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Question

(B) Given $\frac{dy}{dx} = \frac{x + y}{x - y}$. This is a homogeneous differential equation. Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

Vedclass · Accepted Answer

$c(x^2 + y^2)^{\frac{1}{2}} = e^{	an^{-1}(\frac{y}{x})}$, where $c$ is an arbitrary constant

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(B) Given $\frac{dy}{dx} = \frac{x + y}{x - y}$. This is a homogeneous differential equation. Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}$.$x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}$.Separating variables: $\frac{1 - v}{1 + v^2} dv = \frac{dx}{x}$.Integrating both sides: $\int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv = \int \frac{dx}{x}$.$	an^{-1}(v) - \frac{1}{2} \ln(1 + v^2) = \ln|x| + C_1$.$	an^{-1}(\frac{y}{x}) = \ln|x| + \frac{1}{2} \ln(1 + \frac{y^2}{x^2}) + C_1 = \ln|x| + \frac{1}{2} \ln(\frac{x^2 + y^2}{x^2}) + C_1$.$	an^{-1}(\frac{y}{x}) = \ln|x| + \frac{1}{2} \ln(x^2 + y^2) - \ln|x| + C_1 = \ln((x^2 + y^2)^{\frac{1}{2}}) + C_1$.Taking exponential on both sides: $e^{	an^{-1}(\frac{y}{x})} = e^{C_1} (x^2 + y^2)^{\frac{1}{2}}$.Let $e^{C_1} = \frac{1}{c}$, then $c e^{	an^{-1}(\frac{y}{x})} = (x^2 + y^2)^{\frac{1}{2}}$ or $c(x^2 + y^2)^{\frac{1}{2}} = e^{	an^{-1}(\frac{y}{x})}$.

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

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