Understanding the Differential Equation $y'' + y = 0$

The question asks for the solution(s) to the second-order linear ordinary differential equation (ODE):

$ y'' + y = 0 $

This is a classic type of ODE. To find the solutions, we typically use the characteristic equation method.

Solving the Characteristic Equation

For a homogeneous linear ODE with constant coefficients like $ay'' + by' + cy = 0$, we assume a solution of the form $y = e^{rx}$. Substituting this into the ODE gives the characteristic equation:

$ ar^2 + br + c = 0 $

In our specific equation, $y'' + y = 0$, we have $a=1$, $b=0$, and $c=1$. So the characteristic equation is:

$ 1 \cdot r^2 + 0 \cdot r + 1 = 0 $

$ r^2 + 1 = 0 $

Now, we solve for $r$:

$ r^2 = -1 $

$ r = \pm \sqrt{-1} $

$ r = \pm i $

The roots are complex conjugates, $r = 0 \pm 1i$.

Finding the General Solution

When the roots of the characteristic equation are complex conjugates of the form $\alpha \pm \beta i$, the general solution to the ODE is given by:

$ y(x) = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x)) $

In our case, $\alpha = 0$ and $\beta = 1$. Substituting these values, we get:

$ y(x) = e^{0 \cdot x} (C_1 \cos(1 \cdot x) + C_2 \sin(1 \cdot x)) $

Since $e^0 = 1$, the general solution simplifies to:

$ y(x) = C_1 \cos x + C_2 \sin x $

Here, $C_1$ and $C_2$ are arbitrary constants.

Verifying Potential Solutions

The general solution indicates that functions of the form $C_1 \cos x + C_2 \sin x$ satisfy the equation. This means that $\cos x$ (when $C_1=1, C_2=0$) and $\sin x$ (when $C_1=0, C_2=1$) are indeed solutions.

Let's check the other options:

• Option A (cos x): If $y = \cos x$, then $y' = -\sin x$ and $y'' = -\cos x$. Substituting into $y'' + y = 0$: $(-\cos x) + (\cos x) = 0$. This is true.
• Option B (sin x): If $y = \sin x$, then $y' = \cos x$ and $y'' = -\sin x$. Substituting into $y'' + y = 0$: $(-\sin x) + (\sin x) = 0$. This is true.
• Option C (1 + cos x): If $y = 1 + \cos x$, then $y' = -\sin x$ and $y'' = -\cos x$. Substituting into $y'' + y = 0$: $(-\cos x) + (1 + \cos x) = 1 \neq 0$. This is false.
• Option D (1 + sin x): If $y = 1 + \sin x$, then $y' = \cos x$ and $y'' = -\sin x$. Substituting into $y'' + y = 0$: $(-\sin x) + (1 + \sin x) = 1 \neq 0$. This is false.

Conclusion

Based on the derivation of the general solution and verification, the functions $\cos x$ and $\sin x$ are solutions to the differential equation $y'' + y = 0$. Therefore, the correct option includes both A and B.