JEE Main 2023 · 29 January · evening · Single correct · +4 / −1
Let y=y(x) be the solution of the differential equation xlog ₑxdy/dx + y = x²log ₑx,(x > 1). If y(2) = 2, then y(e) is equal to
- (a) (1 + e²)/2
- (b) (1 + e²)/4
- (c) (2 + e²)/2
- (d) (4 + e²)/4
Answer and solution
Answer (d) (4 + e²)/4
Solution
Maths sir · 4 steps
Divide by x logₑ x first, and the integrating factor turns out to be just logₑ x.
- Standard linear form. Divide through by x ln x (non-zero, since x > 1) to get dy/dx + 1/(x ln x) y = x. Here P = 1/(x ln x) and Q = x.
- Integrating factor. Put t = ln x, so ∫ dx/(x ln x) = ∫ dt/t = ln(ln x). The integrating factor is eˡⁿ⁽ˡⁿˣ⁾ = ln x.
y ln x = ∫ x ln x dx = x²/2 ln x - x²/4 + C
- Use y(2) = 2. At x = 2 we get 2 ln 2 = 2 ln 2 - 1 + C, so C = 1.
- So. At x = e, ln e = 1, so y(e) = e²/2 - e²/4 + 1 = (4 + e²)/4, which is option (d).
The idea to keep. When P has x ln x in the denominator, the integrating factor is ln x. Integrate x ln x by parts and never drop the -x²/4 term, because it fixes C.
Answer from the NTA final answer key. Solution by PrepWiser.