Question

The general solution of the differential equation $x^3 \frac{dy}{dx} + 3x^2 y = \cos x$ is

• A. $y = \frac{\sin x}{x^3} + C$
• B. $y = \frac{\sin x}{x^3} + Cx$
• C. $y = \frac{\sin x}{x^2} + C$
• D. $y = \frac{\sin x + C}{x^2}$
• E. $y = \frac{\sin x + C}{x^3}$

Hint:

Before using general methods for linear differential equations (like finding an Integrating Factor), check if the LHS is already the exact derivative of a product. In this case, $x^3$ was already the integrating factor.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Observe the left-hand side of the differential equation. It is in a form that suggests the Product Rule of differentiation.

Step 2: Detailed Explanation:
1. The given equation is:
\[ x^3 \frac{dy}{dx} + 3x^2 y = \cos x \]
2. Notice that $\frac{d}{dx}(x^3 y) = x^3 \frac{dy}{dx} + y \frac{d}{dx}(x^3) = x^3 \frac{dy}{dx} + 3x^2 y$.
3. Therefore, we can rewrite the equation as:
\[ \frac{d}{dx}(x^3 y) = \cos x \]
4. Integrate both sides with respect to $x$:
\[ \int \frac{d}{dx}(x^3 y) dx = \int \cos x dx \]
\[ x^3 y = \sin x + C \]
5. Solve for $y$:
\[ y = \frac{\sin x + C}{x^3} \]

Step 3: Final Answer:
The general solution is $y = \frac{\sin x + C}{x^3}$.