The solution curve, of the differential equation 2 y (d y)/(d x)+3=5 (d y)/(d x), passing through the point (0,1) is a conic, whose vertex lies on the line :

Solution

Maths sir · 5 steps

Collect the dy/dx terms, separate, then complete the square to read off the parabola's vertex.

1. Separate variables. Move both dy/dx terms to one side: (2y - 5)dy/dx = -3. So (2y - 5) dy = -3 dx.
2. Integrate and fix C. Integrating gives y² - 5y = -3x + C. At (0, 1), 1 - 5 = C, so C = -4 and the curve is y² - 5y + 3x + 4 = 0.
3. Complete the square. Add 25/4 to both sides of y² - 5y = -3x - 4. The right side becomes -3x + 9/4.

   (y - 5/2)² = -3(x - 3/4)

4. Read the vertex. This is a parabola of the form Y² = -3X, with vertex where X = 0 and Y = 0, that is at (3/4, 5/2).
5. So. 2x + 3y = 2 × 3/4 + 3 × 5/2 = 3/2 + 15/2 = 9. The vertex lies on 2x + 3y = 9, option (c).

The idea to keep. When a separable equation gives a quadratic in one variable, complete the square and write it as (y - k)² = 4a(x - h). The vertex is (h, k), read straight off the brackets.

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