Maths › Differential equations › First order equations and integrating factors
First order equations and integrating factors
Multiply a stubborn first order equation by exactly the right function and its left side folds into one derivative, ready to integrate whole.
Builds on Solving differential equations and The product, quotient and chain rules.
IN THIS TOPIC
- Recognise the linear form dy/dx + P(x)y = Q(x) and rearrange into it.
- Build the integrating factor e^∫P dx and collapse the left side to one derivative.
- Fix the constant from a boundary condition and check by substituting back.
WHAT YOU PROBABLY THINK
If the variables in a first order equation refuse to separate, the equation cannot be solved.
The multiplier that folds the equation
Separation is a special trick; the workhorse is broader. Any linear equation dy/dx + P(x)y = Q(x), separable or not, gives way when both sides are multiplied by the integrating factor:
The point of this exact multiplier is the product rule read in reverse: after multiplying, the left side is precisely the derivative of IF × y, so integrating both sides needs one step. Non-separable is not unsolvable, which sinks the opening claim.
WORKED EXAMPLE
A full solution, checked
Solve dy/dx + y/x = x for x > 0, given y(1) = 1.
P = 1/x, so IF = e^(ln x) = x.
Multiply through: x dy/dx + y = x², and the left side is d(xy)/dx.
Integrate: xy = x³/3 + c, so y = x²/3 + c/x. With y(1) = 1: 1 = 1/3 + c, giving c = 2/3.
y = x²/3 + 2/(3x). Substituting back: dy/dx + y/x = (2x/3 − 2/3x²) + (x/3 + 2/3x²) = x. It works.
Constant coefficients and the long run
When P is a constant the factor is a plain exponential, and the solution splits into a steady part driven by Q and a transient Ce^(−Px) that dies away. Reading which part is which turns algebra into behaviour: every solution of dy/dx + 3y = 6 slides onto y = 2 whatever its starting value.
WORKED EXAMPLE
Steady state plus transient
Solve dy/dx + 3y = 6.
IF = e3x: the equation becomes d(e3xy)/dx = 6e3x.
Integrate: e3xy = 2e3x + C.
y = 2 + Ce−3x: a fixed level 2, plus a memory of the start that decays with time constant 1/3.
YOUR TURN
An exponential right-hand side
Solve dy/dx + 2y = e−x.
Show the working
IF = e2x, so d(e2xy)/dx = e2x × e−x = ex.
Integrate: e2xy = ex + C.
y = e−x + Ce−2x. Check: dy/dx + 2y = (−e−x − 2Ce−2x) + (2e−x + 2Ce−2x) = e−x.
THE EXAM BIT
- Divide through first so the dy/dx coefficient is 1; the IF formula assumes it.
- No constant is needed inside e^∫P dx; any choice cancels in the working.
- After integrating, divide by the IF before applying the boundary condition.
- Substituting the final answer back into the equation is a one-line check worth doing.
CHECK YOURSELF
Write down the integrating factor for dy/dx + 2xy = x, and the equation it produces.
Show a hint
∫2x dx = x².
Show the answer
I
F
=
e
^
(
x
²
)
.
T
h
e
e
q
u
a
t
i
o
n
b
e
c
o
m
e
s
d
(
e
^
(
x
²
)
y
)
/
d
x
=
x
e
^
(
x
²
)
,
w
h
o
s
e
r
i
g
h
t
s
i
d
e
i
n
t
e
g
r
a
t
e
s
b
y
r
e
c
o
g
n
i
t
i
o
n
t
o
½
e
^
(
x
²
)
+
c
.
Linear first order: multiply by e^∫P dx and the left side becomes d(IF × y)/dx.
Constant P splits solutions into steady state plus a transient Ce^(−Px) that dies away.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.
- Recognise the linear form dy/dx + P(x)y = Q(x) and rearrange into it.
- Build the integrating factor e^∫P dx and collapse the left side to one derivative.
- Fix the constant from a boundary condition and check by substituting back.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.