JEE Main 2021 · 27 July · evening · Single correct · +4 / −1
Let y = y(x) be the solution of the differential
equation (x - x³)dy = (y + yx² - 3x⁴)dx, x > 2. If y(3) = 3, then y(4) is equal to :
- (a) 4
- (b) 12
- (c) 8
- (d) 16
Answer and solution
Answer (b) 12
Solution
Maths sir · 4 steps
Divide by x - x³ to make it linear in y; then the integrating factor makes the right side just 3x².
- Make it linear. Divide both sides by (x - x³) dx = x(1-x²) dx. This gives dy/dx - (1+x²)/x(1-x²) y = -3x³/(1-x²), which has the form dy/dx + Py = Q.
- Integrating factor. Partial fractions give (1+x²)/x(1-x)(1+x) = 1/x + 1/(1-x) - 1/(1+x). So ∫ P dx = -ln x/(x²-1) for x > 2, and the integrating factor is (x²-1)/x.
d/dx(y·(x²-1)/x) = (x²-1)/x·-3x³/(1-x²) = 3x²
- Find the constant. Integrating gives y(x²-1)/x = x³ + C. Put x = 3, y = 3: (3 × 8)/3 = 8 = 27 + C, so C = -19.
- So. At x = 4: 15y/4 = 64 - 19 = 45, so y(4) = 12, option (b).
The idea to keep. For a linear equation, always check what the integrating factor does to Q. Here the 1-x² terms cancel, which tells you the factor was right.
Answer from the NTA final answer key. Solution by PrepWiser.