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The method of undetermined coefficients is a helpful tool for solving linear differential equations, especially when the non-homogeneous part is a simple polynomial, exponential, sine, or cosine function. It involves assuming a form for the particular solution based on the non-homogeneous term. This method works by guessing the form of the solution up to some undetermined constants, which we later calculate.

In our given differential equation, the equation is \[4y'' + 9y = 15\].

Here, the right side is 15, a constant. Therefore, we assume that the particular solution, \( y_p \), is also a constant, \( A \). This assumption guides us through determining the correct value of \( A \) by substituting back into the original equation.- Assume: \( y_p = A \) (a constant form)- Substitute into the equation to solve for \( A \)This straightforward approach simplifies our task, especially when dealing with constant or simple right-hand-side terms.

The homogeneous solution is an essential component of the general solution of a differential equation. It addresses the part of the equation without an external force or term (the right-hand side being zero). For the equation \[4y'' + 9y = 0\],

we derive this homogeneous form by setting the non-homogeneous term to zero.

Finding the characteristic equation helps us solve the homogeneous part.

So, we derive the characteristic equation as \[4r^2 + 9 = 0\]. Solving it gives complex roots \( r = \pm \frac{3i}{2} \).

This result leads us to a sinusoidal solution, reflecting the repeated or oscillatory nature of complex roots:- \( y_h = C_1 \cos\left(\frac{3}{2}x\right) + C_2 \sin\left(\frac{3}{2}x\right) \)The combined sines and cosines reflect solutions typical of systems with stable oscillations.

Finding the particular solution involves forming an educated guess aligned with the non-homogeneous part of the equation. For \[4y'' + 9y = 15\],
our goal is to find a \( y_p \) such that when substituted into the differential equation, the left-hand side equates the non-zero right-hand side (15 in this case).- Since 15 is a constant, we assume \( y_p = A \) (a simple constant).- Substitute in the equation, since the second derivative of a constant is zero, it simplifies to determining the coefficient \( A \).The algebraic substitution yields:- \( 9A = 15 \)- Solving gives \( A = \frac{5}{3} \)Thus, our particular solution is \( y_p = \frac{5}{3} \). This addition to the homogeneous solution completes the general solution, making it suitable for any boundary condition.

The characteristic equation is a pivotal aspect of solving homogeneous linear differential equations. It guides us in identifying the nature of the solutions, whether real, complex, or repeated. For our equation \[4y'' + 9y = 0\],
the characteristic equation is derived from the coefficient of derivatives.- Derived as: \( 4r^2 + 9 = 0 \)- Solving for \( r \) by factoring or using the quadratic formula yields complex roots \( r^2 = -\frac{9}{4} \) thus \( r = \pm \frac{3i}{2} \).These complex roots imply the solution involves sinusoidal functions (sine and cosine). Each pair of complex conjugates (\( \pm \frac{3i}{2} \)) provides us a basis for oscillatory solutions:- \( y_h = C_1 \cos\left(\frac{3}{2}x\right) + C_2 \sin\left(\frac{3}{2}x\right) \)This insight allows us to construct the full solution when combined with a corresponding particular solution. Understanding characteristic equations helps predict system behaviors during oscillations or vibrations.