Question

The solution of the differential equation \(\frac{dy}{dx} = \frac{x+y}{x-y}\) is

• A. \(c(x^2+y^2)^{\frac{1}{2}}+e^{tan^{-1}(\frac{y^2}{x})} = 0\), where c is an arbitrary constant
• B. \(c(x^2+y^2)^{\frac{1}{2}} = e^{tan^{-1}(\frac{y}{x})}\), where c is an arbitrary constant
• C. \(c(x^2-y^2) = e^{tan^{-1}(\frac{y}{x})}\), where c is an arbitrary constant
• D. \(c(x^2+y^2) = e^{tan^{-1}(\frac{y}{x})}\), where c is an arbitrary constant

Hint:

The right side is homogeneous, so put y = vx and separate the variables.

Answer 1

The Correct Option is B

Step 1: Identify the type of equation:
The right side \(\frac{x+y}{x-y}\) has the same degree in \(x\) and \(y\) on top and bottom. Dividing both by \(x\) gives \(\frac{1+y/x}{1-y/x}\), a function of \(y/x\) only. So the equation is homogeneous and the substitution \(y = vx\) works. Then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\).

Step 2: Substitute and separate the variables:
Putting \(y = vx\) gives \(v + x\frac{dv}{dx} = \frac{1+v}{1-v}\). Move \(v\) to the right side:
\[ x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v^2}{1-v} \]
Now separate the variables:
\[ \frac{1-v}{1+v^2}\,dv = \frac{dx}{x} \]

Step 3: Integrate both sides:
Split the left side into two standard integrals:
\[ \int \frac{dv}{1+v^2} - \int \frac{v\,dv}{1+v^2} = \int \frac{dx}{x} \]
\[ \tan^{-1}v - \frac{1}{2}\ln(1+v^2) = \ln|x| + C \]

Step 4: Go back to x and y:
Replace \(v = \frac{y}{x}\). Since \(\frac{1}{2}\ln\left(1+\frac{y^2}{x^2}\right) = \frac{1}{2}\ln(x^2+y^2) - \ln|x|\), the \(\ln|x|\) terms cancel on both sides. This leaves
\[ \tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{2}\ln(x^2+y^2) + C \]
Take the exponential of both sides: \(e^{\tan^{-1}(y/x)} = e^{C}(x^2+y^2)^{1/2}\). Writing the constant as \(c\) gives \(c(x^2+y^2)^{1/2} = e^{\tan^{-1}(y/x)}\).

Step 5: Why the other options are wrong:
Option (A) has \(\frac{y^2}{x}\) inside the inverse tangent and a plus sign, which does not come out of the working. Option (C) has \(x^2-y^2\), but we always got \(x^2+y^2\) from \(1+v^2\). Option (D) has \(x^2+y^2\) without the square root. That would need the coefficient \(1\) instead of \(\frac{1}{2}\) in front of the logarithm, which our integration does not give.

Final Answer:
The general solution is \(c(x^2+y^2)^{1/2} = e^{\tan^{-1}(y/x)}\), which is option (B). \[ \boxed{c(x^2+y^2)^{\frac{1}{2}} = e^{\tan^{-1}(y/x)}} \]