# example of derivative as parameter

For solving the (nonlinear) differential equation

 $\displaystyle x\;=\;\frac{y}{3p}-2py^{2}$ (1)

with  $p=\frac{dy}{dx}$,  according to III in the parent entry (http://planetmath.org/DerivativeAsParameterForSolvingDifferentialEquations), we differentiate both sides in regard to $y$, getting first

 $\frac{1}{p}\;=\;\frac{1}{3p}-\left(\frac{y}{3p^{2}}+2y^{2}\right)\frac{dp}{dy}% -4py.$

Removing the denominators, we obtain

 $2p+(y+6p^{2}y^{2})\frac{dp}{dy}+12p^{3}y=0.$

The left hand side can be factored:

 $\displaystyle(y\frac{dp}{dy}+2p)(1+6p^{2}y)=0$ (2)

Now we may use the zero rule of product; the first factor of the product in (2) yields  $y\frac{dp}{dy}=-2p$, i.e.

 $2\!\int\frac{dy}{y}=-\!\int\frac{dp}{p}+\ln{C},$

whence  $y^{2}=\frac{C}{p}$,  i.e.  $p=\frac{C}{y^{2}}$.  Substituting this into the original equation (1) we get  $\displaystyle x=\frac{y^{3}}{3C}-2C$.  Hence the general solution of (1) may be written

 $y^{3}=3Cx+6C^{2}.$

The second factor in (2) yields  $6p^{2}y=-1$,  which is substituted into (1) multiplied by $3p$:

 $3px=y-(-y)$

Thus we see that  $p=\frac{2y}{3x}$, which is again set into (1), giving

 $x=\frac{y\cdot 3x}{3\cdot 2y}-\frac{4y^{3}}{3x}.$

Finally, we can write it

 $3x^{2}=-8y^{3},$

which (a variant of the so-called semicubical parabola) is the singular solution of (1).

Title example of derivative as parameter ExampleOfDerivativeAsParameter 2013-03-22 18:29:03 2013-03-22 18:29:03 pahio (2872) pahio (2872) 6 pahio (2872) Example msc 34A05 example of solving an ODE