Q67 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
The solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$ is
- A$c(x^2 + y^2)^{\frac{1}{2}} + e^{\tan^{-1}\left(\frac{y^2}{x}\right)} = 0$, where c is an arbitrary constant
- Correct answerB$c(x^2 + y^2)^{\frac{1}{2}} = e^{\tan^{-1}\left(\frac{y}{x}\right)}$, where c is an arbitrary constant
- C$c(x^2 - y^2) = e^{\tan^{-1}\left(\frac{y}{x}\right)}$, where c is an arbitrary constant
- D$c(x^2 + y^2) = e^{\tan^{-1}\left(\frac{y}{x}\right)}$, where c is an arbitrary constant
Solution
The given differential equation is $\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$.
Substituting $y = vx$, we get $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
$v + x\dfrac{dv}{dx} = \dfrac{x+vx}{x-vx} = \dfrac{1+v}{1-v}$
$x\dfrac{dv}{dx} = \dfrac{1+v}{1-v} - v = \dfrac{1+v-v+v^2}{1-v} = \dfrac{1+v^2}{1-v}$
Separating the variables, we get:
$\dfrac{1-v}{1+v^2} dv = \dfrac{dx}{x}$
Integrating both sides:
$\int \dfrac{1}{1+v^2} dv - \int \dfrac{v}{1+v^2} dv = \int \dfrac{dx}{x}$
$\tan^{-1}v - \dfrac{1}{2}\ln(1+v^2) = \ln|x| + C$
Substituting $v = \dfrac{y}{x}$:
$\tan^{-1}\left(\dfrac{y}{x}\right) - \dfrac{1}{2}\ln\left(1+\dfrac{y^2}{x^2}\right) = \ln|x| + C$
$\tan^{-1}\left(\dfrac{y}{x}\right) - \dfrac{1}{2}\ln\left(\dfrac{x^2+y^2}{x^2}\right) = \ln|x| + C$
$\tan^{-1}\left(\dfrac{y}{x}\right) - \dfrac{1}{2}\ln(x^2+y^2) + \ln|x| = \ln|x| + C$
$\tan^{-1}\left(\dfrac{y}{x}\right) - \dfrac{1}{2}\ln(x^2+y^2) = C$
$\ln(x^2+y^2)^{\dfrac{1}{2}} = \tan^{-1}\left(\dfrac{y}{x}\right) - C$
$(x^2+y^2)^{\dfrac{1}{2}} = e^{\tan^{-1}\left(\dfrac{y}{x}\right) - C}$
$e^C (x^2+y^2)^{\dfrac{1}{2}} = e^{\tan^{-1}\left(\dfrac{y}{x}\right)}$
Letting $c = e^C$, we obtain:
$c(x^2+y^2)^{\dfrac{1}{2}} = e^{\tan^{-1}\left(\dfrac{y}{x}\right)}$
Answer: $c(x^2 + y^2)^{\frac{1}{2}} = e^{\tan^{-1}\left(\frac{y}{x}\right)}$, where c is an arbitrary constant
Substituting $y = vx$, we get $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
$v + x\dfrac{dv}{dx} = \dfrac{x+vx}{x-vx} = \dfrac{1+v}{1-v}$
$x\dfrac{dv}{dx} = \dfrac{1+v}{1-v} - v = \dfrac{1+v-v+v^2}{1-v} = \dfrac{1+v^2}{1-v}$
Separating the variables, we get:
$\dfrac{1-v}{1+v^2} dv = \dfrac{dx}{x}$
Integrating both sides:
$\int \dfrac{1}{1+v^2} dv - \int \dfrac{v}{1+v^2} dv = \int \dfrac{dx}{x}$
$\tan^{-1}v - \dfrac{1}{2}\ln(1+v^2) = \ln|x| + C$
Substituting $v = \dfrac{y}{x}$:
$\tan^{-1}\left(\dfrac{y}{x}\right) - \dfrac{1}{2}\ln\left(1+\dfrac{y^2}{x^2}\right) = \ln|x| + C$
$\tan^{-1}\left(\dfrac{y}{x}\right) - \dfrac{1}{2}\ln\left(\dfrac{x^2+y^2}{x^2}\right) = \ln|x| + C$
$\tan^{-1}\left(\dfrac{y}{x}\right) - \dfrac{1}{2}\ln(x^2+y^2) + \ln|x| = \ln|x| + C$
$\tan^{-1}\left(\dfrac{y}{x}\right) - \dfrac{1}{2}\ln(x^2+y^2) = C$
$\ln(x^2+y^2)^{\dfrac{1}{2}} = \tan^{-1}\left(\dfrac{y}{x}\right) - C$
$(x^2+y^2)^{\dfrac{1}{2}} = e^{\tan^{-1}\left(\dfrac{y}{x}\right) - C}$
$e^C (x^2+y^2)^{\dfrac{1}{2}} = e^{\tan^{-1}\left(\dfrac{y}{x}\right)}$
Letting $c = e^C$, we obtain:
$c(x^2+y^2)^{\dfrac{1}{2}} = e^{\tan^{-1}\left(\dfrac{y}{x}\right)}$
Answer: $c(x^2 + y^2)^{\frac{1}{2}} = e^{\tan^{-1}\left(\frac{y}{x}\right)}$, where c is an arbitrary constant
Hence, the correct answer is option: 'B'.