Question

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ t^{2} y^{\prime \prime}-2 y=3 t^{2}-1, \quad t>0 ; \quad y_{1}(t)=t^{2}, \quad y_{2}(t)=t^{-1} $$

Short answer

#tag_title# Step 5: Construct the general solution for the given nonhomogeneous equation#tag_content# We have the two solutions of the homogeneous equation \(y_1(t) = t^2\) and \(y_2(t) = t^{-1}\), and the particular solution of the nonhomogeneous equation \(y_p(t) = \frac{1}{2}\). The general solution is given by the sum of these solutions multiplied by arbitrary constants: $$ y(t) = C_1y_1(t) + C_2y_2(t) + y_p(t) = C_1t^2 + C_2t^{-1} + \frac{1}{2} $$ The general solution for the given nonhomogeneous differential equation is: $$ y(t) = C_1t^2 + C_2t^{-1} + \frac{1}{2} $$

Step-by-step solution

Find the corresponding homogeneous equation

To find the corresponding homogeneous equation, we simply set the right-hand side of the given equation to 0: $$ t^2y'' - 2y = 0 $$

Verify \(y_1(t) = t^2\) satisfies the homogeneous equation

Differentiate \(y_1(t)\) twice to get \(y_1''(t)\): $$ y_1'(t) = 2t \\ y_1''(t) = 2 $$ Now, substitute \(y_1\) and \(y_1''\) into the homogeneous equation: $$ t^2(2) - 2(t^2) = 2t^2 - 2t^2 = 0 $$ Since the homogeneous equation is satisfied for \(y_1(t) = t^2\), it is a solution.

Verify \(y_2(t) = t^{-1}\) satisfies the homogeneous equation

Differentiate \(y_2(t)\) twice to get \(y_2''(t)\): $$ y_2'(t) = -t^{-2} \\ y_2''(t) = 2t^{-3} $$ Now, substitute \(y_2\) and \(y_2''\) into the homogeneous equation: $$ t^2(2t^{-3}) - 2(t^{-1}) = 2t^{-1} - 2t^{-1} = 0 $$ Since the homogeneous equation is satisfied for \(y_2(t) = t^{-1}\), it is a solution.

Find a particular solution of the given nonhomogeneous equation

Let's assume a particular solution of the form \(y_p(t) = At^2 + Bt + C\), where \(A\), \(B\), and \(C\) are constants. Differentiate \(y_p(t)\) twice to get \(y_p''(t)\): $$ y_p'(t) = 2At + B \\ y_p''(t) = 2A $$ Now, substitute \(y_p\) and \(y_p''\) into the nonhomogeneous equation: $$ t^2(2A) - 2(At^2 + Bt + C) = 3t^2 - 1 $$ Simplify this equation: $$ 2At^3 - 2At^2 - 2Bt - 2C = 3t^2 - 1 $$ Comparing coefficients, we get the following system of equations: \\ System equation 1: \(2A = 0\) \\ System equation 2: \(-2A = 3\) \\ System equation 3: \(-2B = 0\) \\ System equation 4: \(-2C = -1\) Solve for \(A\), \(B\), and \(C\): For system equation 1, we get: \(A = 0\) For system equation 3, we get: \(B = 0\) For system equation 4, we get: \(C = \frac{1}{2}\) The particular solution for the nonhomogeneous equation is: $$ y_p(t) = \frac{1}{2} $$

Key concepts

Homogeneous Equation

A homogeneous differential equation is a type of differential equation where the right-hand side is set to zero. This means that all terms of the equation depend solely on the function and its derivatives. For the given problem, the homogeneous equation is written as:

• \(t^2 y'' - 2y = 0\)

When dealing with homogeneous equations, the key task is to verify whether certain functions are solutions to this equation. Solutions to homogeneous equations exhibit a property of linearity, which means any linear combination of solutions is also a solution.
Identifying solutions to a homogeneous equation is essential, as they form the basis when determining the general solution to a nonhomogeneous equation. In this problem, we check the functions \(y_1(t) = t^2\) and \(y_2(t) = t^{-1}\) against the derived homogeneous equation to confirm they satisfy it.

Particular Solution

A particular solution to a differential equation is a solution that satisfies the entire equation, including any non-zero term on the right-hand side. For the given problem, the nonhomogeneous equation is:

• \(t^2 y'' - 2y = 3t^2 - 1\)

Finding a particular solution involves making an educated guess or "ansatz" regarding the form of the solution. Given that the right-hand side includes polynomial terms, we assume the particular solution to be a polynomial of similar or lesser order, specifically:

• \(y_p(t) = At^2 + Bt + C\)

Here, we differentiate this assumed form and substitute into the nonhomogeneous equation to identify the constants \(A\), \(B\), and \(C\) that will satisfy the equation. The solving process yields \(A = 0\), \(B = 0\), and \(C = \frac{1}{2}\), giving a particular solution of \(y_p(t) = \frac{1}{2}\). This solution resolves only the specific nonhomogeneous part of the equation.

Nonhomogeneous Equation

A nonhomogeneous differential equation involves terms that do not solely depend on the function or its derivatives; it includes external forcing functions represented by non-zero terms on the right-hand side. The equation we've focused on is:

• \(t^2 y'' - 2y = 3t^2 - 1\)

This equation can be viewed as having two components: the left side, similar to the homogeneous equation, and the right side, which introduces a "forcing" term. This term requires solving for a particular solution to address the specific external effects on the system represented by the differential equation.
Understanding the structure of nonhomogeneous equations is crucial for finding their solutions as the complete solution is the sum of:

• The general solution of the associated homogeneous equation.
• A particular solution of the nonhomogeneous equation.

These components work together to demonstrate how the system responds under both inherent dynamics and external influences.

Verification of Solutions

Verification is a methodical process to confirm that a proposed solution satisfies a given differential equation. This step is crucial to ensure accuracy and correctness of the solutions applied to both homogeneous and nonhomogeneous equations.
For the homogeneous equation \(t^2 y'' - 2y = 0\), verification involves taking the derivatives of the functions \(y_1(t)\) and \(y_2(t)\), substituting them back into the equation, and checking if they yield zero. This confirms whether each function is indeed a solution.

• For \(y_1(t) = t^2\), after differentiation and substitution, we confirmed that it satisfies the equation.
• Similarly, for \(y_2(t) = t^{-1}\), the process shows it too satisfies the equation.

For the nonhomogeneous equation, the particular solution \(y_p(t) = \frac{1}{2}\) is verified by substitution into the original equation to show that both sides balance to include the effects of the forcing term.
Through verification, clarity is attained on the suitability of the proposed solutions in explaining the behavior described by the differential equations.