Differential Equation: (Z-y) p (x-z) q=(y-x)

Lagrange's Linear Initial Value Problem:
The given differential equation is not a standard form of a linear initial value problem. However, we can rearrange it to resemble the form of a linear first-order ordinary differential equation.

The given equation can be rewritten as:
(Z-y) p (x-z) q + (x-y) = 0

Let's introduce a new dependent variable v = x - y. By doing so, we can rewrite the equation as:
(Z-y) p (v+z) q + v = 0

Now, let's differentiate both sides of the equation with respect to x:
d/dx [(Z-y) p (v+z) q + v] = d/dx [0]

This simplifies to:
[(Z-y) pq + p(v+z)q'] + v' = 0

Simplifying further, we have:
(Z-y) pq + p(v+z)q' + v' = 0

Now, let's substitute v = x - y back into the equation:
(Z-y) pq + p(x+z-y)q' + v' = 0

Since v = x - y, we can differentiate it with respect to x to obtain v':
v' = 1 - y'

Substituting this into the equation, we have:
(Z-y) pq + p(x+z-y)q' + 1 - y' = 0

This equation can be rearranged as:
-(Z-y) pq - p(x+z-y)q' + y' = 1

Explanation:
The given differential equation is not a standard linear initial value problem, but it can be rearranged to resemble one. By introducing a new dependent variable v = x - y and differentiating both sides of the equation, we can rewrite it in a linear form. The rearranged equation involves the derivatives of y and the new variable v, along with other terms. This form allows us to apply Lagrange's linear initial value problem techniques to find a solution.