## Separable differential equation

The first order Differential Equation: [mathjax] $$\frac{dy}{dx}=H(x, y) \text{__eqn(1)}$$ is called Separable differential equation, provided that (H(x, y)) can be written as the product of function (x) and (y): $$\frac{dy}{dx}=g(x)h(y)=\frac{g(x)}{f(y)},$$

where (h(y)=\frac{1}{f(y)}), In this case the variable (x) and (y) can be separated __ by writing informally the equation, $$f(y)dy = g(x)dx$$

which to be concise notation for the differential equation $$f(y)\frac{dy}{dx} =g(x) \quad\text{ _eqn(2)}$$

It is easy to solve this special type of differential equation simply by integration both sides with respect to (x): $$\int{f(y(x))\frac{dy}{dx}dx} = \int{g(x)dx}+C;$$ equivalently, $$\int{f(y)dy}=\int{g(x)dx}+C. \text{__________eqn(3)}$$

All that is required is that the anti derivatives $$F(y)=\int{f(y)dy } \quad\text{ and }G(x)=\int{g(x)dx}$$ can be found. To see that eqns (2) and (3) are equivalent, note that the following consequence of chain rule:

$$D_{x}[F(y(x))]=F^{′}(y(x))y^{′}(x)=f(y)\frac{dy}{dx}=g(x)=D_{x}[G(x)],$$

which in tern is equivalent to $$F(y(x))=G(x)+C,$$ because two functions have the same derivatives on an interval if and only if they differ by a constant on that interval.

**Example: Solve the initial value problem** $$\frac{dy}{dx}=\frac{4-2x}{3y^{2}-5}. \text{ y(1)=3 }$$

**Solution**: Given equation is a Separable Differential Equation, because the variables in it are possible to calculate separately, $$(3y^{2}-5)dy = (4-2x)dx;$$

Now integrating them we get, $$\int{(3y^{2}-5)}dy=\int{(4-2x)}dx$$

integrating the equation gives, $$y^{3}-5y =4x-x^{2}+C \text{_________eqn(4)}$$

eqn(4), is the implicit solution of given differential equation. Now to solve the initial value problem we substitute (x=1, y=3) in eqn(4) which gives (C=9). Thus the desired particular solution (y(x)) is defined implicitly by the equation $$y^{3}-5y =4x-x^{2}+9.$$