Question

x=logp and y=1/p differential equation

Answer 1

To find the differential equation involving the variables x and y, we can express y in terms of x and then differentiate it. Let's proceed with the given equations: 

x = log(p) (Equation 1) y = 1/p (Equation 2) 

First, let's solve Equation 1 for p: x = log(p) p = e^x (where e is the base of the natural logarithm) Now, substitute the value of p in terms of x into Equation 2: y = 1/p y = 1/(e^x) 

To find the differential equation, we differentiate Equation 2 with respect to x: 

dy/dx = d/dx (1/(e^x)) Using the chain rule: dy/dx = -1/(e^x) * d/dx(x) 

Since d/dx(x) is equal to 1, we have: dy/dx = -1/(e^x) 

So, the differential equation involving the variables x and y is: dy/dx = -1/(e^x)

Answer 2

To find the differential equation relating \( x = \log p \) and \( y = \frac{1}{p} \), we can start by expressing \( p \) in terms of \( x \) and \( y \). Since \( x = \log p \), we can rewrite this as \( p = e^x \). Similarly, since \( y = \frac{1}{p} \), we have \( p = \frac{1}{y} \).

Now, we have two expressions for \( p \) in terms of \( x \) and \( y \):
1. \( p = e^x \)
2. \( p = \frac{1}{y} \)

Setting these two expressions equal to each other gives us \( e^x = \frac{1}{y} \).

To find the differential equation relating \( x \) and \( y \), we can differentiate this equation with respect to \( x \) using the chain rule for differentiation:

\[ \frac{d}{dx} (e^x) = \frac{d}{dx} \left( \frac{1}{y} \right) \]
\[ e^x = -\frac{1}{y^2} \frac{dy}{dx} \]

Simplifying, we get:

\[ e^x = -\frac{dy}{dx} \cdot \frac{1}{y^2} \]

\[ \frac{dy}{dx} = -e^x y^{-2} \]

So, the differential equation relating \( x \) and \( y \) is:

\[ \frac{dy}{dx} = -e^x y^{-2} \]