Separable and Linear equation

 

Def 1

A $1$st order Differential Equation of the form

$\frac{\mathrm{d}y}{\mathrm{d} x} = g(x)h(y)$

is said to be separable.

 

 

 

Solution

In this case, we can easily find the solution by collecting $x$ and $y$ terms, and intergrating those terms. So we get

$\frac{1}{h(y)} \mathrm{d}y = g(x) \mathrm{d}x$  $\Rightarrow$  $\int \frac{1}{h(y)} \mathrm{d}y = \int g(x) \mathrm{d}x$

 

 

 

Def 2

A $1$st order DE of the form 

$a_{0}(x) \frac{\mathrm{d}y}{\mathrm{d} x} + a_{1}(x)y = g(x)$

is said to be linear equation in the variable $y$. This is called general form of linear Eq.

A specific form from in the following

$\frac{\mathrm{d}y}{\mathrm{d} x} + P(x)y = f(x)$

is called standard form of linear Eq.

 

 

 

Solution

Idea : multiply $e^{\int P(x) \mathrm{d} x} $

$e^{\int P(x) \mathrm{d} x}[\frac{\mathrm{d}y}{\mathrm{d} x} +P(x)y] = e^{\int P(x) \mathrm{d} x}f(x)$

     $\Rightarrow$  $\frac{\mathrm{d}}{\mathrm{d} x}[e^{\int P(x) \mathrm{d} x} y] = e^{\int P(x) \mathrm{d} x}f(x) $

     $\Rightarrow$  $e^{\int P(x) \mathrm{d} x} y = \int e^{\int P(x) \mathrm{d} x}f(x) \mathrm{d} x + C$     (for some $C \in \mathbb{R}$)

     $\Rightarrow$  $y = C e^{- \int P(x) \mathrm{d} x} + e^{- \int P(x) \mathrm{d} x} \int e^{\int P(x) \mathrm{d} x}f(x) \mathrm{d} x $