First-order linear equations
Chapter 5 - Manufacture a product derivative
A first-order linear equation has the normalized form
The obstacle is the extra (p(t)y) term. An integrating factor (\mu(t)) makes the left side become one product derivative:
Derive the method once
The product rule says
To match (\mu y'+\mu p y), require (\mu'=\mu p), or (\mu'/\mu=p). Therefore
This formula is not a magic insertion. It solves the small design problem "which multiplier makes these two terms a product derivative?"
The reliable sequence
- Normalize the coefficient of (y') to 1.
- Identify (p(t)) and (q(t)).
- Compute (\mu=e^{\int p(t)dt}).
- Multiply every term by (\mu).
- Replace the left side with ((\mu y)').
- Integrate and solve for (y).
- Apply the initial condition and verify.
Worked example: full support
- Read the model. \(y'=6-2y=-2(y-3)\). Before solving, predict movement toward the equilibrium \(y=3\). Starting at 5, the curve should decrease and flatten.
- Identify. \(p(t)=2\), \(q(t)=6\).
- Build the factor. \(\mu=e^{\int 2dt}=e^{2t}\).
- Multiply all terms. \(e^{2t}y'+2e^{2t}y=6e^{2t}\).
- Recognize the product. \((e^{2t}y)'=6e^{2t}\).
- Integrate. \(e^{2t}y=3e^{2t}+C\), hence \(y=3+Ce^{-2t}\).
- Use and check. \(5=3+C\), so \(y=3+2e^{-2t}\). Its graph decreases toward 3, matching the prediction. Substitution gives \(y'+2y=-4e^{-2t}+6+4e^{-2t}=6\).
Backward-faded example
Solve (ty'-y=t^3), (t>0), with (y(1)=2).
Normalize: \(y'-\frac1t y=t^2\).
Thus \(p(t)=-1/t\).
\(\mu=e^{\int -1/t\,dt}=\underline{\hspace{4em}}\).
Then \((\underline{\hspace{3em}}\,y)'=\underline{\hspace{4em}}\).
Integrate, use \(y(1)=2\), and verify in the normalized equation.
Check the missing work
On \(t>0\), \(\mu=e^{-\ln t}=1/t\). Multiplication gives \((y/t)'=t\). Thus \(y/t=t^2/2+C\), so \(y=t^3/2+Ct\). The condition gives \(C=3/2\), hence \(y=(t^3+3t)/2\).
When linear and separable overlap
The homogeneous linear equation (y'+p(t)y=0) is also separable:
For (y'=ty), separation is shorter. For (y'+p(t)y=q(t)) with nonzero (q), the integrating-factor route usually survives while separation fails.
Normalize first. For \(2ty'+6y=t^2\) on \(t>0\), what is \(p(t)\)?
Modeling example: a mixing tank
A 100 L tank receives brine at 2 L/min with concentration 0.1 kg/L. The same volume leaves each minute. If (S(t)) is kilograms of salt, then
In linear form, (S'+0.02S=0.2). Before solving, the zero-rate amount is 10 kg, so every solution should move toward 10. The integrating factor gives (S=10+Ce^{-0.02t}). The constant records the initial displacement from equilibrium.
Error clinic
In \(t y'+2y=1\), the coefficient used in the integrating factor is \(2/t\), not 2. Divide by the coefficient of \(y'\) first and state the interval, such as \(t>0\).
An integrating factor preserves equality only when it multiplies every term. Write the multiplied equation in full before compressing the product derivative.
Practice
- Practice - constant coefficient
5.1 Solve \(y'-3y=6\), \(y(0)=1\). Predict the long-run behavior first and reconcile it with the result.
- Practice - variable coefficient
5.2 Solve \(y'+(2/t)y=t\) for \(t>0\), with \(y(1)=0\).
- Integrated - model
5.3 A room at 20 °C contains an object initially at 80 °C. Newton cooling gives \(T'=-0.25(T-20)\). Solve, then find when the object reaches 30 °C.
- Challenge - reverse engineering
5.4 Find a first-order linear differential equation whose general solution is \(y=t^2+Ce^{-t}\). Explain how your equation separates the forced response from the transient.
Retrieve and discriminate
On blank paper, solve only far enough to expose the key transformation:
- (y'=t(1-y)) by separation;
- (y'+ty=t) by an integrating factor; and
- explain why the surface similarity does not make the methods interchangeable.