Question

Find the solution of the given initial value problem. $$ y^{\prime \prime}+4 y=3 \sin 2 t, \quad y(0)=2, \quad y^{\prime}(0)=-1 $$

Short answer

To find the solution of the given initial value problem, we first solved the homogeneous equation and found the complementary function. Then, we found the particular integral for the non-homogeneous equation. Afterward, we combined the complementary function and the particular integral, resulting in the general solution of the non-homogeneous ODE. Lastly, we used the initial conditions to find the constants and obtained the solution of the initial value problem as: $$ y(t) = 2\cos(2t) - \frac{5}{4}\sin(2t) + \frac{3}{4}\sin(2t) $$

Step-by-step solution

Solve the homogeneous equation

First, we need to find the complementary function by solving the homogeneous equation: $$ y'' + 4y = 0 $$ Its characteristic equation is: $$ r^2 + 4 = 0 $$ Solving for r, we get: $$ r = \pm 2i $$ Since the roots are complex, the complementary function takes the form: $$ y_c(t) = A \cos(2t) + B \sin(2t) $$ where A and B are constants.

Find the particular integral

Next, we find the particular integral for the non-homogeneous equation. The forcing function is \(3\sin(2t)\), so we guess a particular solution of the form: $$ y_p(t) = C\cos(2t) + D\sin(2t) $$ Differentiate this function with respect to t: $$ y_p'(t) = -2C\sin(2t) + 2D\cos(2t) $$ Differentiate it once more: $$ y_p''(t) = -4C\cos(2t) - 4D\sin(2t) $$ Now, substitute \(y_p(t)\) and \(y_p''(t)\) into the non-homogeneous ODE: $$ (-4C\cos(2t) - 4D\sin(2t)) + 4(C\cos(2t) + D\sin(2t)) = 3\sin(2t) $$ Simplifying the equation, we get: $$ (4C - 4D)\sin(2t) = 3\sin(2t) $$ Comparing the coefficients, we find: $$ 4C = 0 \Rightarrow C = 0; \quad 4D - 4D = 3 \Rightarrow D = \frac{3}{4} $$ So, the particular integral is: $$ y_p(t) = \frac{3}{4}\sin(2t) $$

Combine the complementary function and the particular integral

The general solution of the non-homogeneous ODE is the sum of the complementary function and the particular integral: $$ y(t) = y_c(t) + y_p(t) = A\cos(2t) + B\sin(2t) + \frac{3}{4}\sin(2t) $$

Use the initial conditions to find the constants

Apply the initial conditions: \(y(0) = 2\) and \(y'(0) = -1\). From \(y(t)\), calculate \(y'(t)\): $$ y'(t) = -2A\sin(2t) + 2B\cos(2t) + \frac{3}{2}\cos(2t) $$ Now, apply the initial conditions: 1. For \(y(0) = 2\): $$ y(0) = 2 \Rightarrow A\cos(0) + B\sin(0) + \frac{3}{4}\sin(0) = 2 $$ This simplifies to: $$ A = 2 $$ 2. For \(y'(0) = -1\): $$ y'(0) = -1 \Rightarrow -2A\sin(0) + 2B\cos(0) + \frac{3}{2}\cos(0) = -1 $$ Substitute the value of A: $$ 2B + \frac{3}{2} = -1 \Rightarrow B = -\frac{5}{4} $$ Finally, the solution of the initial value problem is: $$ y(t) = 2\cos(2t) - \frac{5}{4}\sin(2t) + \frac{3}{4}\sin(2t) $$

Key concepts

Non-homogeneous differential equation

A non-homogeneous differential equation is a type of differential equation that includes a term known as the "forcing function," which is not dependent on the solution function or its derivatives. In other words, the equation does not equal zero on the right-hand side. This results in solutions that are more complex than homogeneous equations.

For example, consider the differential equation: \[y'' + 4y = 3 \sin 2t\]Here, the term \(3 \sin 2t\) is the non-homogeneous part. It is the reason why the equation is referred to as non-homogeneous, and it impacts how we find the solution. This adds an extra layer of complexity, requiring a specific method to solve it.

To tackle a non-homogeneous equation, we typically split our work into finding two components: the complementary function and the particular integral, which we will cover next.

Complementary function

The complementary function is one of the two main components we look for when solving a non-homogeneous differential equation. It is derived by solving the related homogeneous equation, which is obtained by setting the non-homogeneous part to zero.

In the equation \(y'' + 4y = 0\), the solution involves finding the characteristic roots. These roots, obtained from the characteristic equation, help define the form of the complementary function. For our example, the roots were \(r = \pm 2i\), leading to complex roots.

For complex roots, the complementary function emerges as a combination of sine and cosine functions:

• \(y_c(t) = A \cos(2t) + B \sin(2t)\)

Here, \(A\) and \(B\) are constants that need to be determined either through initial conditions or boundary values. This component captures part of the behavior of the differential equation's solution, independent of the forcing function.

Particular integral

The particular integral is the second essential component of a solution to a non-homogeneous differential equation. This part of the solution accounts for the presence of the non-zero term on the right-hand side of the equation.

In our example, where the equation \(y'' + 4y = 3 \sin 2t\) was given, the particular integral needs to reflect the form of the non-homogeneous component \(3 \sin 2t\). Typically, we guess a trial solution that mimics this form and substitute it back into the equation to determine the coefficients.

Through this process, the particular solution for our equation was found to be:

• \(y_p(t) = \frac{3}{4} \sin(2t)\)

Determining the particular integral allows us to account for the response of the system to the external forces represented by the non-homogeneous term.

Characteristic equation

The characteristic equation is a key tool used in solving both homogeneous and non-homogeneous linear differential equations. It is a polynomial equation derived from the differential equation by assuming a solution of the form \(e^{rt}\), where \(r\) is a constant.

To illustrate, consider the homogeneous equation \(y'' + 4y = 0\). The characteristic equation here is formed as:

• \(r^2 + 4 = 0\)

By solving this quadratic equation, we find the roots \(r = \pm 2i\), indicating complex roots.

The nature of the roots, whether real or complex, affects the form of the complementary function:

• Real roots result in exponential terms.
• Complex roots lead to sine and cosine terms, as in our example with \(\cos(2t)\) and \(\sin(2t)\).

Understanding the characteristic equation helps us determine the necessary functions that combine to form the overall solution.