Question

If the Wronskian of \(f\) and \(g\) is \(t \cos t-\sin t,\) and if \(u=f+3 g, v=f-g,\) find the Wronskian of \(u\) and \(v .\)

Short answer

Answer: \(W(u, v) = t\cos t - \sin t - (f-g)(f'+3g')\)

Step-by-step solution

Find derivatives of u and v

We're given \(u = f + 3g\) and \(v = f - g\). Compute their first derivatives: \(\frac{d}{dt}u = \frac{d}{dt}(f+3g) = \frac{d}{dt}f + 3\frac{d}{dt}g = u'\) \(\frac{d}{dt}v = \frac{d}{dt}(f-g) = \frac{d}{dt}f - \frac{d}{dt}g = v'\)

Find the Wronskian of u and v

Recall the determinant formula for the Wronskian of two functions: \(W(u, v)=\begin{vmatrix} u & v \\ u' & v'\end{vmatrix}\) Now substitute \(u, v, u'\), and \(v'\) into the determinant formula: \(W(u, v) = \begin{vmatrix} f+3g & f-g \\ u' & v' \end{vmatrix}\)

Use the given Wronskian of f and g

We know the Wronskian of \(f\) and \(g\) is \(t \cos t - \sin t\). In the determinant form, the Wronskian of \(f\) and \(g\) is: \(W(f,g) = \begin{vmatrix} f & g \\ f' & g' \end{vmatrix} = t \cos t - \sin t\) We can use the determinant of this matrix to find the missing values \(f'\) and \(g'\) to complete the Wronskian form for \(u\) and \(v\)

Solve for the Wronskian of u and v

We can now expand the determinant of the Wronskian of \(u\) and \(v\) : \(W(u, v) = \begin{vmatrix} f+3g & f-g \\ f'+3g' & f'-g' \end{vmatrix} = (f+3g)(f'-g') - (f-g)(f'+3g')\) Now we use \(W(f,g) = t \cos t - \sin t\) to replace values \((f+3g)(g' - f') = t\cos t -\sin t\). Substitute this into the equation for the Wronskian of \(u\) and \(v\): \(W(u, v) = (t\cos t -\sin t) - (f-g)(f'+3g')\) Thus, we find that the Wronskian of \(u\) and \(v\) is: \(W(u, v) = t\cos t - \sin t - (f-g)(f'+3g')\)

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are fundamental in expressing the relationships that involve rates of change, such as motion, growth, decay, and other physical and natural processes. They can be simple or complex, and come in many different forms, such as ordinary differential equations (ODEs) and partial differential equations (PDEs).

The process of solving a differential equation often involves integration, which, in simple terms, is the reverse operation to differentiation. Solutions to differential equations can be general, representing a family of functions that satisfy the equation, or particular, which satisfy both the equation and specific conditions, known as initial or boundary values. In our example, we work with functions derived from differential equations and examine their behavior through the Wronskian, which is a way to investigate the properties of these solutions.

Linear Independence

In the realm of mathematics, particularly in linear algebra, linear independence is a key concept that tells us about the uniqueness of the set of functions or vectors. A set of functions is said to be linearly independent if no function in the set can be written as a linear combination of the others.

When employing the Wronskian, we use it to test for linear independence of solutions to differential equations. If the Wronskian is not identically zero, it indicates that the functions in question are linearly independent. For instance, in the exercise given, the functions defined by the equations for 'u' and 'v' are a combination of the original functions 'f' and 'g'. By finding the Wronskian of 'u' and 'v', we can interpret whether these newly formed functions maintain linear independence, a property which is crucial for ensuring that the solutions form a valid solution set for their corresponding differential equations.

Determinants

The determinant is a scalar value that can be computed from the elements of a square matrix. It gives important information about the matrix, such as whether the matrix is invertible or singular. In addition, the determinant of a matrix of functions or vectors plays a pivotal role in understanding linear dependence or independence among them.

For the Wronskian, the determinant is specifically used to build a matrix from the functions and their first derivatives. By taking the determinant of this matrix, we receive a single function, the Wronskian, which we can scrutinize to determine the linear independence of the functions. If the determinant (the Wronskian) is nonzero on the interval, then the functions are linearly independent on that interval. This outcome is leveraged in our example to further investigate the relationship between the Wronskian and the functions 'u' and 'v', which ultimately aids in the understanding of the original differential equations associated with 'f' and 'g'.