Question

For the given differential equation, (a) Determine the complementary solution, \(y_{c}(t)=c_{1} y_{1}(t)+c_{2} y_{2}(t)\). (b) Use the method of variation of parameters to construct a particular solution. Then form the general solution. y^{\prime \prime}+4 y=4

Short answer

Answer: The general solution for the given differential equation is \(y(t) = c_1 \cos{(2t)} + c_2 \sin{(2t)} + \cos^2{(2t)} + \frac{1}{2}\sin^2{(2t)}\).

Step-by-step solution

1. Solve the Homogeneous Differential Equation

Solve the homogeneous differential equation \(y'' + 4y = 0\) to find the complementary function, \(y_c(t)\). To solve this, we assume a solution in the form of \(y(t) = e^{rt}\), and plug it into the given homogeneous differential equation: \(r^2 e^{rt} + 4 e^{rt} = 0\) Factoring out the \(e^{rt}\) term, we get: \(r^2+4 = 0\) Now, we solve for the roots, r: \(r^2 = -4\) The roots are \(r_1 = 2i\) and \(r_2 = -2i\). Since we have complex roots, the complementary solution is in the form: \(y_c(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} = c_1 \cos(2t) + c_2 \sin{(2t)}\).

2. Find the Wronskian

To use the method of variation of parameters, we first find the Wronskian, \(W(t)\), of the functions \(y_1(t) = \cos{(2t)}\) and \(y_2(t) = \sin{(2t)}\). The Wronskian is defined as: \(W(t) = \begin{vmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t)\end{vmatrix}\) \(W(t) =\begin{vmatrix} \cos{(2t)} & \sin{(2t)}\\ -2\sin{(2t)} & 2\cos{(2t)}\end{vmatrix}\) Evaluating the determinant, we get: \(W(t) = 2\cos^2{(2t)} + 2\sin^2{(2t)} = 2.\)

3. Calculate the Particular Solution

Next, we calculate the particular solution, \(y_p(t)\), using the following formulas for variation of parameters: \(u_1(t) = -\int \frac{y_2(t)g(t)}{W(t)} dt = -\int \frac{\sin{(2t)}(4)}{2} dt\) \(u_2(t) = \int \frac{y_1(t)g(t)}{W(t)} dt = \int \frac{\cos{(2t)}(4)}{2} dt\) Solve these integrals: \(u_1(t) = -2 \int \sin{(2t)} dt = \cos{(2t)} + C_1\) \(u_2(t) = 2 \int \cos{(2t)} dt = \frac{1}{2}\sin{(2t)} + C_2\) Now, form the particular solution as \(y_p(t) = u_1(t)y_1(t) + u_2(t)y_2(t)\): \(y_p(t) = (\cos{(2t)} + C_1)\cos{(2t)} + (\frac{1}{2}\sin{(2t)} + C_2)\sin{(2t)}\) However, since we are only trying to find a single particular solution, we can let \(C_1\) and \(C_2\) be zero: \(y_p(t) = \cos{(2t)}\cos{(2t)} + \frac{1}{2}\sin{(2t)}\sin{(2t)}\)

4. Write the General Solution

Finally, express the general solution as the sum of the complementary and particular solutions: \(y(t) = y_c(t) + y_p(t) = c_1 \cos{(2t)} + c_2 \sin{(2t)} + \cos^2{(2t)} + \frac{1}{2}\sin^2{(2t)}\). Thus, the general solution for the given differential equation is: \(y(t) = c_1 \cos{(2t)} + c_2 \sin{(2t)} + \cos^2{(2t)} + \frac{1}{2}\sin^2{(2t)}\).

Key concepts

Complementary Solution

In solving differential equations, the concept of a complementary solution is key. This solution is derived from the homogeneous part of the differential equation. In the case of the equation \(y'' + 4y = 0\), we start by proposing that the solution is of the form \(y(t) = e^{rt}\). By substituting this form back into the equation, we find the characteristic equation \(r^2 + 4 = 0\). Solving this gives roots \(r_1 = 2i\) and \(r_2 = -2i\). These are complex conjugates, which indicates an oscillating solution in terms of sine and cosine functions.

Thus, the complementary solution is formulated as:
\(y_c(t) = c_1 \cos(2t) + c_2 \sin(2t)\).
This expression represents the general solution of the homogeneous equation and involves arbitrary constants \(c_1\) and \(c_2\), which are determined by initial conditions.

General Solution

The general solution to a differential equation is a combination of its complementary solution and a particular solution. It essentially covers every possible solution to the equation, encompassing both the natural response and the forced response due to an external input.

In the provided exercise, once the complementary solution is obtained as \(c_1 \cos(2t) + c_2 \sin(2t)\), the next step is to find a particular solution that addresses the non-homogeneous part of the differential equation. The particular solution is found using a specific method like variation of parameters.

The final general solution combines these two parts:
\(y(t) = y_c(t) + y_p(t) = c_1 \cos(2t) + c_2 \sin(2t) + \cos^2(2t) + \frac{1}{2}\sin^2(2t)\).
This solution caters to both the natural behavior of the system (complementary solution) and the influence of external forces (particular solution).

Wronskian

The Wronskian is a very powerful tool in the analysis of differential equations because it helps confirm whether two solutions of a differential equation are linearly independent. Linear independence is crucial for the functions to form a basis for the solution space.

For functions \(y_1(t) = \cos(2t)\) and \(y_2(t) = \sin(2t)\), used in the complementary solution, the Wronskian is calculated as follows:
\[W(t) = \begin{vmatrix} y_1(t) & y_2(t) \ y_1'(t) & y_2'(t)\end{vmatrix} = \begin{vmatrix} \cos(2t) & \sin(2t) \ -2\sin(2t) & 2\cos(2t)\end{vmatrix}\]This determinant simplifies to \(W(t) = 2(\cos^2{(2t)} + \sin^2{(2t)}) = 2\), verifying that our functions are independent.

This result ensures that the set of solutions is indeed a valid basis, allowing the use of methods like variation of parameters confidently.

Method for Solving Differential Equations

Solving differential equations involves using different methods to obtain solutions that describe how a system changes over time. One effective method is the variation of parameters, often used to find a particular solution to non-homogeneous differential equations.

The process typically follows these steps:

• Determine the complementary solution from the homogeneous equation.
• Use the complementary solution to form a basis of solutions, often involving sine and cosine if roots are complex.
• Calculate the Wronskian of these solutions to check for linear independence.
• Use the variation of parameters formulas, \(u_1(t)\) and \(u_2(t)\), derived from \(y_1(t)\) and \(y_2(t)\), to express the particular solution.

The particular solution is found by solving integrals of combinations of these functions and ensuring the use of constants appropriately to meet initial conditions. The final step is to add this particular solution to the complementary solution, forming the overall general solution.