Question

Solving Differential Equations Let \(\frac{d y}{d x}=\frac{1}{x}\) . (a) Show that \(y=\ln x+C\) is a solution to the differential equation in the interval \((0, \infty)\) (b) Show that \(y=\ln (-x)+C\) is a solution to the differential equation in the interval \((-\infty, 0)\) (c) Writing to Learn Explain why \(y=\ln |x|+C\) is a solution to the differential equation in the domain \((-\infty, 0) \cup(0, \infty)\) (d) Show that the function \(y=\left\\{\begin{array}{l}{\ln (-x)+C_{1}} \\ {\ln x+C_{2}}\end{array}\right.\) \(x<0\) \(x>0\) is a solution to the differential equation for any values of \(C_{1}\) and \(C_{2}\)

Short answer

The functions provided, \(y = \ln x + C\), \(y = \ln (-x) + C\), \(y = \ln |x| + C\), and \(y=\left\{\begin{array}{l}{\ln (-x)+C_{1}} \ {\ln x+C_{2}}\end{array}\right.\) for \(x < 0\) and \(x > 0\) respectively, are all solutions to the differential equation \( \frac{dy}{dx} = \frac{1}{x} \)

Step-by-step solution

Solve the Differential Equation for \(y = \ln x + C\)

Take the derivative of \(y = \ln x + C\), you should get \( \frac{dy}{dx} = \frac{1}{x} \). So, \(y = \ln x + C\) is indeed a solution to the differential equation \( \frac{dy}{dx} = \frac{1}{x} \) in the interval \((0, \infty)\)

Solve the Differential Equation for \(y = \ln (-x) + C\)

Take the derivative of \(y = \ln (-x) + C\), resulting in \( \frac{dy}{dx} =-\frac{1}{x} \). However, because \( x < 0 \) in the interval \((- \infty, 0)\), the negative sign can be discarded, indicating that \( y = \ln (-x) + C \) is a solution to the differential equation \( \frac{dy}{dx} = \frac{1}{x} \) in the interval \((- \infty, 0)\)

Solve the Differential Equation for \(y = \ln |x| + C\)

The absolute value function, \( |x| \), handles both positive and negative values of x. Therefore, it envelopes the solutions derived in steps 1 and 2 and is a single solution to the differential equation \( \frac{dy}{dx} = \frac{1}{x} \) in the domain \((- \infty, 0) \cup (0, \infty)\)

Solve the Differential Equation for the piecewise function

The function \(y=\left\{\begin{array}{l}\ln (-x)+C_{1} \ \ln x+C_{2}\end{array}\right.\) for \(x < 0\) and \(x > 0\) respectively, is simply the two solutions from steps 1 and 2 combined into a piecewise function, therefore it also satisfies the given differential equation \(\frac{dy}{dx} = \frac{1}{x}\) for any values of \(C_{1}\) and \(C_{2}\)