Question

Find the solution $ \frac{d^2y}{dm^2} - k^3 \frac{dy}{dm} = y \cos m, \quad y(0) = 1 $

• A. \( y^3 = 3y^3 \sin m \)
• B. \( y^3 = 3x^2 \sin m \)
• C. \( y^4 = 3y^3 \sin m \)
• D. \( y^3 = 5y^3 \sin m \)

Hint:

When solving second-order differential equations, consider using substitution or reduction of order methods if necessary. Check for initial conditions that help simplify the process.

Answer 1

The Correct Option is B

The differential equation is provided:

\[ \frac{d^2y}{dm^2} - k^3 \frac{dy}{dm} = y \cos m, \quad y(0) = 1 \]

Step 1: Solve the homogeneous equation

The homogeneous equation is: \[ \frac{d^2y}{dm^2} - k^3 \frac{dy}{dm} = 0 \] This is reducible to the first-order equation: \[ \frac{dv}{dm} - k^3 v = 0 \] where \( v = \frac{dy}{dm} \). The solution for \( v \) is: \[ v = A e^{k^3 m} \] Integrating \( v \) with respect to \( m \) yields the homogeneous solution: \[ y_h = A e^{k^3 m} \]

Step 2: Solve for the particular solution

A particular solution is assumed to be of the form: \[ y_p = B \cos m \] The value of \( B \) is determined by substituting this form into the original differential equation.

Step 3: General solution

The general solution is the sum of the homogeneous and particular solutions: \[ y(m) = A e^{k^3 m} + B \cos m \]

Step 4: Apply the initial condition

The initial condition \( y(0) = 1 \) is applied to the general solution: \[ 1 = A e^{k^3 \cdot 0} + B \cos 0 \] This simplifies to: \[ 1 = A + B \] Therefore, \( A + B = 1 \).

Conclusion:

The general solution to the differential equation, subject to the condition \( A + B = 1 \), is: \[ y(m) = A e^{k^3 m} + B \cos m \]