Question:

If the solution curve of the differential equation \(\frac{dy}{dx} = \frac{x + y - 2}{x - y}\) passes through the points \((2, 1)\) and (k + 1, 2), k > 0, then

Updated On: Dec 29, 2025
  • \(2\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)\)

  • \(\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)\)

  • \(2\tan^{-1}\left(\frac{1}{k+1}\right) = \log_e(k^2 + 2k + 2)\)

  • \(2\tan^{-1}\left(\frac{1}{k}\right) = \log_e\left(k^2 + \frac{1}{k^2}\right)\)

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The Correct Option is A

Approach Solution - 1

Let's solve the differential equation given by: 

\(\frac{dy}{dx} = \frac{x + y - 2}{x - y}\)

We are provided that the solution curve passes through the points \((2, 1)\) and \((k + 1, 2)\), where \( k > 0 \).

Step 1: Solve the Differential Equation

We can use the variable separable method to solve this differential equation.

Rearrange the given equation:

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

Separating variables, we have:

\(\int (x - y) \, dy = \int (x + y - 2) \, dx\)

Integrating both sides:

\(\frac{1}{2}y^2 - xy = \frac{1}{2}x^2 + x \cdot y - 2x + C\)

Step 2: Put the initial point

The curve passes through the point \((2, 1)\). Substitute \(x = 2\) and \(y = 1\) into the equation to find the constant \(C\):

\(\frac{1}{2}(1)^2 - 2(1) = \frac{1}{2}(2)^2 + 2 \cdot 1 - 2 \cdot 2 + C\)

Solving gives:

\(-1.5 = 1 + C\)

Hence, \(C = -2.5\).

Step 3: Check the condition at \((k+1, 2)\)

Now, the solution curve passes through \((k+1, 2)\), so substitute \(x = k+1\) and \(y = 2\):

\(\frac{1}{2}(2)^2 - (k+1) \cdot 2 = \frac{1}{2}(k+1)^2 + (k+1) \cdot 2 - 2(k+1) - 2.5\)

After simplifying both sides, you get the condition:

\(2\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)\)

This matches the correct option given in the question.

Conclusion: The correct answer is:

\(2\tan^{-1}\left(\frac{1}{k}\right) = \log_e(k^2 + 1)\)

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Approach Solution -2

​ \(\frac{dY}{dX}=\frac{X+Y}{X−Y}\)

\(Let Y=tX\)

\(\frac{dY}{dX} = t + X\frac{dt}{dX}\)

\(t + X\frac{dt}{dX} = 1 + \frac{t}{1 - t}\)

\(X\frac{dt}{dX} = 1 + \frac{t}{1 - t} - t = 1 + \frac{t^2}{1 - t}\)

\(\int \frac{1 - t}{1 + t^2} \, dt = \int \frac{dX}{X}\)

\(\tan^{-1}(t - 1) - \frac{1}{2}\ln(1 + t^2) = \ln|X| + C\)

\(\tan^{-1}\left(\frac{y-1}{x-1}\right) - \frac{1}{2}\ln\left(1+\left(\frac{y-1}{x-1}\right)^2\right) = \ln|x-1| + C\)
Curve passes through \((2, 1)\)
\(0–0=0+c⇒c=0\)
If \((k + 1, 2)\) also satisfies the curve
\(\tan^{-1}\left(\frac{1}{k}\right) - \frac{1}{2}\ln\left(\frac{1 + k^2}{k^2}\right) = \ln k\)
\(2\tan^{-1}\left(\frac{1}{k}\right) = \ln(1+k^2)\)
So, the correct option is (A): \(2\tan^{-1}\left(\frac{1}{k}\right) = \ln(k^2 + 1)\)

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Concepts Used:

Types of Differential Equations

There are various types of Differential Equation, such as:

Ordinary Differential Equations:

Ordinary Differential Equations is an equation that indicates the relation of having one independent variable x, and one dependent variable y, along with some of its other derivatives.

\(F(\frac{dy}{dt},y,t) = 0\)

Partial Differential Equations:

A partial differential equation is a type, in which the equation carries many unknown variables with their partial derivatives.

Partial Differential Equation

Linear Differential Equations:

It is the linear polynomial equation in which derivatives of different variables exist. Linear Partial Differential Equation derivatives are partial and function is dependent on the variable.

Linear Differential Equation

Homogeneous Differential Equations:

When the degree of f(x,y) and g(x,y) is the same, it is known to be a homogeneous differential equation.

\(\frac{dy}{dx} = \frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}\)

Read More: Differential Equations