Question

The solution of the differential equation $x \frac{dy}{dx} + y = x^3 y^6$ is given by}

• A. $x^3 y^5 (2.5 + cx^2) = 1$
• B. $x^5 y^3 (2.5 + cx^2) = 1$
• C. $x^5 y^5 (2.5 + cy^2) = 1$
• D. $x^5 y^5 (2.5 + cx^2) = 1$

Hint:

Notice that the left-hand side is $d(xy) / dx$ after some manipulation. In Bernoulli equations, dividing by the non-linear $y^n$ term is always the first step to reduce it to a standard linear form.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ is a Bernoulli equation, which can be reduced to linear form by a suitable variable substitution.

Step 2: Detailed Explanation:
Given the differential equation: \[ x \frac{dy}{dx} + y = x^3 y^6 \]
Divide the entire equation by $x y^6$: \[ y^{-6} \frac{dy}{dx} + \frac{1}{x} y^{-5} = x^2 \]
Let us introduce a new variable $v$: \[ v = y^{-5} \implies \frac{dv}{dx} = -5 y^{-6} \frac{dy}{dx} \implies y^{-6} \frac{dy}{dx} = -\frac{1}{5} \frac{dv}{dx} \]
Substitute these into the equation: \[ -\frac{1}{5} \frac{dv}{dx} + \frac{1}{x} v = x^2 \implies \frac{dv}{dx} - \frac{5}{x} v = -5 x^2 \]
This is a standard linear first-order differential equation in $v$.
Calculate the Integrating Factor (I.F.): \[ \text{I.F.} = e^{\int -\frac{5}{x} \, dx} = e^{-5 \ln x} = x^{-5} \]
The solution for $v$ is given by: \[ v \cdot x^{-5} = \int -5 x^2 \cdot x^{-5} \, dx = \int -5 x^{-3} \, dx \] \[ v x^{-5} = -5 \left( \frac{x^{-2}}{-2} \right) + c = 2.5 x^{-2} + c \] \[ v = x^5 (2.5 x^{-2} + c) = 2.5 x^3 + c x^5 \]
Substitute $v = y^{-5}$ back into the equation: \[ y^{-5} = x^3 (2.5 + c x^2) \] \[ 1 = x^3 y^5 (2.5 + c x^2) \]

Step 3: Final Answer:
The solution of the differential equation is $x^3 y^5 (2.5 + cx^2) = 1$.