Question

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

• A. $\tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2+y^2} + C$
• B. $\tan^{-1}\left(\frac{x}{y}\right) = \log(x+y) + C$
• C. $x^2 + y^2 = C(x+y)$
• D. $y = x \tan(\log x + C)$

Hint:

For homogeneous differential equations of the form \( \frac{dy}{dx} = f\!\left(\frac{y}{x}\right) \), use the substitution \(y = vx\). This converts the equation into a separable form, which often leads to solutions involving logarithmic and inverse trigonometric functions.

Answer 1

The Correct Option is A

Concept:
The given differential equation is a homogeneous differential equation because the numerator and denominator contain terms of the same degree. Such equations can be solved using the substitution: \[ y = vx \quad \Rightarrow \quad \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Step 1: {Substitute \(y = vx\).} \[ v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} \] \[ = \frac{1+v}{1-v} \]
Step 2: {Simplify the equation.} \[ x\frac{dv}{dx} = \frac{1+v}{1-v} - v \] \[ = \frac{1+v - v + v^2}{1-v} \] \[ = \frac{1+v^2}{1-v} \]
Step 3: {Separate the variables.} \[ \frac{1-v}{1+v^2} dv = \frac{dx}{x} \]
Step 4: {Integrate both sides.} \[ \int \left(\frac{1}{1+v^2} - \frac{v}{1+v^2}\right) dv = \int \frac{dx}{x} \] \[ \tan^{-1} v - \frac{1}{2}\log(1+v^2) = \log x + C \]
Step 5: {Substitute \(v=\frac{y}{x}\).} \[ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\log\left(1+\frac{y^2}{x^2}\right) = \log x + C \] Simplifying the logarithmic terms: \[ \tan^{-1}\left(\frac{y}{x}\right) = \log\left(x\sqrt{1+\frac{y^2}{x^2}}\right) + C \] \[ \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2+y^2} + C \]
Step 6: {Conclusion.} Thus, the general solution of the differential equation is: \[ \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2+y^2} + C \]