Let y=f(x) be the solution of the differential equation y(x+1)dx-x²dy=0,y(1)=e. Then lim _(x → 0 +) f(x) is equal to

Solution

Maths sir · 4 steps

Separate the variables, find y in closed form, then look at what e^(-1/x) does as x → 0⁺.

1. Separate. Rewrite the equation as x² dy = y(x+1) dx. Dividing by x² y gives dy/y = (1/x + 1/x²)dx.
2. Integrate. Integrate both sides to get ln y = ln x - 1/x + C. Put in y(1) = e: 1 = 0 - 1 + C, so C = 2.

   f(x) = x e^(2 - 1/x) = e² · x · e^(-1/x)

3. Take the limit. As x → 0⁺, 1/x → +∞, so e^(-1/x) → 0. The factor x also goes to 0, and e² is a constant, so the whole product goes to 0.
4. So. lim(x → 0⁺) f(x) = 0, which is option (b).

The idea to keep. After separating, always split a fraction like (x+1)/x² into simple terms before integrating. When you take the limit, check the sign of 1/x inside the exponential.

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