Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of [latex]{y}^{\prime }[/latex] be equal to [latex]1[/latex]. To make this happen, we divide both sides by [latex]3{x}^{2}-4[/latex].
This is called the standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation.
Returning to our general definition, we can divide both sides of the equation by [latex]a\left(x\right)[/latex]. This leads to the equation
Now define [latex]p\left(x\right)=\frac{b\left(x\right)}{a\left(x\right)}[/latex] and [latex]q\left(x\right)=\frac{c\left(x\right)}{a\left(x\right)}[/latex]. Then the definition becomes
We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.
standard form for first-order linear differential equations
The standard form of a first-order linear differential equation is:
[latex]y' + p(x)y = q(x)[/latex]
where [latex]p(x)[/latex] and [latex]q(x)[/latex] are functions of [latex]x[/latex] only.
Put each of the following first-order linear differential equations into standard form. Identify [latex]p\left(x\right)[/latex] and [latex]q\left(x\right)[/latex] for each equation.
This is allowable because in the original statement of this problem we assumed that [latex]x>0[/latex]. (If [latex]x=0[/latex] then the original equation becomes [latex]0=2[/latex], which is clearly a false statement.)
In this equation, [latex]p\left(x\right)=-\frac{8}{3x}[/latex] and [latex]q\left(x\right)=-\frac{2}{3x}[/latex].
Subtract [latex]y[/latex] from each side and add [latex]4{x}^{2}-5\text{:}[/latex]
[latex]3y^{\prime} -y=4{x}^{2}-5[/latex].
Next divide both sides by [latex]3\text{:}[/latex]