Question

For the given differential equation, $$ y^{\prime \prime}-4 y^{\prime}+4 y=8+\sin 2 t $$

Short answer

**Answer**: The general solution to the given differential equation is \(y(t) = C_1e^{2t} + C_2te^{2t} + 2 + \frac{1}{4}\sin(2t)\), where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Solve the complementary equation

The complementary equation is a homogeneous version of the given differential equation: $$ y^{\prime \prime} - 4y^{\prime} + 4y = 0 $$ This can be written as a characteristic equation: $$ m^2 - 4m + 4 = 0 $$ Factoring, we get \((m-2)^2 = 0\). Therefore, m has a repeated root: \(m = 2\). The general solution of the complementary equation is given by: $$ y_c(t) = C_1e^{2t} + C_2te^{2t} $$ where \(C_1\) and \(C_2\) are constants.

Choose a trial function for the particular solution

The right-hand side of the given equation is: $$ 8 + \sin 2t $$ We need to choose a trial function that has the same form. Since there are both constant and trigonometric terms, we break this into two separate parts: Part 1: Constant term - We assume a trial function \(y_{p1} = A\), where A is a constant. Part 2: Trigonometric term - We assume a trial function \(y_{p2} = B\cos(2t) + C\sin(2t)\), where B and C are constants.

Find the particular solution

Part 1: We substitute \(y_{p1}\) into the given equation: $$ 0\cdot A - 4 \cdot 0\cdot A + 4\cdot A = 8 $$ Solving for A, we get \(A=2\). So \(y_{p1}(t) = 2\). Part 2: Calculate the first and second derivatives of \(y_{p2}\): $$ y_{p2}^{\prime}(t) = -2B\sin(2t) + 2C\cos(2t) $$ $$ y_{p2}^{\prime \prime}(t) = -4B\cos(2t) - 4C\sin(2t) $$ Substitute \(y_{p2}\) and its derivatives into the given equation: $$ (-4B\cos(2t) - 4C\sin(2t)) - 4(-2B\sin(2t) + 2C\cos(2t)) + 4(B\cos(2t) + C\sin(2t)) = \sin(2t) $$ Comparing coefficients of \(\sin(2t)\) and \(\cos(2t)\) on both sides, we get the following system of equations: $$ \begin{cases} 4B = 0 \\ 4C - 8C + 4C = 1 \end{cases} $$ Solving this system, we find \(B = 0\) and \(C = \frac{1}{4}\). Hence, the trial function for the trigonometric term is \(y_{p2}(t) = \frac{1}{4}\sin(2t)\).

Combine solutions

The general solution to the given equation is the sum of the complementary and particular solutions: $$ y(t) = y_c(t) + y_{p1}(t) + y_{p2}(t) = C_1e^{2t} + C_2te^{2t} + 2 + \frac{1}{4}\sin(2t) $$ That's the general solution of the given differential equation.

Key concepts

Characteristic Equation

In solving differential equations, especially linear constant coefficient equations, the characteristic equation is a crucial tool. This equation emerges when you express a homogeneous differential equation in terms of an algebraic equation. For example, starting with the second-order differential equation \(y'' - 4y' + 4y = 0\) (the homogeneous part), we convert it into the characteristic equation.

• The characteristic equation for this is \(m^2 - 4m + 4 = 0\).
• To solve it, we typically factor, as in \/((m-2)^2 = 0\/).
• Repeating roots indicate solutions like \(e^{mt} and te^{mt}\) based on the discriminant.

This process helps us identify the nature of solutions to differential equations. Using the characteristic roots, we develop a complementary solution that reflects the natural behavior of the system described by the differential equation.

Complementary Function

The complementary function, denoted often as \(y_c(t)\), arises from the solution to the homogeneous part of the differential equation. Fundamentally, it depicts the inherent dynamic behavior without external forces, represented by the equation \(y'' - 4y' + 4y = 0\) in our example.

• Given the characteristic equation is \((m-2)^2 = 0\), we find a repeated root \(m = 2\).
• Repeating roots indicate that our complementary function includes terms like \(C_1 e^{2t} + C_2 t e^{2t}\).
• Here, \(C_1\) and \(C_2\) are constants determined by initial conditions, when available.

Understanding the complementary function is essential as it determines the solution space generated by the system, oblivious to external or forced effects.

Particular Solution

To account for specific external forces, we find the particular solution. This part adapts the solution to fit additional terms in the differential equation, such as non-homogeneous components like \(8 + \sin 2t\).

• A trial function is proposed based on the form of the right-hand side.
• For constants, we use simple forms like \(A = 2\).
• For trigonometric terms, we invoke \(B \cos(2t) + C \sin(2t)\).

The particular solution \(y_p(t)\) becomes \(2 + \frac{1}{4}\sin(2t)\) in this situation, determined via substitution back into the differential equation and solving for unknown coefficients, \(B\) and \(C\). This ensures that the solution satisfies not just inherent dynamics but also external influences.

Homogeneous Equation

The homogeneous equation is pivotal when dealing with linear differential equations, as it forms the foundation for building the complete solution. A homogeneous equation contains only terms with the function and its derivatives, set to zero, like \(y'' - 4y' + 4y = 0\).

• It explores the system's behavior purely under its internal dynamics.
• This equation leads directly to the characteristic equation, revealing eigenvalues which dictate natural frequencies and modes of the system.
• The homogeneous solution is essential for ensuring that the overall solution remains consistent with the system's properties.

By solving the homogeneous equation, we ensure that we capture all natural resonances and behaviors, which then get adjusted with particular solutions to align with external forces. Together, the roll-up of the homogeneous and particular solutions provides a comprehensive view of all possible behaviors in the system.