The solution of the differential equation $p^{2}-5p+6=0$; where $p=\frac{dy}{dx}$ is:

Question

Accepted Answer

To solve the differential equation $ p^2 - 5p + 6 = 0 $, where $ p = \frac{dy}{dx} $, we start by factoring the quadratic equation. The equation can be factored as:

$$
p^2 - 5p + 6 = (p - 2)(p - 3) = 0
$$

This gives us two possible solutions for $ p $:

1. $ p = 2 $
2. $ p = 3 $

These solutions correspond to the differential equations:

1. $ \frac{dy}{dx} = 2 $
2. $ \frac{dy}{dx} = 3 $

Solving these differential equations separately:

1. For $ \frac{dy}{dx} = 2 $, integrating both sides with respect to $ x $ gives:
   $$
   y = 2x + c_1
   $$

2. For $ \frac{dy}{dx} = 3 $, integrating both sides with respect to $ x $ gives:
   $$
   y = 3x + c_2
   $$

The general solution to the original differential equation is the product of the solutions to these two separate differential equations, which is expressed as:

$$
(y - 2x - c_1)(y - 3x - c_2) = 0
$$

This matches Option 1: $(y - 3x - c_1)(y - 2x - c_2) = 0$.

**Explanation of why the other options are incorrect:**

- **Option 2: $(3x + y - c_1)(2y + x - c_2) = 0$**

This option does not correspond to the solutions of the differential equations derived from $ p = 2 $ and $ p = 3 $. The terms $ 3x + y $ and $ 2y + x $ do not match the linear forms derived from integrating the differential equations.

- **Option 3: $(y + 3x - c_1)(y + 2x - c_2) = 0$**

This option incorrectly adds $ 3x $ and $ 2x $ to $ y $, which does not align with the solutions $ y = 3x + c_2 $ and $ y = 2x + c_1 $. The signs are incorrect for the solution derived from the differential equation.

- **Option 4: $(y - x - c_1)(y + x + c_2) = 0$**

This option suggests solutions of the form $ y = x + c_1 $ and $ y = -x - c_2 $, which do not correspond to the solutions $ y = 2x + c_1 $ and $ y = 3x + c_2 $ derived from the differential equations.

Thus, Option 1 is the correct solution as it accurately represents the product of the solutions to the differential equations derived from the given quadratic equation.