Bring the chapter together: write the characteristic equation, tell which kind of roots it has, build the general solution, fit it to a value and a slope, and test a pair of solutions with the Wronskian. No worked example sits above the answers.

A mechanic listening to an engine can tell from the sound alone which of three kinds of trouble it has. The discriminant is that sound.

Decide first what the roots are. For $ay''+by'+cy=0$ solve $ar^{2}+br+c=0$. Two real roots give $C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}$; a repeated root gives $(C_{1}+C_{2}x)e^{rx}$; a complex pair $\alpha\pm i\beta$ gives $e^{\alpha x}(C_{1}\cos\beta x+C_{2}\sin\beta x)$. Then fit the constants to $y(x_{0})$ and $y'(x_{0})$.

• The sign of $b^{2}-4ac$ says which of the three forms to write.
• Two conditions, a value and a slope at one point, fix the two constants.
• Solutions of a homogeneous linear equation add and scale; those of a forced or a nonlinear equation do not.
• $W=y_{1}y_{2}'-y_{1}'y_{2}\ne0$ at one point makes $y_{1}$, $y_{2}$ a basis of the solutions.