Integrating Factor Method for First-Order Linear ODEs

Goal: solve a first-order linear differential equation by using an integrating factor
General form for a linear ODE: $\frac{dy}{dx}+p(x)\,y=f(x)$
Solving means finding a function (y(x)) that satisfies the equation on its domain
When the equation is linear, the integrating factor method is systematic and often the fastest route

Define the integrating factor (\mu(x)) to satisfy the identity
$\mu(x)\,\frac{dy}{dx}+\mu(x)\,p(x)\,y=\frac{d}{dx}\big(\mu(x)\,y\big)$
This requires choosing (\mu) such that
$\mu'(x)=p(x)\,\mu(x) \quad\Rightarrow\quad \mu(x)=\exp\Big(\int p(x)\,dx\Big)$
Multiply the original equation by (\mu(x)):
$\frac{d}{dx}\big(\mu(x)\,y\big)=\mu(x)\,f(x)$
Integrate both sides with respect to (x):
$\mu(x)\,y=\int \mu(x)\,f(x)\,dx + C$
Solve for (y):
$\boxed{\;y(x)=\mu(x)^{-1}\Big(\int \mu(x)\,f(x)\,dx + C\Big)\; }$
Key idea: choosing the correct integrating factor turns the left-hand side into a single derivative, enabling straightforward integration

Step 1: Write the equation in standard form: $\frac{dy}{dx}+p(x)\,y=f(x)$
Step 2: Compute integrating factor: $\mu(x)=\exp\Big(\int p(x)\,dx\Big)$
Step 3: Multiply the equation by (\mu(x)) to get a derivative on the left: $\frac{d}{dx}\big(\mu(x)\,y\big)=\mu(x)\,f(x)$
Step 4: Integrate: $\mu(x)\,y=\int \mu(x)\,f(x)\,dx + C$
Step 5: Solve for (y): \;y(x)=\mu(x)^{-1}\Big(\int \mu(x)\,f(x)\,dx + C\Big)\n

Identify: (p(x)=-\frac{4}{x}), (f(x)=x^{4} e^{x})
Integrating factor: \mu(x)=\exp\Big(\int -\frac{4}{x}\,dx\Big)=\exp(-4\ln x)=x^{-4}
Multiply through: left becomes \frac{d}{dx}\big(x^{-4} y\big) = x^{-4}\,f(x) = x^{-4}\,x^{4} e^{x} = e^{x}
Integrate both sides: x^{-4}\,y = \int e^{x}\,dx = e^{x} + C
Solve for (y): \boxed{\;y(x)=x^{4}\left(e^{x}+C\right) = x^{4} e^{x} + C x^{4}\; }
If an initial value
(y(x0)=y0) is given, substitute to find (C): (C=y0/x0^{4} - e^{x_0})

Consider the IVP: \frac{d a}{d t} + \frac{1}{100}\,a = 6, \quad a(0)=50
Integrating factor: \mu(t)=\exp\Big(\int \frac{1}{100}\,dt\Big)=e^{t/100}
Multiply through and integrate: \frac{d}{dt}\big(e^{t/100} a\big)=6\,e^{t/100} \Rightarrow e^{t/100} a = 600\,e^{t/100} + C
Solve for (a(t)): \boxed{a(t)=600 + C\,e^{-t/100}}
Apply (a(0)=50): (50=600+C \Rightarrow C=-550) so
\boxed{a(t)=600 - 550\,e^{-t/100}}
Transient term: the part that decays with time, here ( -550\,e^{-t/100} )
- As (t\to\infty), the solution tends to the steady-state value (600)
Takeaway: transient terms capture the effect of initial conditions and typically vanish for large (t)

Setup: tank with volume (V=300) gallons, inflow salt water at (3) gal/min with concentration (2) lb/gal, outflow at (3) gal/min; initial salt mass (A(0)=50) lb
Let (A(t)) be pounds of salt in tank at time (t) (minutes)
Inflow salt rate: (3\text{ gal/min} \times 2\text{ lb/gal} = 6) lb/min
Outflow salt rate: concentration in tank is (A(t)/V = A(t)/300) lb/gal, outflow rate is 3 gal/min, so outflow salt rate is 3\times\frac{A(t)}{300}=\frac{A(t)}{100} lb/min
Mass balance (dA/dt):
\frac{dA}{dt} = 6 - \frac{A}{100}
which is equivalent to
\boxed{\frac{dA}{dt} + \frac{1}{100}\,A = 6}
Integrating factor: (\mu(t)=e^{t/100})
Left-hand side becomes the derivative of (\mu A):
\frac{d}{dt}\big(e^{t/100} A\big)=6\,e^{t/100}
Integrate: e^{t/100} A = 600\,e^{t/100} + C \Rightarrow A(t)=600 + C e^{-t/100}
Apply initial condition (A(0)=50): (50=600+C \Rightarrow C=-550) so
\boxed{A(t)=600 - 550\,e^{-t/100}}
Interpretation
- Steady-state amount of salt in tank: 600 lb
- Transient term: (-550\,e^{-t/100}) decays to 0 as (t\to\infty)

