Let x=x(y) be the solution of the differential equation y² d x+(x-1/y) d y=0. If x(1)=1, then x(1/2) is :

Solution

Maths sir · 5 steps

Divide by y² dy and treat x as the unknown: it becomes a linear equation in x.

1. Linear in x. Divide the equation by y² dy to get dx/dy + 1/y²x = 1/y³. This has the form dx/dy + P(y) x = Q(y) with P = 1/y² and Q = 1/y³.
2. Integrating factor. ∫ 1/y² dy = -1/y, so the integrating factor is e^(-1/y). Then x e^(-1/y) = ∫ e^(-1/y)/y³ dy.
3. Do the integral. Put t = -1/y, so dt = dy/y² and 1/y = -t. The integral becomes ∫ -t eᵗ dt = eᵗ(1 - t) = e^(-1/y)(1 + 1/y).

   x = 1 + 1/y + C e^(1/y)

4. Use x(1) = 1. At y = 1: 1 = 1 + 1 + Ce, so C = -1/e. This gives x = 1 + 1/y - e^(1/y - 1).
5. So. At y = 1/2: x = 1 + 2 - e²⁻¹ = 3 - e, which is option (d).

The idea to keep. When the equation has dx multiplied by a function of y alone and x appears only to the first power, solve for x as a function of y using the integrating factor e^(∫ P dy).

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