this will give you
1 + \frac{1}{2}\frac{v^{\prime}}{\sqrt{v}} = \frac{1 - x^2 - x \sqrt{v}}{-x \sqrt{v}} = (x - \frac{1}{x}) \frac{1}{\sqrt{v}} + 1...
You cannot neglect the constant of integration.
if
y^{\prime \prime} = \cos{y}
multiply by y^{\prime}
gives you
y^{\prime} y^{\prime \prime} = y^{\prime} \cos{y}
integrate, and you get
\frac{1}{2}y^{\prime 2} = \sin{y} + C
Note the constant of integration which must not...
Err... if you want ot find out what kind of second-order equations are soluble, you can look here. They also have some solutions of PDEs on other pages.