Let y=y(x) be the solution of the differential equation (2 x log ₑ x) (d y)/(d x)+2 y=3/x log ₑ x, x>0 and y(e⁻¹)=0. Then, y(e) is equal to

Solution

Maths sir · 5 steps

Divide by 2x logₑ x to get the linear form; the integrating factor turns out to be just logₑ x.

1. Standard linear form. Divide every term by 2x logₑ x to get dy/dx + 1/(x logₑ x) y = 3/2x². So P = 1/(x logₑ x) and Q = 3/2x².
2. Integrating factor. Put t = logₑ x, so ∫ dx/(x logₑ x) = ∫ dt/t = logₑ |logₑ x|. The integrating factor is logₑ x. Its sign does not matter, because a minus sign would cancel from both sides.

   y logₑ x = 3/2∫ (logₑ x)/x² dx = -3/2x(logₑ x + 1) + C

3. By parts. Take logₑ x as the first function: ∫ (logₑ x)/x² dx = -(logₑ x)/x + ∫ 1/x² dx = -(logₑ x)/x - 1/x. Multiplying by 3/2 gives the line above.
4. Use the condition. At x = e⁻¹ we have logₑ x = -1 and y = 0. The bracket becomes -1 + 1 = 0, so 0 = 0 + C and C = 0.
5. So. At x = e we have logₑ x = 1, so y(e) × 1 = -3/2e(1 + 1) = -3/e. The answer is (a).

The idea to keep. When P = 1/(x logₑ x), the integrating factor is logₑ x. Then ∫ (logₑ x)/xⁿ dx always goes by parts, with logₑ x as the first function.

Answer from the NTA final answer key. Solution by PrepWiser.