Question

The general solution of the differential equation \(xy'+y=x^{2}\), \(x>0\) is

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

Hint:

Calculus Tip: If you didn't spot the exact differential, you can put it in standard form $y^{\prime} + \frac{1}{x}y = x$. The integrating factor is $e^{\int(1/x)dx} = e^{\ln x} = x$. Multiplying by $x$ brings you right back to Step 1!

Answer 1

Answer: (E)

Concept:
This is a first-order linear differential equation. However, notice that the left side $xy^{\prime} + y$ is actually the exact derivative of a product: $\frac{d}{dx}(xy) = xy^{\prime} + y \cdot 1$. Recognizing this reverse Product Rule allows us to integrate immediately without finding an integrating factor.

Step 1: Recognize the exact differential.
The given equation is: $$x \frac{dy}{dx} + y = x^2$$ Apply the reverse Product Rule to the left side: $$\frac{d}{dx}(x \cdot y) = x^2$$

Step 2: Integrate both sides with respect to x.
Apply the integral operator $\int dx$ to both sides: $$\int \frac{d}{dx}(xy) dx = \int x^2 dx$$

Step 3: Evaluate the integrals.
The integral of a derivative leaves the function itself, and apply the power rule to the right side: $$xy = \frac{x^3}{3} + C$$

Step 4: Isolate y to find the general solution.
Divide the entire equation by $x$ (valid since $x > 0$): $$y = \frac{x^3}{3x} + \frac{C}{x}$$

Step 5: Simplify the final equation.
$$y = \frac{x^2}{3} + \frac{C}{x}$$ Hence the correct answer is (E) $y=\frac{x^{2{3}+\frac{C}{x}$}.