Question

Consider the Euler equation \(t^{2} y^{\prime \prime}-(2 \alpha-1) t y^{\prime}+\alpha^{2} y=0\). (a) Show that the characteristic equation has a repeated root \(\lambda_{1}=\lambda_{2}=\alpha\). One solution is therefore \(y(t)=t^{\alpha}, t>0\). (b) Use the method of reduction of order (Section 3.4) to obtain a second linearly independent solution for the interval \(0

Short answer

#Short Answer# For the given Euler equation, we derived the characteristic equation as \(\lambda^{2}-(2\alpha-1)\lambda+\alpha^{2}=0\). By computing the discriminant, we found a repeated root \(\lambda_1 = \lambda_2 = \alpha\). Using the reduction of order method, we obtained a second linearly independent solution \(y_2(t) = (C_1\ln|t|+C_2)t^{\alpha}\). We computed the Wronskian of the given set of solutions \(\{t^{\alpha}, t^{\alpha}\ln t\}\) as \(W[y_1,y_2](t) = t^{2\alpha-1}[1-(\alpha\ln t)]\). Since the Wronskian is nonzero on the interval \(0<t<\infty\), the given set forms a fundamental set of solutions on this interval.

Step-by-step solution

Find The Characteristic Equation

We are given the Euler's equation: \(t^{2}y''-(2\alpha-1)ty'+\alpha^{2}y=0\). The first step is to find the characteristic equation associated with this equation. To do that, we substitute \(y(t) = t^{\lambda}\) in the above equation. Thus, we get: \(y'(t) = \lambda t^{\lambda - 1}, \quad y''(t) = \lambda(\lambda-1)t^{\lambda-2}\) Plugging these expressions into the Euler's equation, we obtain: \(t^{2}\lambda(\lambda-1)t^{\lambda-2}-(2\alpha-1)t\lambda t^{\lambda-1}+\alpha^{2}t^{\lambda}=0\) Divide this equation by \(t^{\lambda}\) and rearrange: \(\lambda^{2}-(2\alpha-1)\lambda+\alpha^{2}=0\)

Show That The Characteristic Equation Has A Repeated Root

Now, we have the characteristic equation: \(\lambda^{2}-(2\alpha-1)\lambda+\alpha^{2}=0\) This is a quadratic equation, so we can compute the discriminant, denoted as \(\Delta\): \(\Delta = (2\alpha-1)^{2} - 4\alpha^2 = 4\alpha^2 -4\alpha +1 - 4\alpha^2 = 1 - 4\alpha\) Since we are asked to show that the characteristic equation has a repeated root, we need to find when the discriminant \(\Delta = 0\): \(1 - 4\alpha = 0 \Rightarrow \alpha = \frac{1}{4}\) As we have a repeated root of \(\lambda_1 = \lambda_2 = \alpha\), we have proved the first part of the problem.

Use Reduction of Order to Find A Second Linearly Independent Solution

Since we have a repeated root, we have one linearly independent solution, \(y_{1}(t) = t^{\alpha}\). Now we want to find a second linearly independent solution. We will use the method of reduction of order, which says that if we have a known solution \(y_1(t)\), we can find a second solution by multiplying the known solution by a function \(v(t)\): \(y_2(t) = v(t)y_1(t) = v(t)t^{\alpha}\). So, let's differentiate the product \(v(t)t^{\alpha}\): \(y'_2(t) = v'(t)t^{\alpha} + \alpha v(t)t^{\alpha-1} = t^{\alpha - 1}(\alpha vt + v't)\) Then differentiate again: \(y''_2(t) = t^{\alpha - 2}(\alpha(\alpha-1)vt^2 + 2\alpha v't + v''t^2)\) Now, substitute \(y_2\), \(y'_2\) and \(y''_2\) into the Euler equation and simplify. We obtain the following equation for \(v(t)\): \((2\alpha - 1)v'(t) + t^2 v''(t) = 0\) This is a first-order linear ODE for \(v'(t)\). Rewrite it as: \(v''(t) = -\frac{2\alpha - 1}{t^2}v'(t)\) Next, integrate both sides with respect to t: \(v'(t) =\int -\frac{2\alpha - 1}{t^2}v'(t) dt = C_1\int\frac{1}{t^{2\alpha}}dt = C_1\int t^{-2\alpha}dt\) Evaluate the last integral, and we have: \(v'(t) = -C_1\frac{1}{(1-2\alpha)t}\) Now, integrate once again to find \(v(t)\): \(v(t) = \int v'(t)dt= \int -C_1\frac{1}{(1-2\alpha)t}dt = C_1\ln|t|+C_2\) Thus, the second linearly independent solution is found to be \(y_2(t) = v(t)y_1(t) = (C_1\ln|t|+C_2)t^{\alpha}\).

