Question

In each exercise, you are given the general solution of $$ y^{(4)}+a_{3} y^{\prime \prime \prime}+a_{2} y^{\prime \prime}+a_{1} y^{\prime}+a_{0} y=0, $$ where \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\) are real constants. Use the general solution to determine the constants \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\). [Hint: Construct the characteristic equation from the given general solution.] $$ y(t)=c_{1} e^{t}+c_{2} t e^{t}+c_{3} e^{-t}+c_{4} t e^{-t} $$

Short answer

Answer: The values of the constants are \(a_3 = 0\), \(a_2 = 1\), \(a_1 = -1\), and \(a_0 = 0\).

Step-by-step solution

Find the first derivative of y(t)

To find the first derivative of the provided function \(y(t)\), we differentiate each term with respect to t: $$ y'(t) = \frac{d}{dt}(c_1 e^t + c_2 t e^t + c_3 e^{-t} + c_4 t e^{-t}) $$ $$ y'(t) = c_1 e^t + c_2 e^t + c_2 t e^t - c_3 e^{-t} + c_4 e^{-t} - c_4 t e^{-t} $$

Find the second derivative of y(t)

Differentiate \(y'(t)\) once more to obtain the second derivative: $$ y''(t) = \frac{d}{dt}(c_1 e^t + c_2 e^t + c_2 t e^t - c_3 e^{-t} + c_4 e^{-t} - c_4 t e^{-t}) $$ $$ y''(t) = c_1 e^t + 2c_2 e^t + c_2 t e^t + c_3 e^{-t} - 2c_4 e^{-t} + c_4 t e^{-t} $$

Find the third derivative of y(t)

Differentiate \(y''(t)\) to obtain the third derivative: $$ y'''(t) = \frac{d}{dt}(c_1 e^t + 2c_2 e^t + c_2 t e^t + c_3 e^{-t} - 2c_4 e^{-t} + c_4 t e^{-t}) $$ $$ y'''(t) = c_1 e^t + 3c_2 e^t + c_2 t e^t - c_3 e^{-t} - 3c_4 e^{-t} + c_4 t e^{-t} $$

Find the fourth derivative of y(t)

Differentiate \(y'''(t)\) to obtain the fourth derivative: $$ y^{(4)}(t) = \frac{d}{dt}(c_1 e^t + 3c_2 e^t + c_2 t e^t - c_3 e^{-t} - 3c_4 e^{-t} + c_4 t e^{-t}) $$ $$ y^{(4)}(t) = c_1 e^t + 4c_2 e^t + c_2 t e^t + c_3 e^{-t} + 4c_4 e^{-t} + c_4 t e^{-t} $$

Substitute the derivatives into the differential equation and compare coefficients

Substitute the derivatives \(y^{(4)}(t)\), \(y'''(t)\), \(y''(t)\), and \(y'(t)\) into the given equation: $$ y^{(4)}(t) + a_3 y'''(t) + a_2 y''(t) + a_1 y'(t) + a_0 y(t) = 0 $$ Now, compare the coefficients of the like terms on both sides: 1. Coefficients of \(e^t\): \(1 + a_3 + a_2 + a_1 + a_0 = 0\) 2. Coefficients of \(te^t\): \(a_3 + a_2 + a_1 = 0\) 3. Coefficients of \(e^{-t}\): \(1 - a_3 + a_2 - a_1 + a_0 = 0\) 4. Coefficients of \(te^{-t}\): \(-a_3 + a_2 - a_1 = 0\)

Solve for the constants \(a_3\), \(a_2\), \(a_1\), and \(a_0\)

We can solve the four equations simultaneously to find the constants: 1. \(a_3 + a_2 + a_1 = 0\) 2. \(a_0 - a_2 + a_1 = -1\) 3. \(a_3 - a_1 - a_0 = -1\) 4. \(a_3 + a_1 - a_2 = 0\) Solving these equations gives us: \(a_3 = 0\), \(a_2 = 1\), \(a_1 = -1\), \(a_0 = 0\) Therefore, the constants are \(a_3 = 0\), \(a_2 = 1\), \(a_1 = -1\), and \(a_0 = 0\).

Key concepts

Characteristic equation

The **characteristic equation** plays a key role in solving linear differential equations with constant coefficients. It helps determine the solutions of the differential equation by transforming it into an algebraic equation. To form the characteristic equation, we assume the solutions of the differential equation can be expressed in the form of exponential functions. For a differential equation like \( y^{(4)} + a_3 y''' + a_2 y'' + a_1 y' + a_0 y = 0 \), the characteristic equation would typically take the form \( r^4 + a_3 r^3 + a_2 r^2 + a_1 r + a_0 = 0 \).
To construct the characteristic equation from the given general solution \( y(t) = c_1 e^t + c_2 t e^t + c_3 e^{-t} + c_4 t e^{-t} \), we recognize the exponential components as solutions to the characteristic equation, with \( r = 1 \) and \( r = -1 \) being the roots. The nature of these roots (multiplicity, whether real or complex) informs the general solution form.
By knowing the characteristic roots, you can derive the general solution more confidently, and use it to decipher the coefficients \( a_3, a_2, a_1, \) and \( a_0 \) as demonstrated in the problem.

General solution

In the context of differential equations, the **general solution** is the most comprehensive form of solutions that satisfies the equation for any constant values involved. The general solution is derived based on the roots of the characteristic equation. It covers all potential particular solutions for given initial or boundary conditions.
For the fourth-order differential equation provided \( y^{(4)} + a_3 y''' + a_2 y'' + a_1 y' + a_0 y = 0 \), the general solution given is \( y(t) = c_1 e^t + c_2 t e^t + c_3 e^{-t} + c_4 t e^{-t} \). This indicates two pairs of real roots, each occurring with multiplicity two. Hence, we have terms \( e^t \) and \( t e^t \) for the root \( r = 1 \), and \( e^{-t} \) and \( t e^{-t} \) for the root \( r = -1 \).
Understanding how the characteristic equation translates to the general solution helps in managing complex differential equations and tailoring specific solutions to initial conditions.

Higher-order derivatives

Higher-order derivatives involve derivatives beyond the first or second level, which are common in complex differential equations. The task of finding these derivatives is crucial when dealing with higher-order linear differential equations. The fourth derivative, in this case, \( y^{(4)}(t) \), is the key component of the given differential equation.
While solving or substituting into a differential equation, you will typically compute several derivatives step by step. This repeated differentiation process ensures that all terms required to form and solve the differential equation are accurately represented.
Higher-order derivatives can be challenging to manage as they often involve complex combinations of terms. Therefore, it's essential to perform each differentiation step-by-step, simplifying at each stage to maintain clarity and accuracy, as demonstrated in the exercises.

Linear differential equations

**Linear differential equations** are characterized by terms dependent linearly on the unknown function and its derivatives. This linearity implies the superposition principle applies, meaning any linear combination of solutions is also a solution.
In this exercise, the equation \( y^{(4)} + a_3 y''' + a_2 y'' + a_1 y' + a_0 y = 0 \) is a linear differential equation of the fourth order. The coefficients \( a_3, a_2, a_1, \) and \( a_0 \) dictate the behavior and particularities of the solution.
Linear equations are advantageous in mathematical analysis due to their straightforward solution methods, notably through characteristic equations. Converting a differential equation into an algebraic form allows us to explore solutions using known algebraic techniques like root-finding. Understanding linear differential equations' properties is vital for resolving complex mathematical models across various scientific fields.