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:
We have a first-order differential equation. Notice that the left-hand side is exactly the expansion of the derivative of a product.
Step 2: Key Formula or Approach:
Use the product rule backwards: \(\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}\).
Integrate both sides directly.
Step 3: Detailed Explanation:
Given equation:
\[ x^3 \frac{dy}{dx} + 3x^2 y = \cos x \] Observe the left side. It is the exact derivative of the product \((x^3 \cdot y)\):
\[ \frac{d}{dx} (x^3 y) = x^3 \frac{dy}{dx} + y(3x^2) \] So rewrite the equation as:
\[ \frac{d}{dx} (x^3 y) = \cos x \] Integrate both sides with respect to \(x\):
\[ \int \frac{d}{dx} (x^3 y) dx = \int \cos x \, dx \] \[ x^3 y = \sin x + C \] Solve for \(y\) by dividing by \(x^3\):
\[ y = \frac{\sin x + C}{x^3} \] Step 4: Final Answer:
The general solution is \(y = \frac{\sin x + C}{x^3}\).