Linear First Order
Consider the equation x\frac{dy}{dx} + 2y = x^2 with x positive, and identify the general solution after writing it in standard linear form first.
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About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- linear first order
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
y=4x2+x2C
Before any technique applies, this equation must be put into standard linear form by dividing through by x, giving \frac{dy}{dx} + \frac{2}{x}y = x. Now P(x) = \frac{2}{x} and Q(x) = x. The integrating factor is \mu = e^{\int \frac{2}{x},dx} = e^{2\ln x} = x^2, which transforms the left side into the derivative of a product. Multiplying through by x^2 gives x^2\frac{dy}{dx} + 2xy = x^3, and the left side is exactly \frac{d}{dx}(x^2 y). Integrating both sides yields x^2 y = \int x^3,dx = \frac{x^4}{4} + C. Dividing by x^2 gives y = \frac{x^2}{4} + \frac{C}{x^2}. Option y = \frac{x^2}{2} + Cx uses an incorrect integrating factor. Option y = x^2 + \frac{C}{x} stems from misintegrating the right side. Option \frac{x^3}{4} + C forgets to divide by x^2 at the end. This is the canonical linear-equation method emphasized in JEE Advanced. As a final consistency check, substituting y = \frac{x^2}{4} into the original gives x\cdot\frac{x}{2} + 2\cdot\frac{x^2}{4} = \frac{x^2}{2} + \frac{x^2}{2} = x^2, confirming the particular part satisfies the equation.
This medium difficulty mathematics question is from the chapter differential equations, covering the topic of linear first order. It appeared in the 2025 exam.
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