Let y = y(x) be the solution of the differential equation x³dy + (xy - 1)dx = 0,x > 0,y( 1/2 ) = 3 - e. Then y (1) is equal to

Solution

Maths sir · 4 steps

Divide by x³ and it becomes linear in y: find the integrating factor e^(-1/x) and the rest follows.

1. Make it linear. Divide by x³ dx to get dy/dx + 1/x² y = 1/x³. This is linear with P = 1/x² and Q = 1/x³, so the integrating factor is e^(∫ x⁻²dx) = e^(-1/x).
2. Integrate. Then y e^(-1/x) = ∫ e^(-1/x)/x³ dx. Put t = -1/x, so dt = dx/x² and 1/x = -t. The integral becomes ∫ -t eᵗ dt = eᵗ(1 - t) + C = e^(-1/x)(1 + 1/x) + C.

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

3. Fix the constant. At x = 1/2 you get 1 + 2 + C e² = 3 - e. So C e² = -e, which gives C = -e⁻¹.
4. So. y(1) = 1 + 1 - e⁻¹· e¹ = 2 - 1 = 1, which is option (c).

The idea to keep. When you see dy and y dx with no y², divide to reach dy/dx + Py = Q. Then the integrating factor e^(∫ P dx) does all the work.

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