Find the general solution of the differential equation y'' - a*y' + b*y = 0, where a and b are constants.

Step-by-step solution

1. 1

   Derive the characteristic equation. Trying $y = e^{rx}$ gives $y' = re^{rx}$ and $y'' = r^2e^{rx}$, so

   $e^{rx}\left(r^2 - ar + b\right) = 0$

   Since $e^{rx}$ is never zero, the exponent must satisfy

   $r^2 - ar + b = 0$

   Note the sign: the middle coefficient is $-a$, so the characteristic equation carries $-ar$, not $+ar$.

2. 2

   Solve for r and identify the discriminant.

   $r = \frac{a \pm \sqrt{a^2 - 4b}}{2}$

   Everything now depends on the sign of $D = a^2 - 4b$, which is why the answer necessarily comes in three cases.

3. 3

   Case D > 0: two distinct real roots. With $r_{1,2} = \tfrac{a \pm \sqrt{a^2-4b}}{2}$,

   $y = C_1e^{r_1x} + C_2e^{r_2x}$

   The two exponentials are independent because $r_1 \ne r_2$.

4. 4

   Case D = 0: one repeated root. Then $r = \tfrac{a}{2}$, and $e^{rx}$ alone gives only a one-parameter family. The second independent solution is $xe^{rx}$:

   $y = \left(C_1 + C_2x\right)e^{ax/2}$

5. 5

   Case D < 0: complex conjugate roots. Writing $r = \tfrac{a}{2} \pm i\tfrac{\sqrt{4b-a^2}}{2}$ and taking real and imaginary parts:

   $y = e^{ax/2}\left(C_1\cos\!\left(\tfrac{\sqrt{4b-a^2}}{2}x\right) + C_2\sin\!\left(\tfrac{\sqrt{4b-a^2}}{2}x\right)\right)$

   The real part $\tfrac{a}{2}$ controls growth or decay; the imaginary part sets the frequency.

6. 6

   Check a concrete instance. Take $a = 3$, $b = 2$: then $D = 9 - 8 = 1 > 0$, roots $r = 2$ and $r = 1$, and $y = C_1e^{2x} + C_2e^{x}$. Substituting $y = e^{2x}$: $4e^{2x} - 3(2e^{2x}) + 2e^{2x} = 0$. Correct. Take $a = 0$, $b = 4$: $D = -16 < 0$, giving $y = C_1\cos 2x + C_2\sin 2x$, the familiar solution of $y'' + 4y = 0$.