Question

If the equation of the curve which passes through the point \( (1,1) \) satisfies the differential equation: \[ \frac{dy}{dx} = \frac{2x - 5y + 3}{5x + 2y - 3}, \] then the equation of the curve is: \vspace{0.5cm}

• A. \( x^2 + 5xy - y^2 + 3x - 3y - 5 = 0 \)
• B. \( x^2 + 5xy - y^2 + 3x + 3y - 11 = 0 \)
• C. \( x^2 - 5xy - y^2 - 3x - 3y + 11 = 0 \)
• D. \( x^2 - 5xy - y^2 + 3x + 3y - 1 = 0 \) \vspace{0.5cm}

Hint:

For solving exact differential equations, check if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). If true, integrate \( M \) with respect to \( x \) and match terms using \( N \) to determine the function.

Answer 1

The Correct Option is D

Step 1: Solving the Differential Equation Rearrange the given equation: \[ (5x + 2y - 3) dy = (2x - 5y + 3) dx \] Rearrange: \[ \frac{dy}{dx} = \frac{2x - 5y + 3}{5x + 2y - 3} \] This is a first-order linear differential equation. Using the method of separation of variables, we rearrange: \[ (5x + 2y - 3) dy - (2x - 5y + 3) dx = 0 \] This represents an exact differential equation of the form: \[ M(x, y)dx + N(x, y)dy = 0 \] where: \[ M(x, y) = -(2x - 5y + 3), \quad N(x, y) = 5x + 2y - 3 \] Since: \[ \frac{\partial M}{\partial y} = -(-5) = 5, \quad \frac{\partial N}{\partial x} = 5 \] Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact. \vspace{0.5cm} Step 2: Finding the Solution We integrate \( M(x, y) \) with respect to \( x \): \[ F(x, y) = \int (-2x + 5y - 3) dx \] \[ = -x^2 + 5xy - 3x + g(y) \] Differentiating with respect to \( y \): \[ \frac{dF}{dy} = 5x + g'(y) \] Equating with \( N(x, y) \): \[ 5x + g'(y) = 5x + 2y - 3 \] \[ g'(y) = 2y - 3 \] Integrating: \[ g(y) = y^2 - 3y \] Thus, the solution is: \[ F(x, y) = -x^2 + 5xy - 3x + y^2 - 3y = C \] Rearrange: \[ x^2 - 5xy - y^2 + 3x + 3y = C \] Since the curve passes through \( (1,1) \): \[ 1^2 - 5(1)(1) - 1^2 + 3(1) + 3(1) = C \] \[ 1 - 5 - 1 + 3 + 3 = C \] \[ C = 1 \] Thus, the final equation is: \[ x^2 - 5xy - y^2 + 3x + 3y - 1 = 0 \] \vspace{0.5cm}