Question

If \(f, g,\) and \(h\) are differentiable functions, show that \(W(f g, f h)=f^{2} W(g, h)\).

Short answer

Question: Prove that if \(f, g,\) and \(h\) are differentiable functions, then \(W(fg, fh)=f^2W(g, h)\), where \(W(u, v)\) denotes the Wronskian of the functions \(u\) and \(v\). Answer: We computed the Wronskians of the pairs of functions \((fg, fh)\) and \((g, h)\) and showed that \(f^2W(fg, fh) = f^2W(g, h)\), proving that \(W(fg, fh)=f^2W(g, h)\).

Step-by-step solution

Compute the Wronskian of \((fg, fh)\)

We will start by computing the Wronskian of the pair of functions \((fg, fh)\). Using the definition of the Wronskian, we have: $$ W(fg, fh) = \begin{vmatrix} fg & fh \\ (fg)' & (fh)' \end{vmatrix} $$ First, we need to find \((fg)'\) and \((fh)'\) using the Product Rule of differentiation, which states that \((uv)' = u'v + uv'\). Applying the Product Rule to \(fg\) and \(fh\), we have: $$ (fg)' = f'g + fg' \\ (fh)' = f'h + fh' $$ Now, we can substitute these expressions back into the Wronskian definition: $$ W(fg, fh) = \begin{vmatrix} fg & fh \\ f'g + fg' & f'h + fh' \end{vmatrix} = (fg)(f'h + fh') - (fh)(f'g + fg') $$

Compute the Wronskian of \((g, h)\)

Next, we'll compute the Wronskian of the pair of functions \((g, h)\): $$ W(g, h) = \begin{vmatrix} g & h \\ g' & h' \end{vmatrix} = g h' - h g' $$

Relate \(W(fg, fh)\) and \(W(g, h)\)

We are trying to show that \(W(fg, fh) = f^2W(g, h)\). Let's rewrite \(W(fg, fh)\) using the expression we found in Step 1 and multiply it by \(f^2\): $$ f^2W(fg, fh) = f^2[(fg)(f'h + fh') - (fh)(f'g + fg')] = f^2g(f'h + fh') - f^2h(f'g + fg') $$ Now, let's rewrite \(W(g, h)\) using the expression we found in Step 2: $$ W(g, h) = g h' - h g' $$ To show that \(f^2W(fg,fh) = f^2W(g,h)\), we need to show that \( f^2g(f'h + fh') - f^2h(f'g + fg') = f^2(gh' - hg')\). Expanding and simplifying the left-hand side of the equation: $$ f^2g(f'h + fh') - f^2h(f'g + fg') = f^3g'h + f^2gh' - f^3gh' - f^2hg' = f^2(gh' - hg') $$ Thus, we have shown that \(W(fg, fh) = f^2W(g, h)\), as required.

Key concepts

Differentiable Functions

When we talk about differentiable functions, we're referring to those which possess a derivative at each point in their domain. A derivative represents the instantaneous rate of change of the function's value with respect to its input. In layman's terms, it measures how quickly or slowly the function's output is changing at any point.

For a function to be differentiable, it must be smooth and continuous; there can't be any sharp corners or breaks. This requirement is crucial for the mathematics of differential equations, such as the Wronskian, which involves finding the determinants of a matrix composed of functions and their derivatives. A fundamental benefit of differentiability is the ability to apply powerful rules, like the product rule, to simplify complex derivatives of products of functions.

Product Rule of Differentiation

The product rule is a widely-used differentiation technique that allows us to find the derivative of a product of two functions, something we often encounter in calculus. It states that the derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

Expressed mathematically, for two differentiable functions, say, u and v, the rule is represented as \[ (uv)' = u'v + uv' \.\] This rule is especially relevant when working with the Wronskian since it often involves products of functions and their derivatives. When computing the determinant involved in the Wronskian, applying the product rule simplifies the process and makes it possible to handle more complex algebraic manipulations.

Determinants in Differential Equations

In the realm of differential equations, determinants play a significant role, particularly in understanding the behavior of solutions to the equations. The Wronskian is an example of using determinants, which helps in determining whether a set of solutions to a differential equation is linearly independent.

A determinant is essentially a scalar value that can be computed from a square matrix and conveys important properties of the matrix, such as whether the matrix is invertible. When applying this to differential equations, the Wronskian involves a matrix of functions and their derivatives. Calculating the determinant of this matrix, or the Wronskian, we get insight into the linear independence of the functions in question. If the Wronskian is non-zero, it implies that the functions are linearly independent, an essential condition for them to be a fundamental set of solutions for the differential equation.