Linear differential equations (integrating factor)

MATLinear Equations (Integrating Factor)

A first-order linear differential equation has the standard form $\dfrac{dy}{dx}+P(x)\,y=Q(x)$ , where $P,Q$ are functions of $x$ alone. It is solved by multiplying through by the integrating factor $e^{\int P\,dx}$ , which turns the left side into the derivative of a product.

This is one of the most heavily examined methods in ISC, including the variant that is linear in $x$ (treating $x$ as a function of $y$ ).

Standard linear form & solution

\dfrac{dy}{dx}+P(x)\,y=Q(x),\qquad \text{I.F.}=e^{\int P\,dx},\qquad y\cdot(\text{I.F.})=\int Q\,(\text{I.F.})\,dx+C

P,Q

are functions of

x

only;

C

is the constant of integration; the I.F. makes the LHS

\dfrac{d}{dx}\!\big(y\cdot\text{I.F.}\big)

Linear in

x

(treat

x

as function of

y

)

\dfrac{dx}{dy}+P(y)\,x=Q(y),\qquad \text{I.F.}=e^{\int P\,dy},\qquad x\cdot(\text{I.F.})=\int Q\,(\text{I.F.})\,dy+C

P,Q

are functions of

y

only; used when the equation is linear in

x

but not in

y

Worked I.F. example

\dfrac{dy}{dx}+\dfrac{y}{x}=x:\quad \text{I.F.}=e^{\int \frac{1}{x}\,dx}=x,\quad xy=\int x\cdot x\,dx=\dfrac{x^{3}}{3}+C

P=\dfrac{1}{x},\ Q=x

; so

y=\dfrac{x^{2}}{3}+\dfrac{C}{x}

Common integrating-factor shortcuts

e^{\int \frac{n}{x}\,dx}=x^{n},\qquad e^{\int \tan x\,dx}=\sec x,\qquad e^{\int \cot x\,dx}=\sin x

standard simplifications of

e^{\int P\,dx}

; modulus signs are usually dropped on the chosen domain.

First write the equation in standard form so that the coefficient of $\dfrac{dy}{dx}$ is exactly $1$ before identifying $P$ and $Q$ .
The integrating factor is $\text{I.F.}=e^{\int P\,dx}$ ; no constant of integration is added inside this exponent.
After multiplying by the I.F., the left side is exactly $\dfrac{d}{dx}\big(y\cdot\text{I.F.}\big)$ , so integrate both sides directly.
Add the constant $C$ only once, when integrating $\int Q\,(\text{I.F.})\,dx$ .
If the equation is linear in $x$ (e.g. $\dfrac{dx}{dy}-3x=y$ ), switch roles: use $\dfrac{dx}{dy}+P(y)x=Q(y)$ with $\text{I.F.}=e^{\int P\,dy}$ .
Useful identities: $e^{\int \tan x\,dx}=\sec x$ , $e^{\int \cot x\,dx}=\sin x$ and $e^{\int \frac{n}{x}\,dx}=x^{n}$ .
For an initial-value problem, find the general solution first, then substitute the given point to evaluate $C$ .
Always simplify the I.F. (e.g. write $\sec x$ , not $e^{\int\tan x\,dx}$ ) before doing the final integral $\int Q\,(\text{I.F.})\,dx$ .

Where the marks go

Identifying $P$ and $Q$ before normalising the leading coefficient to $1$ — e.g. with $x\dfrac{dy}{dx}+2y=x\ln x$ you must divide by $x$ first.
Forgetting the constant $C$ , or adding a spurious constant inside the exponent of the integrating factor.
Not simplifying the I.F.: leaving $e^{\int \tan x\,dx}$ instead of $\sec x$ makes the final integral much harder.
Applying the $x$ -linear form's I.F. as $e^{\int P\,dx}$ instead of $e^{\int P\,dy}$ when the equation is linear in $x$ with $y$ as the independent variable.

How the board asks it

Numericalstandard linear form and integrating factor
Solve the differential equation $\dfrac{dy}{dx}+2y\tan x=\sin x$ .
Numericalinitial-value problem
Find the particular solution of $\dfrac{dy}{dx}+y\cot x=2x+x^2\cot x$ , given that $y=0$ when $x=\dfrac{\pi}{2}$ .
Numericallinear in $x$ (treat $x$ as a function of $y$ )
Solve the differential equation $\big(1+y^2\big)\,dx=\big(\tan^{-1}y-x\big)\,dy$ .
Numericalnormalising the leading coefficient
Solve the differential equation $x\dfrac{dy}{dx}+2y=x^2\log x$ by first reducing it to standard linear form.
Numericalinitial-value problem with $\text{I.F.}=x$
Solve $\dfrac{dy}{dx}+\dfrac{y}{x}=x^2$ given that $y=1$ when $x=1$ , and hence find $y$ when $x=2$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.