To reduce the differential equation `(dy)/(dx) + P (x)*y = Q (x) * y^(n)` to the linear form , the substitution is

Options

A. `v = 1/(y^(n))`
B. ` v = 1/(y^(n-1))`
C. ` v = y^(n)`
D. `v = y^(n-1)`

Explanation

To reduce the differential equation

\[ \frac{dy}{dx} + P(x) y = Q(x) y^n \]

to a linear form, we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:

\[ \frac{dy}{dx} + P(x) y = Q(x) y^n \]

### Step 2: Isolate the term with \(y^n\)
We can rearrange the equation to isolate the term with \(y^n\):

\[ \frac{dy}{dx} = Q(x) y^n - P(x) y \]

### Step 3: Divide by \(y^n\)
Next, we divide both sides of the equation by \(y^n\):

\[ \frac{1}{y^n} \frac{dy}{dx} = Q(x) - \frac{P(x)}{y^{n-1}} \]

### Step 4: Introduce a substitution
To simplify this equation, we introduce the substitution:

\[ v = y^{1-n} \]

### Step 5: Differentiate the substitution
Now, we differentiate \(v\) with respect to \(x\):

\[ \frac{dv}{dx} = (1-n) y^{-n} \frac{dy}{dx} \]

### Step 6: Substitute back into the equation
Substituting \(y^{1-n}\) and \(\frac{dy}{dx}\) into the equation gives:

\[ \frac{1}{1-n} \frac{dv}{dx} = Q(x) - P(x) v \]

### Step 7: Rearranging to linear form
Now we can rearrange this equation to the standard linear form:

\[ \frac{dv}{dx} + P(x)(1-n)v = (1-n)Q(x) \]

### Conclusion
Thus, the substitution that reduces the given differential equation to a linear form is:

\[ v = y^{1-n} \]