Question

The functions \(u_{1}(t), u_{2}(t)\), and \(u_{3}(t)\) are solutions of the differential equations $$ \begin{aligned} &u_{1}^{\prime \pi}+p(t) u_{1}^{\prime}+q(t) u_{1}=2 e^{t}+1, \quad u_{2}^{\prime \prime}+p(t) u_{2}^{\prime}+q(t) u_{2}=4, \\ &u_{3}^{\prime \prime}+p(t) u_{3}^{\prime}+q(t) u_{3}=3 t . \end{aligned} $$ Use the functions \(u_{1}(t), u_{2}(t)\), and \(u_{3}(t)\) to construct a particular solution of the given differential equation. $$y^{\prime \prime}+p(t) y^{\prime}+q(t) y=t+2$$

Short answer

Answer: The particular solution y(t) is given by \(y(t) = u_1(t) + \frac{1}{2} u_2(t) + u_3(t)\) with the coefficients 1, 1/2, and 1, respectively.

Step-by-step solution

Define the linear combination of functions

We'll start by defining the linear combination of the given functions \(u_1(t)\), \(u_2(t)\), and \(u_3(t)\). Let's consider a function y(t) defined as a linear combination of these functions: $$y(t) = A u_1(t) + Bu_2(t) + Cu_3(t)$$ We'll now find the constants A, B, and C such that the function y(t) satisfies the given differential equation: $$y''(t) + p(t)y'(t) + q(t)y(t) = t + 2$$

Calculate derivatives of y(t)

To proceed, we need to find the first and second derivatives of y(t): $$y'(t) = A u_1'(t) + B u_2'(t) + C u_3'(t)$$ $$y''(t) = A u_1''(t) + B u_2''(t) + C u_3''(t)$$

Plug derivatives into the differential equation

Now we will substitute the derivatives into the given differential equation: $$(A u_1''(t) + B u_2''(t) + C u_3''(t)) + p(t)(A u_1'(t) + B u_2'(t) + C u_3'(t)) + q(t)(A u_1(t) + B u_2(t) + C u_3(t)) = t + 2$$

Use given differential equations of u1(t), u2(t), and u3(t)

We can replace the expressions of the second derivatives and their corresponding functions, since we know the differential equations for each function: $$A(2 e^{t}+1) + B(4) + C(3t) + p(t)(A u_1'(t) + B u_2'(t) + C u_3'(t)) + q(t)(A u_1(t) + B u_2(t) + C u_3(t)) = t + 2$$

Determine A, B, and C

The differential equation holds for all t-values, so we can deduce that the coefficients of the terms must be equal. That is: $$A = 1$$ $$B = \frac{1}{2}$$ $$C = 1$$

Write the particular solution

Now that we have found the values of A, B, and C, we can write the particular solution: $$y(t) = u_1(t) + \frac{1}{2} u_2(t) + u_3(t)$$ Thus, our particular solution y(t) is a linear combination of \(u_1(t)\), \(u_2(t)\), and \(u_3(t)\) with the constants 1, 1/2, and 1, respectively.

Key concepts

Linear Combination of Solutions

Differential equations often have multiple solutions, and an interesting property is that combinations of these solutions can also satisfy the same differential equation. This principle is known as the linear combination of solutions. In the context of a differential equation, a solution refers to a function that satisfies the equation when substituted into it.

When dealing with a set of solutions, say, \( u_1(t)\), \( u_2(t)\), and \( u_3(t)\), any new function formed by taking the sum of these solutions, each multiplied by a constant (which could be zero), is a linear combination. The general form can be expressed as: \( y(t) = A \cdot u_1(t) + B \cdot u_2(t) + C \cdot u_3(t)\), where \(A\), \(B\), and \(C\) are the constants. This new function \(y(t)\) can be a solution to the differential equation if the constants are chosen correctly, as demonstrated in our exercise.

Particular Solution

When it comes to differential equations, especially non-homogeneous ones which include a term that is not a function of the dependent variable or its derivatives only, the particular solution plays a vital role. This solution is specific to the non-homogeneous part of the equation and is not part of the complementary (general) solutions that solve the associated homogeneous equation.

To find a particular solution, one approach is to guess a form of the solution that has the same type of terms as the non-homogeneous part but with undetermined coefficients, and then determine these coefficients. As illustrated in the exercise we reviewed, by setting \( y(t) \) equal to a linear combination of known solutions to differential equations with different non-homogeneous terms, we were able to calculate the coefficients that provided a particular solution to the new equation.

Second-Order Differential Equation

A second-order differential equation involves the second derivative of a function and is a foundational concept in mathematics and physics. These equations often appear in the study of motion and other dynamic processes. The generic form for a linear second-order differential equation is: \( y''(t) + p(t) \cdot y'(t) + q(t) \cdot y(t) = g(t)\), where \( y(t) \) is the unknown function, \( y'(t) \) and \( y''(t) \) are its first and second derivatives, \( p(t) \) and \( q(t) \) are coefficients that can depend on \( t \) but not on \( y(t) \) or its derivatives, and \( g(t) \) is a given function (non-homogeneous term).

Solving these equations typically involves finding the general solution to the associated homogeneous equation (where \( g(t) = 0 \) ) and then finding a particular solution that satisfies the original non-homogeneous equation. The complete solution is the sum of the general and particular solutions.