It is important to know how to solve differential equations.

In this section ONLY First Order Linear Differential equations will be mentioned.

Common form

${\displaystyle \frac{dy}{dx}+P(x)y=Q(x)}$

this is the typical form for all the First Order Linear Differential equations,in which y and its first derivative appear linearly. By choosing different functions ${\displaystyle P(x)}$ and ${\displaystyle Q(x)}$, this form represents any first-order linear differential equation.

${\displaystyle P(x)}$ and ${\displaystyle Q(x)}$ are two functions only about variable ${\displaystyle x}$.

Solution

To solve first order linear equation, it is important to introduce a integral factor ${\displaystyle \mu }$.

Integrating factor

We first try to solve the equation like indefinite integral, but obviously, it is impossible to calculate ${\displaystyle \int [\frac{dy}{dx}+P(x)y]dx}$ in most cases, so, just as how we think about integration by parts, we turn to the fomula of the derivative of the product, known as

${\displaystyle (xy{)}^{\prime }=x^{\prime }y+x{y}^{\prime }}$,

we try to make the left side of the equation to derivative of the product of two functions so that it will be simple to do the indefinite integral by two sides of the equation.

But how to make ${\displaystyle y^{\prime }+P(x)y}$ to the form of the derivative of the product of two functions? Since ${\displaystyle P(x)}$ is a function about x, suppose that we have a function ${\displaystyle \mu }$ which is also about x, we multiply the both sides of the equation by ${\displaystyle \mu }$, so the left side becomes

${\displaystyle \mu y^{\prime }+\mu P(x)y}$,

suppose that it can be modified to the form ${\displaystyle \mu y^{\prime }+{\mu }^{\prime }y}$, it is simple to find out that ${\displaystyle {\mu }^{\prime }}$ must equals to ${\displaystyle \mu P(x)}$, shown as

${\displaystyle {\mu }^{\prime }=\mu P(x)}$,

we move ${\displaystyle \mu }$ and ${\displaystyle {\mu }^{\prime }}$ together and do the indefinite integral, it is

${\displaystyle \int \frac{{\mu }^{\prime }}{\mu }dx=\int P(x)dx}$,

which actually can be written as

${\displaystyle \ln \mu =\int P(x)dx+C}$,

then we make bothe sides the exponent of e, it is

${\displaystyle \mu =e^{\int P(x)dx}}$

and that is our integrating factor.

We may notice that in the exponent of e, the indefinite integral produces a constant C, but since it is in the exponent, it will be eliminated at last.

How to solve

We have integral factor ${\displaystyle \mu }$ now, simply multiply it to the both sides of the function, do the indefinite integral with x and y separately by each sides, and we have the final equation

${\displaystyle \mu y=\int \mu Q(x)dx}$,

symplify it and replace ${\displaystyle \mu }$ by ${\displaystyle e^{\int P(x)dx}}$, we have

${\displaystyle y=e^{-\int P(x)dx}·\int e^{\int P(x)dx}Q(x)dx}$

which is the general solution of the first order linear equation.

BE CAREFUL that in the right side of the equation, the constant C of the indefinite integral should be multiplied by ${\displaystyle \mu }$