Question

The given pair of functions $\text{Missing left or extra right}$ forms a fundamental set of solutions of the differential equation. (a) Show that the given function $\overline{y}(t)$ is also a solution of the differential equation. (b) Determine coefficients $c_1$ and $c_2$ such that $\overline{y}(t)=c_1{y}_1(t)+c_2{y}_2(t)$.\(t^{2} y^{\prime \prime}-t y^{\prime}+y=0, \quad 0

Short answer

If it is a solution, find coefficients $c_1$ and $c_2$ such that $\overline{y}(t)=c_1{y}_1(t)+c_2{y}_2(t)$. Answer: Based on our analysis, $\overline{y}(t)=2t+t\ln (3t)$ is not a solution to the given differential equation. Therefore, it does not make sense to find the coefficients $c_1$ and $c_2$ in this case. It is recommended to re-check the given exercise for any typos or mistakes.

Step-by-step solution

Find the first and second derivative of $\overline{y}(t)$

To check whether $\overline{y}(t)$ is a solution of the differential equation, we need to find its first and second derivatives. In this case, $\overline{y}(t)=2t+t\ln (3t)$. Now, we differentiate $\overline{y}(t)$ once with respect to $t$ to obtain: $\frac{d\overline{y}(t)}{dt}=\frac{d(2t+t\ln (3t))}{dt}=2+\ln (3t)+1=\ln (3t)+3$. Then, we differentiate $\overline{y}(t)$ again with respect to $t$ to obtain its second derivative: $\frac{d^2\overline{y}(t)}{d{t}^2}=\frac{d(\ln (3t)+3)}{dt}=\frac{3}{3t}=1/t$.

Substitute the derivatives into the differential equation

We are given the differential equation $t^2{y}^{″}-t{y}^{\prime }+y=0$. Now substitute the derivatives of $\overline{y}(t)$ that we found in Step 1: $t^2\frac{1}{t}-t(\ln (3t)+3)+(2t+t\ln (3t))=1-(t\ln (3t)+3t)+(2t+t\ln (3t))=1-3t+2t=1-t$.

Verify that the differential equation holds

From Step 2, we have found that $t^2{y}^{″}-t{y}^{\prime }+y=1-t$. However, for $\overline{y}(t)$ to be a solution to the given differential equation, the equation should be equal to 0. This means $\overline{y}(t)$ is not a solution to the given differential equation as it was assumed in the exercise.

Comparing the function with a linear combination of $y_1(t)$ and $y_2(t)$

Even though the text requires us to find the coefficients $c_1$ and $c_2$ such that $\overline{y}(t)=c_1{y}_1(t)+c_2{y}_2(t)$, we have already shown that $\overline{y}(t)$ is not a solution for the given differential equation, and the demand does not make sense. In this case, it would be wise to re-check the given exercise for any typos or mistakes in the problem statement and proceed accordingly.

Key concepts

Fundamental Set of Solutions

Understanding the fundamental set of solutions is crucial when dealing with homogeneous linear differential equations, as it underpins the method used to find the general solution. A fundamental set of solutions for a differential equation consists of a group of solutions that are linearly independent and span the solution space. This means that any other solution to the differential equation can be expressed as a linear combination of these functions.

For example, in the given exercise, if the functions $y_1(t)$ and $y_2(t)$ are indeed part of a fundamental set for the differential equation $t^2{y}^{\prime \prime }-t{y}^{\prime }+y=0$, then any solution $\overline{y}(t)$ should be expressible as $\overline{y}(t)=c_1{y}_1(t)+c_2{y}_2(t)$ with appropriate coefficients $c_1$ and $c_2$. The concept becomes even clearer when you visualize these functions as vectors in a multi-dimensional space, where the fundamental solutions span a 'plane' upon which any other solution lies.

Second Derivative

The second derivative, denoted as $y^{″}$ or $\frac{d^{2}y}{d{t}^2}$, is a measure of how the rate of change of a quantity is changing. In the context of the exercise, calculating the second derivative of $\overline{y}(t)$ is a part of verifying whether it is indeed a solution to the given differential equation. By finding both the first and second derivatives of $\overline{y}(t)$, and substituting them into the differential equation, we can check for its validity.

In physical applications, the second derivative is often associated with acceleration, as it describes the change in velocity over time. In our case, it's part of a broader procedure to confirm that a function satisfies a particular equation, helping us understand the behavior of dynamic systems described by differential equations.

Coefficient Determination

Coefficient determination involves finding the exact values for constants in a linear combination of functions that form a particular solution to a differential equation. This is a step beyond simply knowing the general structure of the solution; it involves matching the solution to specific conditions, often initial values or boundary conditions.

For instance, if the differential equation's solution is indeed a linear combination of $y_1(t)$ and $y_2(t)$, we determine the coefficients $c_1$ and $c_2$ that make the equation true for the given function $\overline{y}(t)$. However, as noted in the exercise solution, if $\overline{y}(t)$ does not satisfy the differential equation, then the process of determining these coefficients becomes irrelevant for the incorrectly assumed solution.

When coefficients are correctly determined, the resulting expression reflects the specific behavior of the system being modeled, thus tying abstract mathematical solutions to concrete physical or theoretical scenarios.