The solution curve, of the differential equation \( 2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx} \), passing through the point \( (0, 1) \), is a conic, whose vertex lies on the line:
- \( 2x + 3y = 9 \)
- \( 2x + 3y = -9 \)
- \( 2x + 3y = -6 \)
- \( 2x + 3y = 6 \)
The Correct Option is A
Approach Solution - 1
To solve the given differential equation and find the line on which the vertex of the resulting conic lies, let's start by simplifying and solving the given differential equation:
The differential equation provided is: \(2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx}\)
Rearrange the terms to separate the variables:
\(2y \frac{dy}{dx} - 5 \frac{dy}{dx} = -3\)
Factor out \(\frac{dy}{dx}\):
\(\left( 2y - 5 \right) \frac{dy}{dx} = -3\)
Solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{-3}{2y - 5}\)
This is a separable differential equation. To solve this, separate the variables:
\((2y - 5) \, dy = -3 \, dx\)
Integrate both sides:
\(\int (2y - 5) \, dy = -\int 3 \, dx\)
This gives:
\(y^2 - 5y = -3x + C\)
Rearrange to get the equation of the conic:
\(y^2 - 5y + 3x = C\)
Use the initial condition that the curve passes through the point \((0, 1)\) to find the constant:
Substitute \((x, y) = (0, 1)\) in the equation:
\(1^2 - 5 \times 1 + 3 \times 0 = C\)
Calculate:
\(1 - 5 = C \Rightarrow C = -4\)
Thus, the equation of the conic becomes:
\(y^2 - 5y + 3x = -4\)
This is the equation of a conic. To find the vertex, consider completing the square for the equation in terms of \(y\):
\(y^2 - 5y = -(3x + 4)\)
Complete the square for the \(y\) terms:
\(y^2 - 5y + \left(\frac{5}{2}\right)^2 = \left(\frac{5}{2}\right)^2 - (3x + 4)\)
Which simplifies to:
\(\left(y - \frac{5}{2}\right)^2 = \frac{25}{4} - 3x - 4\)
Further simplify:
\(\left(y - \frac{5}{2}\right)^2 = \frac{9}{4} - 3x\)
The vertex of this conic section is at \((x, y) = (h, \frac{5}{2})\). For the equation to be zero at the vertex:
\(\frac{9}{4} = 3h\)
Solve for \(h\):
\(h = \frac{3}{4}\)
Thus, the vertex lies on the line \(y = \frac{5}{2}\).
Plug known values into the equation of the line to cross-check:
- \(2x + 3y = 9\) is tested for \((x, y) = \left(\frac{3}{4}, \frac{5}{2}\right)
Substitute \(x = \frac{3}{4}, y = \frac{5}{2}\):
\(2 \times \frac{3}{4} + 3 \times \frac{5}{2} = 9\)
Calculate:
\(\frac{3}{2} + \frac{15}{2} = 9\)
\(9 = 9\)
Thus, the correct line on which the vertex lies is \(2x + 3y = 9\).
Approach Solution -2
Solution:
\((2y - 5)\frac{dy}{dx} = -3\)
\((2y - 5)dy = -3dx\)
\(2\cdot\frac{y^2}{2} - 5y = -3x + \lambda\)
Given: Curve passes through \((0, 1)\)
\(\implies \lambda = -4\)
Curve Equation:
\(\left(\frac{y - 5}{2}\right)^2 = -3\left(x - \frac{3}{4}\right)\)
Vertex of the Parabola:
\(\left(\frac{3}{4}, \frac{5}{2}\right)\)
Final Equation: \(2x + 3y = 9\)
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