Understanding the Differential Equation

The given equation is a second-order linear homogeneous differential equation with constant coefficients:

\(4 \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} + 5y = 0\)

This type of differential equation has the general form \(ay'' + by' + cy = 0\).

Solving the Characteristic Equation

To find the solutions, we first write the characteristic equation associated with the differential equation. The characteristic equation is obtained by replacing \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with 1:

\(4r^2 + 4r + 5 = 0\)

This is a quadratic equation in terms of \(r\). We can solve for \(r\) using the quadratic formula:

\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a = 4\), \(b = 4\), and \(c = 5\). Substituting these values into the formula:

\(r = \frac{-4 \pm \sqrt{4^2 - 4(4)(5)}}{2(4)}\)

\(r = \frac{-4 \pm \sqrt{16 - 80}}{8}\)

\(r = \frac{-4 \pm \sqrt{-64}}{8}\)

\(r = \frac{-4 \pm 8i}{8}\)

Simplifying, we get the roots:

\(r = \frac{-4}{8} \pm \frac{8i}{8}\)

\(r = -\frac{1}{2} \pm i\)

Identifying the Nature of the Roots

The roots of the characteristic equation are complex conjugates: \(r_1 = -\frac{1}{2} + i\) and \(r_2 = -\frac{1}{2} - i\). They are in the form \(\alpha \pm i\beta\), where \(\alpha = -\frac{1}{2}\) and \(\beta = 1\).

Finding the Linearly Independent Solutions

For a second-order linear homogeneous differential equation with constant coefficients whose characteristic equation has complex conjugate roots \(\alpha \pm i\beta\), the two linearly independent solutions are given by:

• \(y_1(x) = e^{\alpha x} \cos(\beta x)\)
• \(y_2(x) = e^{\alpha x} \sin(\beta x)\)

Substituting the values \(\alpha = -\frac{1}{2}\) and \(\beta = 1\) into these forms:

• \(y_1(x) = e^{(-1/2)x} \cos(1x) = e^{-x/2} \cos x\)
• \(y_2(x) = e^{(-1/2)x} \sin(1x) = e^{-x/2} \sin x\)

Therefore, the linearly independent solutions of the given differential equation are \(e^{-x/2} \cos x\) and \(e^{-x/2} \sin x\).

Confirming the Linearly Independent Solutions

The two functions \(y_1(x) = e^{-x/2} \cos x\) and \(y_2(x) = e^{-x/2} \sin x\) are indeed linearly independent because the ratio \(\frac{y_2(x)}{y_1(x)} = \frac{e^{-x/2} \sin x}{e^{-x/2} \cos x} = \tan x\) is not a constant.

The general solution would be a linear combination of these two solutions: \(y(x) = c_1 e^{-x/2} \cos x + c_2 e^{-x/2} \sin x\), where \(c_1\) and \(c_2\) are arbitrary constants. The question asks for the linearly independent solutions themselves.