Question

Find the Wronskian of two solutions of the given differential equation without solving the equation. \((\cos t) y^{\prime \prime}+(\sin t) y^{\prime}-t y=0\)

Short answer

Question: Find the Wronskian of two solutions \(y_1(t)\) and \(y_2(t)\) of the given differential equation \((\cos t) y^{\prime \prime}+(\sin t) y^{\prime}-t y=0\) without actually solving the equation. Answer: The Wronskian of the two solutions to the given differential equation is \(W(y_1, y_2) = 0\).

Step-by-step solution

Write down the definition of the Wronskian

The Wronskian of two functions \(y_1(t)\) and \(y_2(t)\) is given by: \[ W(y_1, y_2) = \begin{vmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \end{vmatrix} = y_1(t)y_2'(t) - y_2(t)y_1'(t). \]

Apply properties of the second order linear differential equation to calculate the Wronskian

We are given the differential equation: \[ (\cos t) y^{\prime \prime}+(\sin t) y^{\prime}-t y=0. \] Let \(y_1(t)\) and \(y_2(t)\) be two solutions of this differential equation. We can then use the property that, for second order linear ordinary differential equations, the Wronskian can be found by differentiating the equation with respect to \(t\). Taking the time derivative of the original equation and applying the product rule to the derivatives of \(y\), we get: \[ -\sin(t)y^{\prime \prime}+\cos(t)y^{(3)}+\cos(t)y^{\prime \prime}+\sin(t)y^{(3)}-ty^{\prime}-y=0. \] Substitute \(y = y_1\) and \(y = y_2\) in this modified equation to obtain: \[ -\sin(t)y_1^{\prime \prime}+\cos(t)y_1^{(3)}+\cos(t)y_1^{\prime \prime}+\sin(t)y_1^{(3)}-ty_1^{\prime}-y_1=0 \\ -\sin(t)y_2^{\prime \prime}+\cos(t)y_2^{(3)}+\cos(t)y_2^{\prime \prime}+\sin(t)y_2^{(3)}-ty_2^{\prime}-y_2=0. \] Now, form a new equation by multiplying the first equation by \(y_2\) and the second equation by \(y_1\): \[y_1y_2[2\cos(t)y^{\prime \prime}+2\sin(t)y^{(3)}-ty^{\prime}-y]=0.\] Integrate both sides of the equation: \[ \int{y_1y_2[2\cos(t)y^{\prime \prime}+2\sin(t)y^{(3)}-ty^{\prime}-y] dt} = C. \] Now we can use the formula for the Wronskian: \[ W(y_1, y_2) = y_1(t)y_2'(t) - y_2(t)y_1'(t). \] We have: \[ W(y_1, y_2) = \int (\cancelto{0}{y_1y_2[2\cos(t)y^{\prime \prime}+2\sin(t)y^{(3)}-ty^{\prime}-y]}) dt. \] As the integrand is zero, the integral itself will also be zero. Therefore, the Wronskian of the two solutions to the given differential equation is: \[ W(y_1, y_2) = 0. \]

Key concepts

Linear Ordinary Differential Equations

Linear ordinary differential equations (ODEs) are equations that involve the derivatives of a function and are linear in the unknown function and its derivatives. This means that each term of the equation is either the unknown function itself, one of its derivatives, or a constant, and these terms are only added, subtracted, or multiplied by a function that does not depend on the unknown function. For example, a general second-order linear ordinary differential equation can be written in the form
\[a_2(t)y''+a_1(t)y'+a_0(t)y=g(t),\]
where \(y'\) and \(y''\) are the first and second derivatives of the unknown function \(y\), respectively, and \(a_0(t)\), \(a_1(t)\), \(a_2(t)\), and \(g(t)\) are known functions of the independent variable \(t\). One of the significant properties of linear ODEs is the principle of superposition, which states that if \(y_1(t)\) and \(y_2(t)\) are solutions to the homogeneous equation (where \(g(t) = 0\)), any linear combination of these solutions is also a solution.

Second Order Differential Equation

Second order differential equations are widely used in physics and engineering to model systems with two time-dependent states. The equation given in the exercise is a particular kind of second order differential equation where the coefficients are functions of \(t\), and it lacks a term without derivatives:
\[(\cos t) y'' + (\sin t) y' - t y = 0.\]
This format implies that the solution involves second derivatives of the function \(y(t)\). To solve such equations, various methods can be used, including the characteristic equation approach for constant coefficients, reduction of order, and variation of parameters for non-constant coefficients. The presence of specific functions as coefficients, like sine and cosine in this case, often suggests the use of Special functions or series solutions due to the periodic nature of the coefficients.

Determinant

The determinant is a scalar value that can be computed from the elements of a square matrix. It provides critical information about the matrix, such as whether the matrix is invertible, and the volume transformation factor when the matrix is viewed as a linear transformation. The simplest non-trivial determinant is that of a 2x2 matrix:
\[\begin{vmatrix}a & b\c & d\end{vmatrix} = ad - bc.\]
The determinant of a 2x2 matrix is used in the Wronskian to study the existence and uniqueness of solutions of systems of linear differential equations. The Wronskian itself is the determinant of a special matrix composed of the solutions to the ODE and their derivatives. When the Wronskian is not equal to zero, it indicates that the solutions are linearly independent and thus form a fundamental set of solutions for the differential equation. In the provided exercise, the Wronskian was zero, suggesting a particular relationship between the solutions, which could imply dependency or special conditions imposed by the form of the differential equation.