Recent content by Matthew Rodman

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...

This is a simple Bernoulli equation. You may want to google it's solution, or refer to any Calc textbook.

Is this a research question, or just homework? If it's serious, I may be able to get somewhere with an analytic (explicit) solution... possibly.

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...

Sorry that should read: y(x) = \frac{1}{\alpha}\int{exp(Ei^{-1}(\alpha x + \beta)) dx}

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...

You can solve the cubic v equation with Vieta's Substitution. {Wolfram.com link}

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...

Rearrange to get \frac{f^{\prime}}{f} = - \frac{x}{x^2 - a^2} now integrate to get \ln{f} = -\frac{1}{2} \ln{(x^2 - a^2)} + C