Solve differential equation: y 16y 0 y1 cos4x
[latex]
\[
\textbf{Q: Solve the differential equation:} \quad y” = 16y
\]
\[
\textbf{Step 1: Solve the characteristic equation}
\]
The characteristic equation for the differential equation \( y” = 16y \) is:
\[
r^2 – 16 = 0
\]
Solving for \( r \), we get:
\[
r^2 = 16 \quad \Rightarrow \quad r = \pm 4
\]
So, the general solution to the differential equation is:
\[
y(x) = C_1 e^{4x} + C_2 e^{-4x}
\]
\[
\textbf{Step 2: Apply the initial conditions}
\]
We are given the initial conditions:
\[
y(0) = 1, \quad y'(0) = \cos(4x)
\]
Substituting \( y(0) = 1 \) into the general solution:
\[
y(0) = C_1 e^{4(0)} + C_2 e^{-4(0)} = C_1 + C_2 = 1
\]
So, we have the first equation:
\[
C_1 + C_2 = 1
\]
Next, differentiate the general solution to find \( y'(x) \):
\[
y'(x) = 4C_1 e^{4x} – 4C_2 e^{-4x}
\]
Substitute \( y'(0) = \cos(4x) \) at \( x = 0 \):
\[
y'(0) = 4C_1 e^{4(0)} – 4C_2 e^{-4(0)} = 4C_1 – 4C_2 = \cos(4(0)) = 1
\]
So, we have the second equation:
\[
4C_1 – 4C_2 = 1 \quad \Rightarrow \quad C_1 – C_2 = \frac{1}{4}
\]
\[
\textbf{Step 3: Solve the system of equations}
\]
Now, solve the system of equations:
1. \( C_1 + C_2 = 1 \)
2. \( C_1 – C_2 = \frac{1}{4} \)
Adding these equations:
\[
( C_1 + C_2 ) + ( C_1 – C_2 ) = 1 + \frac{1}{4}
\]
\[
2C_1 = \frac{5}{4} \quad \Rightarrow \quad C_1 = \frac{5}{8}
\]
Substitute \( C_1 = \frac{5}{8} \) into the first equation:
\[
\frac{5}{8} + C_2 = 1 \quad \Rightarrow \quad C_2 = 1 – \frac{5}{8} = \frac{3}{8}
\]
\[
\textbf{Final Solution:}
\]
The solution to the differential equation is:
\[
y(x) = \frac{5}{8} e^{4x} + \frac{3}{8} e^{-4x}
\]