Question

If the Wronskian \(W\) of \(f\) and \(g\) is \(3 e^{4 t}\), and if \(f(t)=e^{2 t},\) find \(g(t)\)

Short answer

Question: Determine the function g(t) given that the Wronskian W of the functions f(t) and g(t) is 3e^(4t) and the function f(t) is e^(2t). Answer: g(t) = 3te^(2t) + Ce^(2t), where C is an arbitrary constant.

Step-by-step solution

Write down the given information

We are given the following: - Wronskian, \(W = 3e^{4t}\) - Function \(f(t) = e^{2t}\) Our task is to find the function \(g(t)\).

Compute the derivative of \(f(t)\)

In order to use the Wronskian formula, we need to find the derivative of \(f(t)\). Differentiating \(f(t) = e^{2t}\) with respect to \(t\), we get: \(f'(t) = 2e^{2t}\)

Write down the Wronskian formula and substitute the given information

We know that the Wronskian is given by the formula: \(W(f,g) = fg' - f'g\) Substitute the given information into the formula: \(3e^{4t} = (e^{2t})(g'(t)) - (2e^{2t})(g(t))\)

Solve the differential equation for \(g(t)\)

Now, we need to solve the differential equation for \(g(t)\). This is a first-order linear inhomogeneous differential equation. Let's first divide through by \(e^{2t}\): \(\frac{3e^{4t}}{e^{2t}}=g'(t)-2g(t)\) Simplify the equation: \(3e^{2t}=g'(t)-2g(t)\) Let's write it in the form of a first-order linear differential equation: \(g'(t) - 2g(t) = 3e^{2t}\)

Use the integrating factor method to solve the differential equation

We can solve this differential equation using the integrating factor method. First, we calculate the integrating factor (\(IF\)): \(IF = e^{\int -2dt} = e^{-2t}\) Now, multiply both sides of the differential equation by the integrating factor: \(e^{-2t}g'(t) - 2e^{-2t}g(t) = 3e^{2t}e^{-2t}\) This simplifies to: \((e^{-2t}g(t))' = 3\) Now, integrate both sides with respect to \(t\): \(\int(e^{-2t}g(t))' dt = \int 3 dt\) This gives: \(e^{-2t}g(t) = 3t + C\) Now, multiply both sides by the integrating factor's inverse, which is \(e^{2t}\): \(g(t) = 3te^{2t} + Ce^{2t}\) Here, \(g(t)\) is the function we were looking for, and \(C\) is an arbitrary constant.