Understanding the Differential Equation Solution

We are asked to find the nature of the solution for the given differential equation.

The differential equation is:

\(\frac{{{d^2}y}}{{d{x^2}}} - y = 0\)

This is a **second order** linear homogeneous differential equation with **constant coefficients**. To find the **differential equation solution**, we first form the characteristic equation.

Forming the Characteristic Equation

For a linear homogeneous differential equation with constant coefficients in the form \(a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = 0\), the **characteristic equation** is given by \(am^2 + bm + c = 0\).

In our case, the equation is \(\frac{{{d^2}y}}{{d{x^2}}} - y = 0\), which fits the form with \(a=1\), \(b=0\), and \(c=-1\).

So, the **characteristic equation** is:

\(m^2 - 1 = 0\)

Solving the Characteristic Equation

We need to find the roots of the characteristic equation:

\(m^2 - 1 = 0\)

\(m^2 = 1\)

Taking the square root of both sides, we get:

\(m = \pm \sqrt{1}\)

\(m_1 = 1\)

\(m_2 = -1\)

The roots are real and distinct (\(m_1 \neq m_2\)).

Writing the General Differential Equation Solution

For a **second order** linear homogeneous differential equation with real and distinct roots \(m_1\) and \(m_2\), the general **differential equation solution** is given by:

\(y(x) = C_1 e^{m_1 x} + C_2 e^{m_2 x}\)

Substituting our roots \(m_1 = 1\) and \(m_2 = -1\):

\(y(x) = C_1 e^{1 \cdot x} + C_2 e^{-1 \cdot x}\)

\(y(x) = C_1 e^x + C_2 e^{-x}\)

This is the general **differential equation solution** for the given equation.

Determining the Nature of the Solution

The general solution \(y(x) = C_1 e^x + C_2 e^{-x}\) consists of terms involving \(e^x\) and \(e^{-x}\).

Functions of the form \(e^{kx}\) are exponential functions.

Therefore, the nature of this **differential equation solution** is exponential.

Let's compare this with the given options:

• Oscillatory solutions typically involve trigonometric functions like \(\sin(kx)\) or \(\cos(kx)\), which arise from complex conjugate roots of the characteristic equation. Our roots are real.
• Exponential solutions involve terms like \(e^{kx}\), which arise from real roots of the characteristic equation. Our solution \(C_1 e^x + C_2 e^{-x}\) fits this description.
• Trigonometric solutions are the same as oscillatory solutions.

Based on the form of the general solution \(y(x) = C_1 e^x + C_2 e^{-x}\), which is a sum of exponential terms, the nature of the solution is Exponential. This aligns with the characteristic equation having real roots.