Solutions to Some ODEs
Differential Equations · Section 1.2 · generated practice set
The constant-coefficient first-order template, solved once as a constant plus an exponential, and what the sign of the rate decides about the long run.
Practice this set → Fresh numbers on every attempt. No account needed.
Method
Everything in 1.2 is the single template $y' = ay + b$ with $a \neq 0$. Learn one move and you own the section.
The move
Factor the right side so the equilibrium is visible: $$y' = a\left(y + \tfrac{b}{a}\right).$$ Substituting $u = y + b/a$ gives $u' = au$, whose solution is $u = Ce^{at}$. Undo the substitution:
$$y(t) = -\frac{b}{a} + Ce^{at}, \qquad C = y_0 + \frac{b}{a}.$$
Read the structure: the constant $-b/a$ is the equilibrium, $C$ measures how far the initial value sits from it, and the sign of $a$ decides whether that gap shrinks or explodes.
On an exam, if a first-order problem has constant coefficients, write the answer in the form $A + Be^{rt}$ and identify the three numbers. It is faster than an integrating factor and far faster than separating.
Definitions and theorems
Worked example
Solve $y' = 3y - 12$, $y(0) = 2$, and describe the behavior as $t \to \infty$.
- Factor: $y' = 3(y - 4)$, so the equilibrium is $y = 4$.
- Let $u = y - 4$: $u' = 3u \Rightarrow u = Ce^{3t}$, so $y = 4 + Ce^{3t}$.
- $y(0) = 2$: $2 = 4 + C \Rightarrow C = -2$.
- $y(t) = 4 - 2e^{3t}$.
- Since $a = 3 > 0$ and $C = -2 < 0$, $y \to -\infty$. The initial value started below the unstable equilibrium and ran away downward.
Common mistakes
- Sign of the equilibrium. For $y' = ay + b$ the equilibrium is $-b/a$, not $b/a$. Check it by plugging back in.
- Applying the IC before solving for $C$ correctly. $C$ is the gap $y_0 - y^*$, not $y_0$.
- Dropping the constant of integration. $u' = au$ integrates to $\ln|u| = at + c$, and the $c$ becomes the multiplicative $C$ after exponentiating.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.