First-order linear equations

Chapter 5 - Manufacture a product derivative

A first-order linear equation has the normalized form

$$y'+p(t)y=q(t).$$

The obstacle is the extra $p(t)y$ term. An integrating factor $\mu(t)$ makes the left side become one product derivative:

$$\mu y'+\mu p y=(\mu y)'.$$

Derive the method once

The product rule says

$$(\mu y)'=\mu y'+\mu' y.$$

To match $\mu y'+\mu p y$, require $\mu'=\mu p$, or $\mu'/\mu=p$. Therefore

$$\boxed{\mu(t)=e^{\int p(t),dt}}.$$

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

1. Normalize the coefficient of $y'$ to 1.
2. Identify $p(t)$ and $q(t)$.
3. Compute $\mu=e^{\int p(t)dt}$.
4. Multiply every term by $\mu$.
5. Replace the left side with $(\mu y)'$.
6. Integrate and solve for $y$.
7. Apply the initial condition and verify.

Worked example: full support

Backward-faded example

Solve $ty'-y=t^3$, $t\gt 0$, with $y(1)=2$.

When linear and separable overlap

The homogeneous linear equation $y'+p(t)y=0$, with no external input on the right, is also separable:

$$\frac{y'}{y}=-p(t).$$

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.

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

$$S'=\underbrace{(2)(0.1)}$${\text{rate in}}-\underbrace{(2)\frac{S}{100}}{\text{rate out}} =0.2-0.02S.

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

Practice

1. Practice - constant coefficient

   5.1 Solve $y'-3y=6$, $y(0)=1$. Predict the long-run behavior first and reconcile it with the result.

   Detailed solution

2. Practice - variable coefficient

   5.2 Solve $y'+(2/t)y=t$ for $t\gt 0$, with $y(1)=0$.

   Detailed solution

3. 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.

   Detailed solution

4. 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.

   Detailed solution

5. Practice - extension

   5.5 Solve $y'+2y=e^{-t}$, $y(0)=0$, using an integrating factor. For $t\ge0$, determine when the solution reaches its maximum and find that maximum. Check both the equation and the initial condition.

   Detailed solution

6. Challenge - transfer

   5.6 Solve the variable-volume tank model $S'+S/(100+t)=6$, $S(0)=0$, for $t\ge0$, with $S$ in grams and $100+t$ in liters. Find the concentration $S/(100+t)$ and its limit. Explain why salt amount has no finite limit even though concentration does.

   Detailed solution

Retrieve and discriminate

On blank paper, solve only far enough to expose the key transformation:

1. $y'=t(1-y)$ by separation;
2. $y'+ty=t$ by an integrating factor; and
3. show that these are the same equation, so both methods apply; then explain why replacing the right side of the second equation by $t^2$ generally removes separability but preserves linearity.

I derived the integrating factor from the product rule and completed the faded solution.

Next: Use phase lines and feedback to understand autonomous equations.