Solving the Differential Equation \( y” + y = 0 \)
Problem: Solve the differential equation \( y” + y = 0 \).
Explanation:
The equation is a second-order linear homogeneous differential equation with constant coefficients. Its general form is \( y” + y = 0 \).
First, we assume a solution of the form \( y(x) = e^{rx} \).
Upon differentiating, we find:
- First derivative: \( y'(x) = r \cdot e^{rx} \)
- Second derivative: \( y”(x) = r^2 \cdot e^{rx} \)
Substituting into the original equation, we get \( (r^2 + 1) \cdot e^{rx} = 0 \).
Since \( e^{rx} \) is never zero, \( r^2 + 1 = 0 \), which leads to \( r = \pm i \) where \( i \) is the imaginary unit.
Therefore, the general solution becomes \( y(x) = A \cos(x) + B \sin(x) \).
About Linear Independence:
The functions \( \cos(x) \) and \( \sin(x) \) are linearly independent because there is no constant \( C \) such that \( \cos(x) = C \times \sin(x) \) for all \( x \).
Solving the Differential Equation \( y” + 5y + 12 = 0 \)
Problem: Solve the differential equation \( y” + 5y + 12 = 0 \).
Explanation:
The equation is a second-order linear homogeneous differential equation with constant coefficients. Its general form is \( y” + 5y + 12 = 0 \).
First, we assume a solution of the form \( y(x) = e^{rx} \).
Upon differentiating, we find:
- First derivative: \( y'(x) = r e^{rx} \)
- Second derivative: \( y”(x) = r^2 e^{rx} \)
Substituting into the original equation, we get \( r^2 e^{rx} + 5 e^{rx} + 12 = 0 \).
Since \( e^{rx} \) is never zero, we get \( r^2 + 17 = 0 \), leading to \( r = \pm i\sqrt{17} \).
Therefore, the general solution becomes \( y(x) = A \cos(\sqrt{17}x) + B \sin(\sqrt{17}x) \).