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. $$y^{\prime \prime \prime}+p(t) y^{\prime}+q(t) y=e^{t}+t+1$$

Short answer

Answer: The particular solution of the given differential equation is \(y(t) = \frac{1}{2}u_1(t) + \frac{1}{3}u_3(t)\).

Step-by-step solution

Find a combination of given functions

To find a particular solution, we should look for a function \(y(t) = c_1u_1(t) + c_2u_2(t) + c_3u_3(t)\), where \(c_1, c_2,\) and \(c_3\) are constants to be determined.

Differentiate the linear combination and substitute into the given DE

Now, we need to differentiate the function \(y(t)\) twice: $$y'(t) = c_1u_1'(t) + c_2u_2'(t) + c_3u_3'(t)$$ $$y''(t) = c_1u_1''(t) + c_2u_2''(t) + c_3u_3''(t)$$ And substitute these expressions into the given DE: $$y^{\prime\prime \prime} + p(t)y^{\prime} + q(t)y = (c_1u_1^{\prime\prime \prime} + c_2u_2^{\prime\prime \prime} + c_3u_3^{\prime\prime \prime}) + p(t)(c_1u_1^{'} + c_2u_2^{'}+c_3u_3^{'}) + q(t)(c_1u_1+c_2u_2+c_3u_3) = e^{t}+t+1.$$

Use the given DEs for \(u_1(t),u_2(t),\) and \(u_3(t)\)

Now, substitute the given DEs for \(u_1(t),u_2(t),\) and \(u_3(t)\) to relate the left-hand side with the right-hand side: $$c_1(2e^t + 1) + c_2(4) + c_3(3t) = e^t + t + 1$$

Equate coefficients and solve for constants

Now, compare the coefficients of the different terms $$c_1(2e^t) + c_2(0) + c_3(0) = e^t$$ $$c_1(1) + c_2(4) + c_3(3t) = t + 1$$ From the first equation, \(c_1 = \frac{1}{2}\). Now, plug this into the second equation and solve for \(c_2\) and \(c_3\): $$\frac{1}{2} + 4c_2 + 3tc_3 = t + 1$$ $$1 + 4c_2 = 1 \Rightarrow c_2 = 0$$ $$3tc_3=t \Rightarrow c_3 = \frac{1}{3}$$

Write the particular solution

Now that we have the constants, we can write the particular solution of the given DE: $$y(t) = \frac{1}{2}u_1(t) + 0u_2(t) + \frac{1}{3}u_3(t) = \frac{1}{2}u_1(t) + \frac{1}{3}u_3(t)$$

Key concepts

Particular Solution

In differential equations, finding a particular solution is a critical step when dealing with non-homogeneous equations. A particular solution is a specific solution that satisfies the entire differential equation, including the non-homogeneous part (the right-hand side). Each differential equation can have an infinite number of solutions, forming a family of curves. The particular solution refers to the one specific curve that passes through the constraints defined by the equation.

In the given exercise, the equation is complex because of its non-homogeneous components like \(e^{t}\), \(t\), and constant terms. To find the particular solution, we need a systematic approach. Using the trial solution method, where we assume a linear combination of known solutions (like \(y(t) = c_1u_1(t) + c_2u_2(t) + c_3u_3(t)\)), helps us in matching the non-homogeneous part. From here, it's about substituting back into the equation and comparing coefficients to solve for the unknown constants \(c_1\), \(c_2\), and \(c_3\). This process leads to a particular solution that uniquely fits the non-homogenous equation.

Linear Differential Equations

Linear differential equations are equations that involve an unknown function and its derivatives, with no terms multiplying them. They are significant because they can model a wide range of real-world phenomena. In the linear form, these equations are expressed generally as \(a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + ... + a_1(x)y' + a_0(x)y = g(x)\), where \(g(x)\) is the non-homogeneous part.

For linear equations, the superposition principle applies. This means if two functions are solutions of the linear differential equation, their sum is also a solution. This property makes linear differential equations more manageable and predictable compared to non-linear equations. In our exercise, the provided functions \(u_1(t)\), \(u_2(t)\), and \(u_3(t)\) already solve individual linear differential equations. By combining these with appropriate coefficients, we form a solution for the main equation, utilizing the linearity property.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique to find a particular solution to linear non-homogeneous differential equations. When the non-homogeneous term \(g(t)\) in an equation is a linear combination of known functions, guess a solution form resembling \(g(t)\) but with undetermined coefficients.

The process involves the following steps:

• Identify the form of \(g(t)\) — it could be polynomials, exponentials, or trigonometric functions.
• Formulate a potential solution with undetermined coefficients corresponding to \(g(t)\).
• Substitute this potential solution back into the differential equation.
• Adjust the coefficients by equating with similar terms on the equation's both sides.

In the exercise, after determining \(c_1\), \(c_2\), and \(c_3\), these coefficients align \(u_1(t)\), \(u_2(t)\), and \(u_3(t)\) with \(e^t \) and \(t\) from the main equation. Thus, by solving for the coefficients, we effectively find the particular solution tailored to the given equation.