Forming the Differential Equation from General Solution $y = ae^{-2x} \cos(3x + b)$

The problem asks us to find the differential equation that corresponds to the general solution $y = ae^{-2x} \cos(3x + b)$. This general solution contains two arbitrary constants, '$a$' and '$b$'. A differential equation formed from a general solution with '$n$' arbitrary constants is typically of order '$n$'. Therefore, we expect to find a second-order differential equation (involving $y''$).

Understanding the Structure of the General Solution

The given general solution is of the form $y = e^{\alpha x} (A \cos(\beta x + \phi))$ or variations thereof. Specifically, our solution $y = ae^{-2x} \cos(3x + b)$ has the form $y = e^{\alpha x} \cos(\beta x + b)$, where $\alpha = -2$ and $\beta = 3$.

Solutions of this type arise from linear homogeneous differential equations with constant coefficients where the characteristic equation has complex conjugate roots of the form $r = \alpha \pm i\beta$.

Identifying the Roots of the Characteristic Equation

From the structure $y = ae^{-2x} \cos(3x + b)$, we can infer that the roots of the characteristic equation of the differential equation are complex conjugates:

• The term $e^{-2x}$ suggests the real part of the roots is $\alpha = -2$.
• The term $\cos(3x + b)$ suggests the imaginary part of the roots is $\beta = 3$.

Therefore, the roots of the characteristic equation are:

$$r_1 = -2 + 3i$$

$$r_2 = -2 - 3i$$

Constructing the Characteristic Equation

If the roots of a characteristic equation are $r_1$ and $r_2$, the equation can be written as $(r - r_1)(r - r_2) = 0$. Substituting our roots:

$$ (r - (-2 + 3i))(r - (-2 - 3i)) = 0 $$

Let's simplify this expression:

$$ (r + 2 - 3i)(r + 2 + 3i) = 0 $$

This is in the form $(X - Y)(X + Y) = X^2 - Y^2$, where $X = r + 2$ and $Y = 3i$.

$$ (r + 2)^2 - (3i)^2 = 0 $$

$$ (r^2 + 4r + 4) - (9i^2) = 0 $$

Since $i^2 = -1$, we have:

$$ (r^2 + 4r + 4) - (-9) = 0 $$

$$ r^2 + 4r + 4 + 9 = 0 $$

$$ r^2 + 4r + 13 = 0 $$

This is the characteristic equation.

Deriving the Differential Equation

For a linear homogeneous differential equation with constant coefficients, the characteristic equation $ar^2 + br + c = 0$ corresponds to the differential equation $ay'' + by' + cy = 0$.

Comparing our characteristic equation $r^2 + 4r + 13 = 0$ to the standard form, we have $a=1$, $b=4$, and $c=13$.

Therefore, the corresponding differential equation is:

$$ 1 \cdot y'' + 4 \cdot y' + 13 \cdot y = 0 $$

$$ y'' + 4y' + 13y = 0 $$

Conclusion

The differential equation formed from the general solution $y = ae^{-2x} \cos(3x + b)$ is $y'' + 4y' + 13y = 0$. This matches one of the provided options.