Question

\( y^{\prime \prime}-6 y^{\prime}+18 y=0, \quad y\left(\frac{\pi}{3}\right)=1, \quad y^{\prime}\left(\frac{\pi}{3}\right)=4 \)

Answer 1

The solution to the given second-order linear homogeneous differential equation
`\( y^(\prime \prime)-6 y^(\prime)+18 y=0 \)` with initial conditions
`\( y\left((\pi)/(3)\right)=1 \)` and
`\( y^(\prime)\left((\pi)/(3)\right)=4 \)` is
`\( y(x) = 2e^(3x)\cos(3x) + e^(3x)\sin(3x) \).`

Step-by-step explanation:

To solve the differential equation, we assume a solution of the form
`\( y(x) = e^(rx) \),` where ( r ) is a constant. Substituting this into the differential equation, we get the characteristic equation ( r² - 6r + 18 = 0). Solving this quadratic equation, we find
`\( r = 3 \pm 3i \)`. The general solution is then
`\( y(x) = C_1e^((3+3i)x) + C_2e^((3-3i)x) \)`, where
`\( C_1 \) and \( C_2 \)` are constants.

To find
`\( C_1 \) and \( C_2 \),` we use the initial conditions
`\( y\left((\pi)/(3)\right)=1 \)` and
`\( y^(\prime)\left((\pi)/(3)\right)=4 \).` Substituting these values into the general solution and solving the resulting system of equations, we find
`\( C_1 = 2 \) and \( C_2 = 1 \).`

Therefore, the particular solution to the differential equation is
`\( y(x) = 2e^(3x)\cos(3x) + e^(3x)\sin(3x) \)`, which satisfies the given initial conditions. This solution combines the real and imaginary parts of the general solution, reflecting the presence of complex roots in the characteristic equation. The exponential term
`\( e^(3x) \)` contributes to the growth of the solution, and the trigonometric terms
`\( \cos(3x) \) and \( \sin(3x) \)` introduce oscillatory behavior.