The gradient of the curve passing through `(4,0)` is given by `(dy)/(dx) - (y)/(x) + (5x)/( (x+2) (x-3))=0` if the point ( 5,a) lies on the curve then the value of a is

Options

A. `(67)/(12)`
B. `5 sin"" (7)/(12) `
C. `5 log ""(7)/(12) `
D. None of these

Answer

To solve the problem, we need to follow these steps:

### Step 1: Rewrite the given differential equation
The gradient of the curve is given by:
\[ \frac{dy}{dx} - \frac{y}{x} + \frac{5x}{(x+2)(x-3)} = 0 \]
Rearranging this, we get:
\[ \frac{dy}{dx} = \frac{y}{x} - \frac{5x}{(x+2)(x-3)} \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
From the equation, we can identify:
- \( p(x) = -\frac{1}{x} \)
- \( q(x) = -\frac{5x}{(x+2)(x-3)} \)

### Step 3: Find the integrating factor
The integrating factor \( I(x) \) is given by:
\[ I(x) = e^{\int p(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln |x|} = \frac{1}{x} \]

### Step 4: Multiply the differential equation by the integrating factor
Multiplying the entire differential equation by \( \frac{1}{x} \):
\[ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = -\frac{5}{(x+2)(x-3)} \]

### Step 5: Rewrite the left-hand side
The left-hand side can be rewritten as:
\[ \frac{d}{dx}\left(\frac{y}{x}\right) = -\frac{5}{(x+2)(x-3)} \]

### Step 6: Integrate both sides
Integrating both sides with respect to \( x \):
\[ \int \frac{d}{dx}\left(\frac{y}{x}\right) \, dx = \int -\frac{5}{(x+2)(x-3)} \, dx \]
The left-hand side simplifies to:
\[ \frac{y}{x} = \int -\frac{5}{(x+2)(x-3)} \, dx + C \]

### Step 7: Solve the right-hand side integral
To solve the integral on the right side, we can use partial fractions:
\[ -\frac{5}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3} \]
Solving for \( A \) and \( B \), we find:
\[ A = 5/5 = 1, \quad B = -1 \]
Thus, the integral becomes:
\[ \int \left( \frac{1}{x+2} - \frac{1}{x-3} \right) \, dx = \ln|x+2| - \ln|x-3| + C \]

### Step 8: Substitute back
Substituting back, we have:
\[ \frac{y}{x} = \ln\left(\frac{x+2}{x-3}\right) + C \]
Thus, multiplying through by \( x \):
\[ y = x \ln\left(\frac{x+2}{x-3}\right) + Cx \]

### Step 9: Use the point (4, 0) to find \( C \)
Substituting \( (4, 0) \):
\[ 0 = 4 \ln\left(\frac{4+2}{4-3}\right) + 4C \]
This simplifies to:
\[ 0 = 4 \ln(6) + 4C \implies C = -\ln(6) \]

### Step 10: Substitute \( C \) back into the equation
Now we have:
\[ y = x \ln\left(\frac{x+2}{x-3}\right) - x \ln(6) \]

### Step 11: Find the value of \( a \) when \( x = 5 \)
Substituting \( x = 5 \):
\[ y = 5 \ln\left(\frac{5+2}{5-3}\right) - 5 \ln(6) \]
This simplifies to:
\[ y = 5 \ln\left(\frac{7}{2}\right) - 5 \ln(6) = 5 \left( \ln\left(\frac{7}{2}\right) - \ln(6) \right) \]
Using properties of logarithms:
\[ y = 5 \ln\left(\frac{7}{12}\right) \]

### Final Answer
Thus, the value of \( a \) is:
\[ a = 5 \ln\left(\frac{7}{12}\right) \]