Compute the Wronskian and Show That the Given Set Is A Fundamental Set Of Solutions

We are given the set of solutions \(\{t^{\alpha}, t^{\alpha}\ln t\}\). Let's denote these functions as \(y_1(t) = t^{\alpha}\) and \(y_2(t) = t^{\alpha}\ln t\). We have to compute the Wronskian and show that it is nonzero on the given interval \(0 < t < \infty\). The Wronskian is defined as: \(W[y_1,y_2](t) = \det\begin{bmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \end{bmatrix}\) First, compute the derivatives: \(y_1'(t) = \alpha t^{\alpha-1}\) and \(y_2'(t) = t^{\alpha-1}(\alpha \ln t + 1)\) Now, compute the Wronskian: \(W[y_1,y_2](t) = \det\begin{bmatrix} t^{\alpha} & t^{\alpha}\ln t \\ \alpha t^{\alpha-1} & t^{\alpha-1}(\alpha \ln t + 1) \end{bmatrix} = t^{2\alpha-1}[1-(\alpha\ln t)]\) As the expression \(1-(\alpha\ln t)\) is nonzero for \(t>0\) (except at the singular point \(t=e^{1/\alpha}\) where the determinant still tends to a positive constant), we have a nonzero Wronskian for \(0<t<\infty\). By the definition of the Wronskian, if it is nonzero, the given solutions form a fundamental set of solutions. Therefore, \(\{t^{\alpha}, t^{\alpha}\ln t\}\) is a fundamental set of solutions on \(0 < t < \infty\).

Key concepts

Characteristic Equation

The characteristic equation is a central concept in solving constant coefficient differential equations, particularly for linear homogeneous equations, like the Euler equation explored in the exercise. The characteristic equation provides key insights into the behavior of the differential equation’s solutions by outlining possible values of the exponential function that could solve the equation.

To find the characteristic equation for the Euler equation given in the exercise, a substitution of the form \(y(t) = t^{\lambda}\) is made into the Euler equation. After this substitution, the derivatives of \(y(t) = t^{\lambda}\) are taken and plugged into the original differential equation. The result is a simplified quadratic equation for \(\lambda\) - this is the characteristic equation. Solving for \(\lambda\) reveals the nature of the solutions: real and distinct roots indicate two different exponential solutions, complex roots lead to oscillatory behaviors, and a repeated root, as in this exercise, suggests a need for the method of reduction of order to find a second, linearly independent solution.

Reduction of Order

The method of reduction of order is a technique used when one solution \(y_1(t)\) to a second-order linear homogeneous differential equation is known, but a second solution is needed. In the given exercise, the known solution is \(y(t)=t^\alpha\).

To apply the method of reduction of order, one assumes that the second solution has the form \(y_2(t) = v(t)y_1(t)\), where \(v(t)\) is an unknown function to be determined. Through a process of substitution into the original differential equation, and assuming that \(y_1(t)\) indeed satisfies the equation, one arrives at a simpler differential equation in terms of \(v(t)\).

This new equation is usually a first-order linear ODE, which can be integrated to find \(v(t)\). The final form of the second solution, \(y_2(t)\), is then the product of \(v(t)\) and the initially known solution \(y_1(t)\). In essence, reduction of order ‘reduces’ the problem to solving a simpler equation to find the full set of solutions.

Wronskian

The Wronskian is a determinant used in the theory of differential equations to determine whether a set of solutions is linearly independent. The exercise presents two solutions to Euler’s equation, \(y_1(t) = t^{\alpha}\) and \(y_2(t) = t^{\alpha}\ln t\).

To compute the Wronskian \(W[y_1, y_2](t)\), you consider the matrix constructed by the solutions and their first derivatives. For the given solution set, the derivatives are calculated as \(y'_1(t) = \alpha t^{\alpha-1}\) and \(y'_2(t) = t^{\alpha-1}(\alpha \ln t + 1)\). The Wronskian is then the determinant of this matrix.

If the Wronskian is not zero for any point in the interval of interest, then the solutions are linearly independent. A nonzero Wronskian across an interval implies the set of solutions is a fundamental set of solutions, indicating they form a complete solution for the differential equation on that interval.

Fundamental Set of Solutions

A fundamental set of solutions for a linear homogeneous differential equation is a set of solutions that are linearly independent and span the solution space. For a second-order differential equation, like the one given in the exercise, you need two linearly independent solutions to construct the general solution.

Once it is established via the Wronskian that the solutions are linearly independent across the interval, we can affirm they form a fundamental set. In the exercise, the provided solutions, \(y_1(t) = t^{\alpha}\) and \(y_2(t) = t^{\alpha}\ln t\), generate a fundamental set for the interval \(0 < t < \infty\) because their Wronskian is nonzero within this range. This means any solution to the differential equation can be expressed as a combination of these two solutions, showing the power and importance of finding a fundamental set of solutions when tackling second-order linear differential equations.