### 6 The integrating factor

The equation

$\phantom{\rule{2em}{0ex}}{x}^{2}\frac{dy}{dx}+3x\phantom{\rule{1em}{0ex}}y={x}^{3}$

is not exact. However, if we multiply it by $x$ we obtain the equation

$\phantom{\rule{2em}{0ex}}{x}^{3}\frac{dy}{dx}+3{x}^{2}y={x}^{4}.$

This can be re-written as

$\phantom{\rule{2em}{0ex}}\frac{d}{dx}\left({x}^{3}y\right)={x}^{4}$

which is an exact equation with solution

$\phantom{\rule{2em}{0ex}}{x}^{3}y=\int {x}^{4}dx$

$\phantom{\rule{2em}{0ex}}\text{so}\phantom{\rule{2em}{0ex}}{x}^{3}y=\frac{1}{5}{x}^{5}+C$

and hence

$\phantom{\rule{2em}{0ex}}y=\frac{1}{5}{x}^{2}+\frac{C}{{x}^{3}}.$

The function by which we multiplied the given differential equation in order to make it exact is called an integrating factor . In this example the integrating factor is simply $x$ .

Which of the following differential equations can be made exact by multiplying by ${x}^{2}$ ?

1. $\frac{dy}{dx}+\frac{2}{x}y=4$
2. $x\phantom{\rule{1em}{0ex}}\frac{dy}{dx}+3y={x}^{2}$
3. $\frac{1}{x}\phantom{\rule{1em}{0ex}}\frac{dy}{dx}-\frac{1}{{x}^{2}}y=x$

4. $\frac{1}{x}\phantom{\rule{1em}{0ex}}\frac{dy}{dx}+\frac{1}{{x}^{2}}y=3.$

Where possible, write the exact equation in the form $\frac{d}{dx}\left(f\left(x\right)\phantom{\rule{1em}{0ex}}y\right)=g\left(x\right)$ .

1. Yes. ${x}^{2}\frac{dy}{dx}+2xy=4{x}^{2}$ becomes $\frac{d}{dx}\left({x}^{2}y\right)=4{x}^{2}$ .
2. Yes. ${x}^{3}\frac{dy}{dx}+3{x}^{2}y={x}^{4}$ becomes $\frac{d}{dx}\left({x}^{3}y\right)={x}^{4}$ .
3. No. This equation is already exact as it can be written in the form $\frac{d}{dx}\left(\frac{1}{x}\phantom{\rule{1em}{0ex}}y\right)=x$ .
4. Yes. $x\frac{dy}{dx}+y=3{x}^{2}$ becomes $\frac{d}{dx}\left(xy\right)=3{x}^{2}$ .