First-Order Linear Equations
Differential Equations · Section 2.1 · generated practice set
Integrating factors in the right order: standard form first, then the factor, then the product rule, then one integration.
Practice this set → Fresh numbers on every attempt. No account needed.
Method
The integrating-factor method, in the order you should write it on paper.
- Standard form. Divide until the coefficient of $y'$ is exactly 1: $y' + p(t)y = g(t)$. Do this first, every time. Most lost points start here.
- Integrating factor. $\mu(t) = e^{\int p(t)\,dt}$. No constant of integration needed — any antiderivative works.
- Recognize the product rule. $\mu y' + \mu p y = (\mu y)'$. So the equation becomes $(\mu y)' = \mu g$.
- Integrate once. $\mu y = \int \mu g\,dt + C$. This is where $C$ enters — exactly once.
- Divide by $\mu$ and then apply the initial condition.
Two integrating factors worth memorizing
$p = a$ constant $\Rightarrow \mu = e^{at}$. $p = a/t \Rightarrow \mu = e^{a\ln t} = t^a$.
The second one is the whole reason problems appear as $ty' + ay = \ldots$.
Definitions and theorems
Worked example
Solve $ty' + 2y = t^3$, $y(1) = 2$, for $t > 0$.
- Standard form: divide by $t$ → $y' + \frac{2}{t}y = t^2$.
- $\mu = e^{\int 2/t\,dt} = e^{2\ln t} = t^2$.
- Multiply: $t^2y' + 2ty = t^4$, and the left side is $(t^2y)'$.
- Integrate: $t^2y = \frac{t^5}{5} + C$, so $y = \frac{t^3}{5} + \frac{C}{t^2}$.
- $y(1) = 2$: $2 = \frac15 + C \Rightarrow C = \frac95$. So $y = \frac{t^3}{5} + \frac{9}{5t^2}$.
Common mistakes
- Skipping standard form. Computing $\mu$ from $ty' + 2y$ without dividing by $t$ gives $e^{2t}$ — wrong, and everything after it is wrong too.
- Adding $C$ inside $\mu$. The constant belongs to the single integration in step 4, not to $\int p\,dt$.
- Forgetting to divide the $C$ by $\mu$. The answer is $\frac{1}{\mu}\left(\int\mu g + C\right)$, so $C$ carries a factor $1/\mu$ — here $C/t^2$, not a bare $C$.
- Sign errors in $\mu$ when $p < 0$. $y' - 3y = \ldots$ has $\mu = e^{-3t}$.
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.