Find the general solution of a*y'' + b*y = 0, where a and b are constants and a is not zero.

Step-by-step solution

1. 1

   Normalise the equation. Dividing by $a$ (legitimate since $a \ne 0$) gives

   $y'' + \frac{b}{a}y = 0$

   Only the single ratio $\tfrac{b}{a}$ matters — the individual values of $a$ and $b$ never appear again.

2. 2

   Form the characteristic equation. Substituting $y = e^{rx}$ and cancelling the non-zero $e^{rx}$:

   $r^2 + \frac{b}{a} = 0 \quad\Longrightarrow\quad r^2 = -\frac{b}{a}$

   There is no linear term in $r$ because the equation has no $y'$ term — this is the undamped case, so the roots are either purely imaginary or purely real, never both.

3. 3

   Case b/a > 0: purely imaginary roots. Then $r = \pm i\sqrt{b/a}$ and

   $y = C_1\cos\!\left(\sqrt{\tfrac{b}{a}}\,x\right) + C_2\sin\!\left(\sqrt{\tfrac{b}{a}}\,x\right)$

   This is simple harmonic motion with angular frequency $\omega = \sqrt{b/a}$ — the mass-spring equation $m\ddot y + ky = 0$ is exactly this case.

4. 4

   Case b/a < 0: two real roots. Then $r = \pm\sqrt{-b/a}$ and

   $y = C_1e^{\sqrt{-b/a}\,x} + C_2e^{-\sqrt{-b/a}\,x}$

   One solution grows and one decays; equivalently $y = A\cosh\!\left(\sqrt{-b/a}\,x\right) + B\sinh\!\left(\sqrt{-b/a}\,x\right)$.

5. 5

   Case b = 0: a double root at zero. The equation collapses to $y'' = 0$, whose solutions are the straight lines

   $y = C_1x + C_2$

   This is the $r = 0$ repeated-root case, with $e^{0x} = 1$ and $xe^{0x} = x$ as the two independent solutions.

6. 6

   Verify one case. With $a = 2$, $b = 8$ we get $\tfrac{b}{a} = 4$ and $y = C_1\cos 2x + C_2\sin 2x$. Substituting $y = \cos 2x$: $y'' = -4\cos 2x$, so $2(-4\cos 2x) + 8\cos 2x = 0$. Correct.