Solving first order linear ordinary differential equations

Integrating factors are useful for solving ordinary differential equations that can be expressed in the form

${\displaystyle y'+P(x)y=Q(x)}$

The basic idea is to find some function, say ${\displaystyle M(x)}$, called the "integrating factor", which we can multiply through our differential equation in order to bring the left-hand side under a common derivative. For the canonical first-order linear differential equation shown above, the integrating factor is ${\displaystyle e^{\int P(x)\,dx}}$.

Note that it is not necessary to include the arbitrary constant in the integral, or absolute values in case the integral of ${\displaystyle P(x)}$ involves a logarithm. Firstly, we only need one integrating factor to solve the equation, not all possible ones; secondly, such constants and absolute values will cancel out even if included. For absolute values, this can be seen by writing ${\displaystyle |f(x)|=f(x)\operatorname {sgn} f(x)}$, where ${\displaystyle \operatorname {sgn} }$ refers to the sign function, which will be constant on an interval if ${\displaystyle f(x)}$ is continuous. As ${\displaystyle \ln |f(x)|}$ is undefined when ${\displaystyle f(x)=0}$, and a logarithm in the antiderivative only appears when the original function involved a logarithm or a reciprocal (neither of which are defined for 0), such an interval will be the interval of validity of our solution.

To derive this, let ${\displaystyle M(x)}$ be the integrating factor of a first order linear differential equation such that multiplication by ${\displaystyle M(x)}$ transforms a partial derivative into a total derivative, then:

1. ${\displaystyle M(x){\underset {\text{partial derivative}}{(\underbrace {y'+P(x)y} )}}}$
2. ${\displaystyle M(x)y'+M(x)P(x)y}$
3. ${\displaystyle \underbrace {M(x)y'+M'(x)y} _{\text{total derivative}}}$

Going from step 2 to step 3 requires that ${\displaystyle M(x)P(x)=M'(x)}$, which is a separable differential equation, whose solution yields ${\displaystyle M(x)}$ in terms of ${\displaystyle P(x)}$:

4. ${\displaystyle M(x)P(x)=M'(x)}$
5. ${\displaystyle P(x)={\frac {M'(x)}{M(x)}}}$
6. ${\displaystyle \int P(x)\,dx=\ln M(x)+c}$
7. ${\displaystyle M(x)=Ce^{\int P(x)\,dx}}$

To verify, multiplying by ${\displaystyle M(x)}$ gives

${\displaystyle M(x)y'+P(x)M(x)y=Q(x)M(x)}$

By applying the product rule in reverse, we see that the left-hand side can be expressed as a single derivative in ${\displaystyle x}$

${\displaystyle M(x)y'+P(x)M(x)y=M(x)y'+M'(x)y={\frac {d}{dx}}(M(x)y)}$

We use this fact to simplify our expression to

${\displaystyle {\frac {d}{dx}}\left(M(x)y\right)=Q(x)M(x)}$

Integrating both sides with respect to ${\displaystyle x}$

${\displaystyle Ce^{\int P(x)\,dx}y=\int Q(x)Ce^{\int P(x)\,dx}dx}$

${\displaystyle e^{\int P(x)\,dx}y=\left(\int Q(x)e^{\int P(x)\,dx}\,dx\right)+C}$

where ${\displaystyle C}$ is a constant.

Moving the exponential to the right-hand side, the general solution to Ordinary Differential Equation is:

${\displaystyle y=e^{-\int P(x)\,dx}\left(\int Q(x)e^{\int P(x)\,dx}\,dx\right)+Ce^{-\int P(x)\,dx}}$

In the case of a homogeneous differential equation, ${\displaystyle Q(x)=0}$ and the general solution to Ordinary Differential Equation is:

${\displaystyle y=Ce^{-\int P(x)\,dx}}$.

for example, consider the differential equation

${\displaystyle y'-{\frac {2y}{x}}=0.}$

We can see that in this case ${\displaystyle P(x)={\frac {-2}{x}}}$

${\displaystyle M(x)=e^{\int _{1}^{x}P(x)dx}}$

${\displaystyle M(x)=e^{\int _{1}^{x}{\frac {-2}{x}}\,dx}=e^{-2\ln x}={\left(e^{\ln x}\right)}^{-2}=x^{-2}}$

${\displaystyle M(x)={\frac {1}{x^{2}}}.}$

Multiplying both sides by ${\displaystyle M(x)}$ we obtain

${\displaystyle {\frac {y'}{x^{2}}}-{\frac {2y}{x^{3}}}=0}$

The above equation can be rewritten as

${\displaystyle {\frac {d(x^{-2}y)}{dx}}=0}$

By integrating both sides with respect to x we obtain

${\displaystyle x^{-2}y=C}$

or

${\displaystyle y=Cx^{2}}$

The same result may be achieved using the following approach

${\displaystyle {\frac {y'}{x^{2}}}-{\frac {2y}{x^{3}}}=0}$

${\displaystyle {\frac {y'x^{3}-2x^{2}y}{x^{5}}}=0}$

${\displaystyle {\frac {x(y'x^{2}-2xy)}{x^{5}}}=0}$

${\displaystyle {\frac {y'x^{2}-2xy}{x^{4}}}=0.}$

Reversing the quotient rule gives

${\displaystyle \left({\frac {y}{x^{2}}}\right)'=0}$

or

${\displaystyle {\frac {y}{x^{2}}}=C,}$

or

${\displaystyle y=Cx^{2}.}$

where ${\displaystyle C}$ is a constant.

Solving second order linear ordinary differential equations

The method of integrating factors for first order equations can be naturally extended to second order equations as well. The main goal in solving first order equations was to find an integrating factor ${\displaystyle M(x)}$ such that multiplying ${\displaystyle y'+p(x)y=h(x)}$ by it would yield ${\displaystyle (M(x)y)'=M(x)h(x)}$, after which subsequent integration and division by ${\displaystyle M(x)}$ would yield ${\displaystyle y}$. For second order linear differential equations, if we want ${\displaystyle M(x)=e^{\int p(x)\,dx}}$ to work as an integrating factor, then

${\displaystyle (M(x)y)''=M(x)\left(y''+2p(x)y'+\left(p(x)^{2}+p'(x)\right)y\right)=M(x)h(x)}$

This implies that a second order equation must be exactly in the form ${\displaystyle y''+2p(x)y'+\left(p(x)^{2}+p'(x)\right)y=h(x)}$ for the integrating factor to be usable.

Solving nth order linear differential equations

Integrating factors can be extended to any order, though the form of the equation needed to apply them gets more and more specific as order increases, making them less useful for orders 3 and above. The general idea is to differentiate the function ${\displaystyle M(x)y}$ ${\displaystyle n}$ times for an ${\displaystyle n}$th order differential equation and combine like terms. This will yield an equation in the form

${\displaystyle M(x)F\!\left(y,y',y'',\ldots ,y^{(n)}\right)}$

If an ${\displaystyle n}$th order equation matches the form ${\displaystyle F\!\left(y,y',y'',\ldots ,y^{(n)}\right)}$ that is gotten after differentiating ${\displaystyle n}$ times, one can multiply all terms by the integrating factor and integrate ${\displaystyle h(x)M(x)}$ ${\displaystyle n}$ times, dividing by the integrating factor on both sides to achieve the final result.