Q85 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { The general solution of the differential equation } \dfrac{d y}{d x}=\dfrac{x y}{x^2+y^2} \text { is }$
- A$y=c e^{\left(-\dfrac{x^2}{2 y^2}\right)}$
- B$y=c e^{\left(\dfrac{x^2}{y^2}\right)}$
- C$y=c e^{\left(\dfrac{x^2}{3 y^2}\right)}$
- Correct answerD$y=c e^{\left(\dfrac{x^2}{2 y^2}\right)}$
Solution
The given differential equation is $\dfrac{dy}{dx} = \dfrac{xy}{x^2 + y^2}$.
Taking the reciprocal, we have $\dfrac{dx}{dy} = \dfrac{x^2 + y^2}{xy} = \dfrac{x}{y} + \dfrac{y}{x}$.
Let $x = vy$, then $\dfrac{dx}{dy} = v + y \dfrac{dv}{dy}$.
Substituting into the equation: $v + y \dfrac{dv}{dy} = v + \dfrac{1}{v}$.
$y \dfrac{dv}{dy} = \dfrac{1}{v}$.
$v dv = \dfrac{dy}{y}$.
Integrating both sides: $\int v dv = \int \dfrac{dy}{y} \Rightarrow \dfrac{v^2}{2} = \ln|y| + C$.
Substituting $v = \dfrac{x}{y}$: $\dfrac{x^2}{2y^2} = \ln|y| + C$.
Rearranging for $\ln|y|$: $\ln|y| = \dfrac{x^2}{2y^2} - C$.
Taking the exponential of both sides: $y = e^{\dfrac{x^2}{2y^2} - C} = e^{-C} e^{\dfrac{x^2}{2y^2}}$.
Letting $e^{-C} = c$, we obtain $y = c e^{\dfrac{x^2}{2y^2}}$.
Answer: $y=c e^{\left(\dfrac{x^2}{2 y^2}\right)}$
Taking the reciprocal, we have $\dfrac{dx}{dy} = \dfrac{x^2 + y^2}{xy} = \dfrac{x}{y} + \dfrac{y}{x}$.
Let $x = vy$, then $\dfrac{dx}{dy} = v + y \dfrac{dv}{dy}$.
Substituting into the equation: $v + y \dfrac{dv}{dy} = v + \dfrac{1}{v}$.
$y \dfrac{dv}{dy} = \dfrac{1}{v}$.
$v dv = \dfrac{dy}{y}$.
Integrating both sides: $\int v dv = \int \dfrac{dy}{y} \Rightarrow \dfrac{v^2}{2} = \ln|y| + C$.
Substituting $v = \dfrac{x}{y}$: $\dfrac{x^2}{2y^2} = \ln|y| + C$.
Rearranging for $\ln|y|$: $\ln|y| = \dfrac{x^2}{2y^2} - C$.
Taking the exponential of both sides: $y = e^{\dfrac{x^2}{2y^2} - C} = e^{-C} e^{\dfrac{x^2}{2y^2}}$.
Letting $e^{-C} = c$, we obtain $y = c e^{\dfrac{x^2}{2y^2}}$.
Answer: $y=c e^{\left(\dfrac{x^2}{2 y^2}\right)}$
Hence, the correct answer is option: 'D'.