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Differential Equation Giving a Pair of Lines | JEE

JEE Maths question with a full step-by-step solution.

Question

The solution of

\dfrac{dy}{dx} = \dfrac{x - 2y + 3}{2x - y + 5}

f(x, y, c) = 0

. For suitable values of

c

f(x, y, c) = 0

represents a pair of lines whose point of intersection is:

\left(-\dfrac{7}{3}, \dfrac{2}{3}\right)

\left(-\dfrac{7}{3}, \dfrac{4}{3}\right)

\left(-\dfrac{7}{3}, \dfrac{1}{3}\right)

correct

\left(\dfrac{7}{3}, -\dfrac{1}{3}\right)

Solution

Step 1: Cross-multiply and group:

(x - 2y + 3)\,dx - (2x - y + 5)\,dy = 0

\Rightarrow x\,dx + y\,dy - 2(y\,dx + x\,dy) + 3\,dx - 5\,dy = 0.

Step 2: Integrate term by term:

\frac{x^2}{2} + \frac{y^2}{2} - 2xy + 3x - 5y = c' \Rightarrow x^2 + y^2 - 4xy + 6x - 10y + c = 0.

Step 3: For a pair of lines, the point of intersection is the centre, found from

\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0

2x - 4y + 6 = 0 \Rightarrow x - 2y + 3 = 0,

-4x + 2y - 10 = 0 \Rightarrow 2x - y + 5 = 0.

Step 4: Solve: from the first,

x = 2y - 3

; substituting,

2(2y - 3) - y + 5 = 0 \Rightarrow 3y = 1 \Rightarrow y = \dfrac{1}{3}

x = -\dfrac{7}{3}

. Correct answer: (3)

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