Miscellaneous Remarks

In Section 7.4, we have learned to solve the linear equations. There are many other equations, though not linear, which are also amicable for solving. Below, we consider a few classes of equations which can be solved. In this section or in the sequel, $p$ denotes $$\frac{dy}{dx}$$ or $y^\prime.$ A word of caution is needed here. The method described below are more or less ad hoc methods.

1. EQUATIONS SOLVABLE FOR Y:
   Consider an equation of the form

   $$y = f(x,p).$$ (7.5.1)

   
   
   Differentiating with respect to $x,$ we get

   $$\frac{dy}{dx} = p = \frac{\partial f(x,p)}{\partial x} + \frac{\... ...dot \frac{dp}{dx} \; {\mbox{ of equivalently }} \;\;p = g(x,p, \frac{dp}{dx}).$$ (7.5.2)

   
   
   The Equation (7.5.2) can be viewed as a differential equation in $p$ and $x.$ We now assume that (7.5.2) can be solved for $p$ and its solution is

   $$h(x, p, c) = 0.$$ (7.5.3)

   
   
   If we are able to eliminate $p$ between (7.5.1) and (7.5.3), then we have an implicit solution of the (7.5.1).

   Solve $y = 2 p x - x p^2.$
   Solution: Differentiating with respect to $x$ and replacing $$\frac{dy}{dx}$$ by $p,$ we get

   $$p = 2p - p^2 + 2x \frac{dp}{dx} - 2x p \frac{dp}{dx} \;\; {\mbox{ or }} \;\; (p + 2 x \frac{dp}{dx}) (1-p) = 0.$$
   So, either
   $$p + 2 x \frac{dp}{dx} = 0 \;\; {\mbox{ or }} \;\; p = 1.$$
   That is, either $p^2 x = c$ or $p=1.$ Eliminating $p$ from the given equation leads to an explicit solution
   $$y = 2 x \sqrt{\frac{c}{x}} - c \; {\mbox{ or }} \;\; y = x.$$

   The first solution is a one-parameter family of solutions, giving us a general solution. The latter one is a solution but not a general solution since it is not a one parameter family of solutions.

2. EQUATIONS IN WHICH THE INDEPENDENT VARIABLE MATHEND000# IS MISSING:
   These are equations of the type $f(y, p) = 0.$ If possible we solve for $y$ and we proceed. Sometimes introducing an arbitrary parameter helps. We illustrate it below.

   Solve $y^2 + p^2 = a^2$ where $a$ is a constant.
   Solution: We equivalently rewrite the given equation, by (arbitrarily) introducing a new parameter $t$ by

   $$y = a \sin t, \;\; p = a \cos t$$
   from which it follows
   $$\frac{dy}{dt} = a \cos t; \;\; p = \frac{dy}{dx}= \frac{dy}{dt}\biggl/\frac{dx}{dt}\biggr.$$
   and so
   $$\frac{dx}{dt} = \frac{1}{p} \frac{dy}{dt} = 1\;\; {\mbox{ or }} \;\; x = t + c.$$
   Therefore, a general solution is
   $$y = a \sin (t+c).$$

3. EQUATIONS IN WHICH MATHEND000# (DEPENDENT VARIABLE OR THE UNKNOWN) IS MISSING:
   We illustrate this case by an example.

   Find the general solution of $x = p^3 - p - 1.$
   Solution: Recall that $p = \frac{dy}{dx}.$ Now, from the given equation, we have

   $$\frac{dy}{dp}= \frac{dy}{dx}\cdot\frac{dx}{dp}= p (3p^2 - 1).$$
   Therefore,
   $$y = \frac{3}{4} p^4 - \frac{1}{2} p^2 + c$$
   (regarding $p$ as a parameter). The desired solution in this case is in the parametric form, given by
   $$x = t^3 - t - 1 \;\; {\mbox{ and }} y = \frac{3}{4} t^4 - \frac{1}{2} t^2 + c$$
   where $c$ is an arbitrary constant.
   Remark 7.5.1   The readers are again informed that the methods discussed in $1), 2), 3)$ are more or less ad hoc methods. It may not work in all cases.

EXERCISE 7.5.2  

1. Find the general solution of $y = (1 + p ) x + p^2.$
   Hint: Differentiate with respect to $x$ to get $$\frac{dx}{dp} = - (x + 2 p)$$ ( a linear equation in $x$ ). Express the solution in the parametric form
   $$y(p) = (1 + p) x + p^2, \;\; x(p) = 2(1-p) + c e^{-p}.$$

2. Solve the following differential equations:
   1. $8 y = x^2 + p^2.$
   2. $y + x p = x^4 p^2.$
   3. $y^2 \log y - p^2 = 2 x y p.$
   4. $2 y + p^2 + 2 p = 2 x (p+1).$
   5. $2 y = 2 x^2 + 4 p x + p^2.$