JEE Main 2020 · 3 September · evening · Single correct · +4 / −1

If x³dy + xy dx = x²dy + 2y dx; y(2) = e and
x > 1, then y(4) is equal to :

  1. (a) √e/2
  2. (b) 1/2 + √e
  3. (c) 3/2 + √e
  4. (d) 3/2√e
Answer and solution

Answer (d) 3/2√e

Solution

Maths sir · 4 steps

Put all the dy terms on one side and all the dx terms on the other, and the variables separate at once.

  1. Separate variables. Collect terms: (x³ - x²) dy = (2 - x) y dx. Divide by y x²(x-1), which is non-zero because x > 1.

    dy/y = (2 - x)/x²(x-1) dx = (-1/x - 2/x² + 1/(x-1))dx

  2. Partial fractions. Write 2 - x = A x(x-1) + B(x-1) + Cx². Putting x = 0 gives B = -2, putting x = 1 gives C = 1, and matching the x² terms gives A + C = 0, so A = -1.
  3. Integrate. ln y = -ln x + 2/x + ln(x-1) + c = ln(x-1)/x + 2/x + c. Putting y(2) = e gives 1 = ln 1/2 + 1 + c, so c = ln 2.
  4. So. At x = 4: ln y = ln 3/4 + 1/2 + ln 2 = ln 3/2 + 1/2. So y(4) = 3/2 e^(1/2) = 3/2√e, which is option (d).

The idea to keep. When dy and dx terms are mixed, group them first and factor. A rational function of x on the other side then calls for partial fractions before you integrate.

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

Study Differential Equations

Asked about 200 times in JEE Main since 2020. This question is from 2020.