# 4.5 First-order linear equations  (Page 2/10)

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Examples of first-order nonlinear differential equations include

$\begin{array}{ccc}\hfill {\left(y\prime \right)}^{4}-{\left(y\prime \right)}^{3}& =\hfill & \left(3x-2\right)\left(y+4\right)\hfill \\ \hfill 4y\prime +3{y}^{3}& =\hfill & 4x-5\hfill \\ \hfill {\left(y\prime \right)}^{2}& =\hfill & \text{sin}\phantom{\rule{0.1em}{0ex}}y+\text{cos}\phantom{\rule{0.1em}{0ex}}x.\hfill \end{array}$

These equations are nonlinear because of terms like ${\left({y}^{\prime }\right)}^{4},{y}^{3},$ etc. Due to these terms, it is impossible to put these equations into the same form as [link] .

## Standard form

Consider the differential equation

$\left(3{x}^{2}-4\right){y}^{\prime }+\left(x-3\right)y=\text{sin}\phantom{\rule{0.1em}{0ex}}x.$

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 ${y}^{\prime }$ be equal to $1.$ To make this happen, we divide both sides by $3{x}^{2}-4.$

${y}^{\prime }+\left(\frac{x-3}{3{x}^{2}-4}\right)\phantom{\rule{0.1em}{0ex}}y=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{3{x}^{2}-4}$

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 [link] , we can divide both sides of the equation by $a\left(x\right).$ This leads to the equation

${y}^{\prime }+\frac{b\left(x\right)}{a\left(x\right)}y=\frac{c\left(x\right)}{a\left(x\right)}.$

Now define $p\left(x\right)=\frac{b\left(x\right)}{a\left(x\right)}$ and $q\left(x\right)=\frac{c\left(x\right)}{a\left(x\right)}.$ Then [link] becomes

${y}^{\prime }+p\left(x\right)y=q\left(x\right).$

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.

## Writing first-order linear equations in standard form

Put each of the following first-order linear differential equations into standard form. Identify $p\left(x\right)$ and $q\left(x\right)$ for each equation.

1. $y\prime =3x-4y$
2. $\frac{3xy\prime }{4y-3}=2$ (here $x>0\right)$
3. $y=3y\prime -4{x}^{2}+5$
1. Add $4y$ to both sides:
$y\prime +4y=3x.$

In this equation, $p\left(x\right)=4$ and $q\left(x\right)=3x.$
2. Multiply both sides by $4y-3,$ then subtract $8y$ from each side:
$\begin{array}{ccc}\hfill \frac{3xy\prime }{4y-3}& =\hfill & 2\hfill \\ \hfill 3xy\prime & =\hfill & 2\left(4y-3\right)\hfill \\ \hfill 3xy\prime & =\hfill & 8y-6\hfill \\ \hfill 3xy\prime -8y& =\hfill & -6.\hfill \end{array}$

Finally, divide both sides by $3x$ to make the coefficient of $y\prime$ equal to $1\text{:}$
$y\prime -\frac{8}{3x}y=-\frac{2}{3x}.$
This is allowable because in the original statement of this problem we assumed that $x>0.$ (If $x=0$ then the original equation becomes $0=2,$ which is clearly a false statement.)
In this equation, $p\left(x\right)=-\frac{8}{3x}$ and $q\left(x\right)=-\frac{2}{3x}.$
3. Subtract $y$ from each side and add $4{x}^{2}-5\text{:}$
$3y\prime -y=4{x}^{2}-5.$

Next divide both sides by $3\text{:}$
$y\prime -\frac{1}{3}y=\frac{4}{3}{x}^{2}-\frac{5}{3}.$

In this equation, $p\left(x\right)=-\frac{1}{3}$ and $q\left(x\right)=\frac{4}{3}{x}^{2}-\frac{5}{3}.$

Put the equation $\frac{\left(x+3\right)y\prime }{2x-3y-4}=5$ into standard form and identify $p\left(x\right)$ and $q\left(x\right).$

$y\prime +\frac{15}{x+3}y=\frac{10x-20}{x+3};p\left(x\right)=\frac{15}{x+3}$ and $q\left(x\right)=\frac{10x-20}{x+3}$

## Integrating factors

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:

$y\prime +p\left(x\right)y=q\left(x\right).$

The first term on the left-hand side of [link] is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from the Differentiation Rules section. If we multiply [link] by a yet-to-be-determined function $\mu \left(x\right),$ then the equation becomes

$\mu \left(x\right){y}^{\prime }+\mu \left(x\right)p\left(x\right)y=\mu \left(x\right)q\left(x\right).$

The left-hand side [link] can be matched perfectly to the product rule:

$\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]={f}^{\prime }\left(x\right)g\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right).$

Matching term by term gives $y=f\left(x\right),g\left(x\right)=\mu \left(x\right),$ and ${g}^{\prime }\left(x\right)=\mu \left(x\right)p\left(x\right).$ Taking the derivative of $g\left(x\right)=\mu \left(x\right)$ and setting it equal to the right-hand side of ${g}^{\prime }\left(x\right)=\mu \left(x\right)p\left(x\right)$ leads to

${\mu }^{\prime }\left(x\right)=\mu \left(x\right)p\left(x\right).$

This is a first-order, separable differential equation for $\mu \left(x\right).$ We know $p\left(x\right)$ because it appears in the differential equation we are solving. Separating variables and integrating yields

$\begin{array}{ccc}\hfill \frac{{\mu }^{\prime }\left(x\right)}{\mu \left(x\right)}& =\hfill & p\left(x\right)\hfill \\ \hfill \int \frac{{\mu }^{\prime }\left(x\right)}{\mu \left(x\right)}dx& =\hfill & \int p\left(x\right)\phantom{\rule{0.1em}{0ex}}dx\hfill \\ \hfill \text{ln}|\mu \left(x\right)|& =\hfill & \int p\left(x\right)\phantom{\rule{0.1em}{0ex}}dx+C\hfill \\ \hfill {e}^{\text{ln}|\mu \left(x\right)|}& =\hfill & {e}^{\int p\left(x\right)\phantom{\rule{0.1em}{0ex}}dx+C}\hfill \\ \hfill |\mu \left(x\right)|& =\hfill & {C}_{1}{e}^{\int p\left(x\right)\phantom{\rule{0.1em}{0ex}}dx}\hfill \\ \hfill \mu \left(x\right)& =\hfill & {C}_{2}{e}^{\int p\left(x\right)\phantom{\rule{0.1em}{0ex}}dx}.\hfill \end{array}$

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