Classification of Differential Equations
Differential Equations · Section 1.3 · generated practice set
Order, linear versus nonlinear, and the substitutions that turn a proposed solution into a characteristic polynomial.
Practice this set → Fresh numbers on every attempt. No account needed.
Method
Two questions get asked here, and a third that looks different but is not.
Order
The highest derivative of the unknown that appears. $y^{(4)}$ is order 4 — the parentheses mean a derivative count, not a power.
Linear or not
Linear means the equation can be written $$a_n(t)y^{(n)} + \cdots + a_1(t)y' + a_0(t)y = g(t).$$ Test it this way: every appearance of $y$ and its derivatives must be first power, never multiplied together, never inside a function. Coefficients may be as ugly as you like as long as they involve only $t$.
- $t^2y'' + e^t y = \ln t$ → linear (ugly in $t$, clean in $y$).
- $y'' + \sin y = 0$ → nonlinear ($y$ inside $\sin$).
- $(1 + y^2)y'' = t$ → nonlinear ($y$ in a coefficient).
- $y'y'' = t$ → nonlinear (two derivatives multiplied).
Verifying a proposed solution
Substitute and cancel. For $y = e^{rt}$ in a constant-coefficient equation you always land on the characteristic polynomial; for $y = t^r$ in an Euler equation $t^2y'' + \alpha ty' + \beta y = 0$ you land on $r(r-1) + \alpha r + \beta = 0$.
Definitions and theorems
Worked example
Classify $t^2y''' + (\cos t)y' - 5y = e^t$, then find all $r$ with $y = e^{rt}$ solving $y'' - y' - 6y = 0$.
- Highest derivative is $y'''$, so the order is 3.
- Each of $y'''$, $y'$, $y$ appears to the first power with coefficients in $t$ only, so it is linear.
- For the second equation substitute $y = e^{rt}$: $y' = re^{rt}$, $y'' = r^2e^{rt}$.
- $e^{rt}(r^2 - r - 6) = 0$, and $e^{rt} \neq 0$, so $r^2 - r - 6 = 0$.
- $(r-3)(r+2) = 0 \Rightarrow r = 3, -2$.
Common mistakes
- Counting $y^{(4)}$ as a power. It is the fourth derivative; the equation is order 4 and can still be linear.
- Calling an equation nonlinear because $t$ appears nonlinearly. $t^2$, $\cos t$, $\ln t$ as coefficients are all fine.
- Forgetting to divide out $e^{rt}$. You must state $e^{rt} \neq 0$ to justify setting the polynomial to zero — graders look for it.
- Mixing the two substitutions. $e^{rt}$ is for constant coefficients; $t^r$ is for Euler equations with $t^2, t$ coefficients.
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.