Equations Reducible to Separable Form

There are many equations which are not of the form 7.2.1, but by a suitable substitution, they can be reduced to the separable form. One such class of equation is

$$y^\prime = \frac{g_1(x,y)}{g_2(x,y)} \;\; {\mbox{ or equivalently }} \;\; y^\prime = g(\frac{y}{x})$$

where $g_1$ and $g_2$ are homogeneous functions of the same degree in $x$ and $y,$ and $g$ is a continuous function. In this case, we use the substitution, $y = x u(x)$ to get $y^\prime = x u^\prime + u.$ Thus, the above equation after substitution becomes

$$x u^\prime + u(x) = g(u),$$

which is a separable equation in $u.$ For illustration, we consider some examples.

EXAMPLE 7.2.2  

1. Find the general solution of $2 x y y^\prime - y^2 + x^2 = 0.$
   Solution: Let $I$ be any interval not containing $0.$ Then
   $$2 \frac{y}{x} y^\prime - (\frac{y}{x})^2 + 1 = 0.$$
   Letting $y = x u(x),$ we have
   $$2 u (u^\prime x + u) - u^2 + 1 = 0 \;\; {\mbox{ or }} \;\; 2 x u u^\prime + u^2 + 1 = 0 \;\; {\mbox{or equivalently}}$$
   $$\frac{2u}{1 + u^2} \frac{du}{dx} = - \frac{1}{x}.$$
   On integration, we get
   $$1 + u^2 = \frac{c}{x}$$
   or
   $$x^2 + y^2 - c x = 0.$$
   The general solution can be re-written in the form
   $$(x - \frac{c}{2})^2 + y^2 = \frac{c^2}{4}.$$
   This represents a family of circles with center $(\frac{c}{2}, 0)$ and radius $\frac{c}{2}.$
2. Find the equation of the curve passing through $(0,1)$ and whose slope at each point $(x,y)$ is $- \frac{x}{2y}.$
   Solution: If $y$ is such a curve then we have
   $$\frac{dy}{dx} = - \frac{x}{2y} {\mbox{ and }} y(0) = 1.$$
   Notice that it is a separable equation and it is easy to verify that $y$ satisfies $x^2 + 2 y^2 = 2.$
3. The equations of the type
   $$\frac{dy}{dx} = \frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}$$
   can also be solved by the above method by replacing $x$ by $x + h$ and $y$ by $y+k,$ where $h$ and $k$ are to be chosen such that
   $$a_1 h + b_1 k + c_1 = 0 = a_2 h + b_2 k + c_2.$$
   This condition changes the given differential equation into $$\frac{dy}{dx} = \displaystyle \frac{a_1 x + b_1 y}{a_2 x + b_2 y}.$$ Thus, if $x \neq 0$ then the equation reduces to the form $y^\prime = g(\frac{y}{x})$ .

EXERCISE 7.2.3  

1. Find the general solutions of the following:
   1. $$\frac{dy}{dx} = - x (\ln x) (\ln y).$$
   2. $y^{-1} \cos^{-1} + (e^x + 1) \displaystyle \frac{dy}{dx} = 0.$
2. Find the solution of
   1. $(2 a^2 + r^2) = r^2 \cos \displaystyle \frac{d\theta}{dr}, \; r(0) = a.$
   2. $x e^{x + y} = \displaystyle \frac{dy}{dx}, \;\; y(0) = 0.$
3. Obtain the general solutions of the following:
   1. $\{ y - x {\mbox{cosec }}(\displaystyle \frac{y}{x})\} = x \displaystyle \frac{dy}{dx}.$
   2. $x y^\prime = y + \sqrt{x^2 + y^2}.$
   3. $$\frac{dy}{dx} = \frac{x - y + 2}{- x + y + 2}.$$
4. Solve $y^\prime = y - y^2$ and use it to determine $\lim\limits_{x \longrightarrow \infty} y.$ This equation occurs in a model of population.