Question

Solve the following differential equations. \(y^{\prime \prime}+9 y=0\)

Short answer

y(x) = c_1 cos(3x) + c_2 sin(3x)

Step-by-step solution

Identify the Type of Differential Equation

Observe that the given differential equation is a second-order linear homogeneous differential equation of the form y'' + ay = 0y'' + 9y = 0

Write the Characteristic Equation

For a differential equation of the form y'' + ay = 0, the characteristic equation is given by: r^2 + a = 0Here, a = 9 so the characteristic equation is: r^2 + 9 = 0

Solve the Characteristic Equation

The characteristic equation r^2 + 9 = 0 can be solved for r.r^2 = -9r = ±√(-9)Since the square root of a negative number involves imaginary numbers: r = ±3i

Write the General Solution

When the characteristic roots are imaginary, the general solution to the differential equation y'' + 9y = 0 is: y(x) = c_1 cos(3x) + c_2 sin(3x)

Key concepts

Linear Homogeneous Differential Equations

In differential equations, we have different types. One important type is the linear homogeneous differential equation. These are equations where you have the highest derivative of a function and all the terms involve the function or its derivatives. The general form of such an equation is:
\[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + ... + a_1 y' + a_0 y = 0 \]
Note that there is no addition of any non-zero functions on the right side. Everything equals zero.
The equation given in the problem,
\[ y'' + 9y = 0, \]
is a classic example of a second-order linear homogeneous differential equation. Here, the highest-order derivative is the second one, denoted by \( y'' \). Such equations often appear in physics when modeling systems like springs or circuits. Understanding how to solve them is essential for many scientific applications.

Characteristic Equation

To solve linear homogeneous differential equations, we use what's called the characteristic equation. This helps us determine the roots, which play a crucial role in forming the final solution.
For the differential equation given,
\[ y'' + 9y = 0, \]
you can write the characteristic equation by changing the terms involving the derivatives to terms with \( r \). You replace \( y'' \) with \( r^2 \) and any term involving \( y' \) with \( r \), then set everything equal to zero. So here, it becomes:
\[ r^2 + 9 = 0. \]
We now need to solve for \( r \). Since \( r^2 = -9 \), taking the square root gives us the imaginary roots \( r = \pm 3i \). These roots are crucial for the next step in finding the solution.

Imaginary Roots

When solving characteristic equations, sometimes the roots we find are imaginary. Imaginary roots occur when we have to take the square root of a negative number.
For the characteristic equation
\[ r^2 + 9 = 0, \]
solving this equation gives us the roots \( r = \pm 3i \). The 'i' here indicates an imaginary number (\( i \) is the square root of -1).
When solving differential equations, these imaginary roots tell us that the general solution involves both sine and cosine functions. So for our example equation, the general solution is:
\[ y(x) = c_1 \cos(3x) + c_2 \sin(3x), \]
where \( c_1 \) and \( c_2 \) are constants determined by initial conditions. The functions \( \cos(3x) \) and \( \sin(3x) \) come from the imaginary parts of the roots and reflect the oscillatory nature usually associated with imaginary roots in physical systems. Understanding this form of solution is essential for correctly interpreting the behavior of systems modeled by such differential equations.