Modified setup: inflow still (3) gal/min at concentration (2) lb/gal; volume now grows as (V(t)=300+t) (due to higher inflow than outflow)
Outflow rate remains (3) gal/min; however, concentration in tank is now (A(t)/V(t) = A(t)/(300+t))
Mass balance becomes
\frac{dA}{dt} = 6 - 3\cdot\frac{A}{300+t} \quad\Rightarrow\quad \frac{dA}{dt} + \frac{3}{300+t}\,A = 6
Integrating factor:
\mu(t) = \exp\Big(\int \frac{3}{300+t}\,dt\Big) = (300+t)^3
Multiply through and use product rule:
\frac{d}{dt}\big((300+t)^3 A\big) = 6(300+t)^3
Integrate:
(300+t)^3 A = \int 6(300+t)^3 dt = \frac{3}{2}(300+t)^4 + C
Solve for (A(t)):
\boxed{A(t)=\frac{3}{2}(300+t) + \frac{C}{(300+t)^3}}
Apply initial condition (A(0)=50) to find (C):
- (50 = \frac{3}{2}\cdot 300 + \frac{C}{300^3} = 450 + \frac{C}{300^3})
- This yields (C = (50-450)\cdot 300^3 = -400\cdot 300^3) (a large negative constant)
- Thus the explicit solution is
  \boxed{A(t)=\frac{3}{2}(300+t) - \frac{400\cdot 300^3}{(300+t)^3}}
Important practical note:
- If the tank volume grows without bound (or until capacity is reached), the model changes (overflow, capacity limits, etc.) and the formula must be adjusted accordingly
- In such a case, the solution shows a linear growth in the main term (due to increasing volume) plus a decaying transient term

Integrating factor is the central tool for linear first-order ODEs; it is chosen as
\mu(x)=\exp\Big(\int p(x)\,dx\Big)
The left-hand side becomes the derivative of a product: \frac{d}{dx}\big(\mu(x)\,y\big)
After integrating, the constant of integration (C) encodes the initial condition
Transient term: the portion of the solution that decays to zero as the independent variable grows (e.g., an exponential term like (e^{-\lambda t})); it captures the influence of initial conditions and dies out over time
Practical modeling tip: always check units (dimension analysis) to ensure consistency, especially in mixing problems where rates, concentrations, and volumes interact
In problems with time-varying volume, the outflow term often becomes (\text{outflow rate} \times \text{concentration} = \text{(flow rate)} \times \left( \frac{A(t)}{V(t)}\right)), leading to a variable-coefficient differential equation

Integrating factor: \mu(x)=\exp\Big(\int p(x)\,dx\Big)
Solution form: \boxed{ y(x)=\mu(x)^{-1}\Big(\int \mu(x)\,f(x)\,dx + C\Big) }
Example solution (constant volume): for \frac{dA}{dt}+\frac{1}{100}A=6 $, with (A(0)=50):$ A(t)=600-550e^{-t/100}
Variable volume form (volume (V(t))): \frac{dA}{dt}+\frac{3}{V(t)}A=6, \quad V(t)=300+t \quad\Rightarrow\quad A(t)=\frac{3}{2}\,V(t) + \frac{C}{V(t)^3} $$

The recipe is a powerful tool: when you recognize a linear first-order ODE, apply the integrating factor to transform the problem into a direct integration
Always verify the left-hand side becomes a derivative of a product by using the product rule as a quick check
For IVPs, determine the constant C from the given initial condition; this yields a unique solution on the interval of validity
The “transient term” concept helps interpret how solutions settle to a steady state in real-world processes (e.g., mixing tanks, chemical reactors), and it highlights the impact (and eventual fading) of initial conditions in dynamic systems