Recipe

Write and use the general solution of a linear homogeneous equation

Use when: you have two solutions $y_1,y_2$ of $y''+p(x)y'+q(x)y=0$ and need the complete solution set, possibly pinned down by initial conditions

Goal: the general solution $y=C_1y_1+C_2y_2$ , and — if conditions are given — the one specific solution matching them

Taught in The general solution of a linear homogeneous equation

Quick recipe

Confirm y1, y2 form a fundamental set
Write the general solution
If given initial conditions, substitute them into y and y'
Solve the resulting 2x2 linear system for C1, C2

1
Check that both functions actually solve the equation, then confirm they're independent by computing $W(y_1,y_2)$ and checking it's nonzero at one point.

Example

For $y''+y'-2y=0$ : $y_1=e^x$ and $y_2=e^{-2x}$ each solve the equation, and $W=e^x(-2e^{-2x})-e^x(e^{-2x})=-3e^{-x}$ , nonzero for every $x$ .

Why: this section's theorem — that $y=C_1y_1+C_2y_2$ accounts for every solution — only applies once $y_1,y_2$ are confirmed to be an actual fundamental set, not just any two solutions.
2
With a confirmed fundamental set, the general solution is $y=C_1y_1+C_2y_2$ , with $C_1,C_2$ left as arbitrary constants.

Example

$y=C_1e^{x}+C_2e^{-2x}$

Why: this section's theorem states that this two-parameter family accounts for every solution of the equation, with nothing else left to find.
3
Check whether the problem hands you a starting value and slope to pin the constants down, or just asks for the general family.

Example

For $y(0)=3$ , $y'(0)=-3$ : $y(0)=C_1+C_2=3$ , and since $y'=C_1e^{x}-2C_2e^{-2x}$ , $y'(0)=C_1-2C_2=-3$ .
- two conditions are given (e.g. $y(x_0)$ , $y'(x_0)$ ) — proceed to the system below.
- no conditions are given — the general solution above is already the final answer; stop here.
Why: two arbitrary constants need exactly two independent conditions to be pinned down to a single, specific solution.
4
Solve the pair of equations for $C_1,C_2$ , then substitute back into $y$ .

Example

Subtracting the second equation from the first: $3C_2=6$ , so $C_2=2$ , and then $C_1=1$ . $\boxed{y=e^{x}+2e^{-2x}}$

Why: the general solution is a whole family of curves; the conditions pick out the one curve that actually matches the problem.

Common mistakes

Writing $y=C_1y_1+C_2y_2$ without ever checking $W\ne0$ — two solutions that happen to be dependent don't form a fundamental set, and the theorem doesn't apply to them.
Solving for $C_1,C_2$ using only $y(x_0)$ — one equation can't pin down two constants; $y'(x_0)$ is required too.
Forgetting to differentiate the general solution before substituting $x_0$ into the second condition.
Treating the general solution as one particular answer instead of the entire two-parameter family it actually is.