JEE Main 2021 · 27 August · evening · Single correct · +4 / −1
If the solution curve of the differential equation (2x - 10y³)dy + ydx = 0, passes through the points (0, 1) and (2, β), then β is a root of the equation :
- (a) y⁵ - 2y - 2 = 0
- (b) 2y⁵ - 2y - 1 = 0
- (c) 2y⁵ - y² - 2 = 0
- (d) y⁵ - y² - 1 = 0
Answer and solution
Answer (d) y⁵ - y² - 1 = 0
Solution
Maths sir · 4 steps
Do not chase dy/dx here: flip it to dx/dy and the equation becomes linear in x.
- Treat x as dependent. The 10y³ term rules out separating the variables. Divide through by y dy instead, and x appears only to the first power.
dx/dy + 2/y x = 10y²
- Integrating factor. The factor is e^(∫ 2/y dy) = y². Multiplying by it gives d/dy(xy²) = 10y⁴, so xy² = 2y⁵ + C.
- Use (0, 1). Putting x = 0, y = 1 gives 0 = 2 + C, so C = -2. The curve is xy² = 2y⁵ - 2.
- So. At (2, β), 2β² = 2β⁵ - 2. Dividing by 2 gives β⁵ - β² - 1 = 0, so β is a root of option (d).
The idea to keep. If dy/dx will not behave, look at dx/dy. When x appears only to the first power, the equation is linear in x and the integrating factor does the work.
Answer from the NTA final answer key. Solution by PrepWiser.